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+---
+id: quant-algebra-sequences-functions
+section: quant
+topic: algebra
+subtopic: sequences-functions
+title: "Sequences & Functions"
+tags: [sequences, arithmetic-sequence, geometric-sequence, recursion, functions, function-notation]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Two closely related ideas show up together on GMAT Focus Quant: **sequences** (ordered lists of
+numbers built by a rule) and **functions** (a rule that turns an input into an output). Both reward
+the same discipline — read the rule exactly, mind the indexing, and don't do more arithmetic than the
+question needs. The traps are almost always **off-by-one** errors and **misreading the rule**, not
+hard computation.
+
+## Core concepts
+
+**Arithmetic sequences** add a fixed **common difference** \(d\) each step. If the first term is
+\(a_1\), the \(n\)th term is
+
+\[a_n = a_1 + (n-1)\,d.\]
+
+The multiplier is \((n-1)\), **not** \(n\): you take \(d\) one fewer time than the term number,
+because the first term takes zero steps. To **sum** the first \(n\) terms, average the first and last
+term and multiply by how many there are:
+
+\[S_n = \frac{n(a_1 + a_n)}{2}.\]
+
+**Geometric sequences** multiply by a fixed **common ratio** \(r\) each step:
+
+\[a_n = a_1 \, r^{\,n-1}.\]
+
+Again the exponent is \(n-1\), not \(n\). Terms grow (or shrink) by a constant *factor*, so they
+change far faster than an arithmetic sequence's constant *steps*.
+
+**Recursive sequences** define each term from the previous one, e.g. \(a_{n+1} = 2a_n - 1\) with a
+given starting value. There is no shortcut unless you spot a pattern — just apply the rule one step at
+a time, and **stop at the right term** (writing out \(a_1, a_2, a_3, \dots\) with labels prevents the
+usual off-by-one slip).
+
+**Functions** are rules written as \(f(x)\). To evaluate \(f(\text{something})\), substitute that
+"something" for **every** \(x\) in the definition. For a **composition** \(f(f(x))\), work
+**inside-out**: compute the inner \(f\) first, then feed its result back into \(f\).
+
+## Worked examples
+
+**Arithmetic term.** First term \(a_1 = 7\), common difference \(d = 4\). The 20th term is
+\(a_{20} = 7 + (20-1)\cdot 4 = 7 + 76 = 83\) — not \(7 + 20\cdot 4 = 87\). The \((n-1)\) is the whole
+game.
+
+**Geometric term.** First term \(3\), ratio \(2\). The 6th term is \(3\cdot 2^{6-1} = 3\cdot 32 = 96\),
+not \(3\cdot 2^6 = 192\).
+
+**Recursion.** With \(a_1 = 2\) and \(a_{n+1} = 2a_n - 1\): \(a_2 = 3,\; a_3 = 5,\; a_4 = 9,\;
+a_5 = 17\). Label each step so you hand back \(a_5\), not \(a_4\) or \(a_6\).
+
+**Composition.** If \(f(x) = 2x - 3\), then \(f(f(4))\): first \(f(4) = 2\cdot 4 - 3 = 5\), then
+\(f(5) = 2\cdot 5 - 3 = 7\). Inner first, then outer — and don't drop the \(-3\) either time.
+
+## Common traps
+
+- **Off-by-one in the term formula.** Using \(a_1 + n\,d\) or \(a_1 r^{\,n}\) instead of the
+  \((n-1)\) version overshoots by exactly one step. This is the single most common error here.
+- **Forgetting the first term.** Computing \((n-1)d\) or \(r^{\,n-1}\) but not adding/multiplying the
+  first term.
+- **Dropping ÷2 in the arithmetic sum,** or using the wrong count of terms (the number of terms from
+  \(a\) to \(b\) with step \(d\) is \(\frac{b-a}{d} + 1\), not \(\frac{b-a}{d}\)).
+- **Answering the wrong quantity.** A problem that gives two terms and asks for the *first term* is
+  easy to finish by reporting \(d\) instead. Reread what's being asked.
+- **Composition outside-in or stopping early.** \(f(f(x))\) is not \(f(x)\), and it is not
+  \([f(x)]^2\); evaluate the inner function first, then the outer.
+
+## Key takeaways
+
+- Arithmetic: \(a_n = a_1 + (n-1)d\); sum \(= \dfrac{n(a_1+a_n)}{2}\).
+- Geometric: \(a_n = a_1 r^{\,n-1}\); terms change by a constant factor, so they move fast.
+- Recursive sequences: apply the rule step by step and **stop at the right index** — label your terms.
+- Functions: substitute for every \(x\); compose **inside-out** and keep the constant term.
+- Nearly every miss here is an **off-by-one** or a **misread rule**, not hard arithmetic — slow down on
+  the indexing.
diff --git a/content/lessons/quant/quant-arithmetic-exponents-roots.md b/content/lessons/quant/quant-arithmetic-exponents-roots.md
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+---
+id: quant-arithmetic-exponents-roots
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+title: "Exponents & Roots"
+tags: [exponents, roots, radicals, powers]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Most GMAT Focus exponent questions reward you for **rewriting everything in the same base** and
+then just tracking the exponents. Almost nothing here needs a calculator — the whole game is a
+handful of rules applied cleanly and, above all, **not confusing multiplication of bases with
+addition of exponents**.
+
+## Core concepts
+
+**The power rules.** For any nonzero base \(a\):
+
+\[a^m \cdot a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn}\]
+
+Read them carefully: you **add** exponents when you *multiply* like bases, and you **multiply**
+exponents when you *raise a power to a power*. Mixing these two up is the single most common
+exponent error on the test.
+
+**Zero and negative exponents.**
+
+\[a^0 = 1 \ (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}\]
+
+A negative exponent means "reciprocal," not "negative number": \(2^{-3} = \tfrac{1}{8}\), not \(-8\).
+
+**Fractional exponents are roots.**
+
+\[a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}\]
+
+So \(27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9\). Take the root first when it keeps the numbers
+small.
+
+**Roots simplify by pulling out perfect squares.** \(\sqrt{ab} = \sqrt{a}\,\sqrt{b}\), so
+
+\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\]
+
+You can only **add or subtract** radicals when the part under the root matches:
+\(5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\). You may **never** merge \(\sqrt{a} + \sqrt{b}\) into
+\(\sqrt{a+b}\).
+
+**Factoring out a common power** cracks the hardest questions. Terms like \(2^n + 2^n\) are not a
+new base — they collapse:
+
+\[2^n + 2^n = 2 \cdot 2^n = 2^{n+1}, \qquad 3^{k+1} - 3^{k} = 3^{k}(3 - 1) = 2\cdot 3^{k}\]
+
+## Worked examples
+
+**Same-base equation.** Solve \(2^{x+1} = 8^{x-1}\). Rewrite \(8 = 2^3\), so the right side is
+\(2^{3(x-1)} = 2^{3x-3}\). Same base ⟹ equal exponents: \(x + 1 = 3x - 3\), giving \(2x = 4\) and
+\(x = 2\).
+
+**Collapse a sum of equal powers.** If \(2^n + 2^n + 2^n + 2^n = 2^{12}\), the left side is
+\(4 \cdot 2^n = 2^2 \cdot 2^n = 2^{n+2}\). So \(n + 2 = 12\) and \(n = 10\).
+
+**Ratio of powers.** With \(x = 3^{20}\) and \(y = 3^{18}\), \(\dfrac{x}{y} = 3^{20-18} = 3^2 = 9\).
+
+## Common traps
+
+- **Multiplying exponents when you should add them** (and vice versa). \(2^5 \cdot 2^3 = 2^8\), not
+  \(2^{15}\).
+- **Reading a negative exponent as a negative number.** \(3^{-2} = \tfrac{1}{9}\), and \(3^0 = 1\)
+  (not \(0\)).
+- **Merging unlike radicals.** \(\sqrt{50} + \sqrt{18} \neq \sqrt{68}\); simplify each first, then add.
+- **Treating a sum of powers as a new base.** \(2^n + 2^n\) is \(2^{n+1}\), not \(4^n\) or \(2^{2n}\).
+
+## Key takeaways
+
+- Get everything into a **common base**, then just manipulate exponents.
+- Multiply like bases → **add** exponents; raise a power to a power → **multiply** exponents.
+- \(a^0 = 1\), \(a^{-n} = 1/a^n\), and \(a^{m/n} = \sqrt[n]{a^m}\).
+- Simplify radicals by pulling out perfect squares; only combine radicals with the same radicand.
+- Sums of equal powers **factor**: \(k\) copies of \(a^n\) is \(k \cdot a^n\).
diff --git a/content/lessons/quant/quant-counting-probability-combinatorics.md b/content/lessons/quant/quant-counting-probability-combinatorics.md
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+---
+id: quant-counting-probability-combinatorics
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+title: "Combinatorics: Counting Arrangements and Selections"
+tags: [combinatorics, permutations, combinations, factorial, counting]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Combinatorics is the art of counting outcomes **without listing them all**. On GMAT Focus
+Quant the whole game is deciding a single question: **does order matter?** Get that right and
+pick the matching tool, and most counting problems collapse into one line of arithmetic.
+
+## Core concepts
+
+**The factorial.** The number of ways to arrange \(n\) distinct objects in a row is
+\(n!\) (read "\(n\) factorial"):
+
+\[n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\]
+
+So \(5! = 5\cdot4\cdot3\cdot2\cdot1 = 120\). By convention \(0! = 1\).
+
+**Permutations — order matters.** The number of ways to arrange \(k\) objects chosen from
+\(n\) distinct objects, where the order of the chosen objects matters, is
+
+\[P(n,k) = \frac{n!}{(n-k)!} = n(n-1)\cdots(n-k+1)\]
+
+Use permutations for rankings, seatings, passwords, or codes — anything where "A then B"
+differs from "B then A."
+
+**Combinations — order does NOT matter.** The number of ways to *choose* \(k\) objects from
+\(n\), ignoring order, is
+
+\[C(n,k) = \binom{n}{k} = \frac{n!}{k!\,(n-k)!}\]
+
+Use combinations for committees, teams, handshakes, or any unordered selection. Notice
+\(C(n,k) = \dfrac{P(n,k)}{k!}\): a combination is a permutation with the \(k!\) reorderings of
+the same group divided out.
+
+**Two governing principles.**
+
+- **Multiplication (AND).** If a task is a sequence of independent stages, multiply the counts.
+  Choosing 3 seniors *and* 2 juniors: multiply \(C(5,3)\times C(4,2)\).
+- **Addition (OR).** If outcomes fall into mutually exclusive cases, add the counts.
+
+**Identical items.** When some objects repeat, divide by the factorial of each repeat count.
+Distinct arrangements of a word with letter counts \(n_1, n_2, \dots\) totalling \(n\) letters:
+
+\[\frac{n!}{n_1!\,n_2!\cdots}\]
+
+**Circular arrangements.** Seating \(n\) people around a round table gives \((n-1)!\)
+arrangements, because rotations of the same circle are not distinct — fix one person, then
+arrange the rest.
+
+## Worked examples
+
+**Combination.** A book club picks 3 of its 8 members to lead a discussion. Order does not
+matter, so \(C(8,3) = \dfrac{8\cdot7\cdot6}{3!} = \dfrac{336}{6} = 56\) ways.
+
+**Permutation.** Those same 8 members award a gold, silver, and bronze badge — now order
+matters, so \(P(8,3) = 8\cdot7\cdot6 = 336\) ways. Same people, \(3! = 6\) times as many
+outcomes because the badges are distinguishable.
+
+**Complement ("at least").** From 4 men and 3 women, choose a 3-person team with at least one
+woman. Count the total and subtract the unwanted case: \(C(7,3) - C(4,3) = 35 - 4 = 31\).
+
+## Common traps
+
+- **Order confusion.** The single biggest error. A committee is unordered (combination); a
+  lineup or ranking is ordered (permutation). Using \(P\) where \(C\) belongs overcounts by
+  exactly \(k!\).
+- **Adding instead of multiplying.** "3 seniors AND 2 juniors" multiplies: \(C(5,3)\times C(4,2)\),
+  not \(C(5,3)+C(4,2)\).
+- **Forgetting repeated items.** Arranging the letters of a word with repeats requires dividing
+  \(n!\) by the factorial of *each* repeated letter's count — miss one and you overcount.
+- **"At least one" head-on.** Counting cases directly often double-counts. Use the complement:
+  total minus "none."
+- **Blocks that must stay together.** Glue the block into one unit, arrange the units, then
+  multiply by the internal arrangements of the block.
+
+## Key takeaways
+
+- Ask first: **does order matter?** Yes → permutation \(P(n,k)\); no → combination \(C(n,k)\).
+- \(C(n,k) = \dfrac{P(n,k)}{k!}\) — dividing out the reorderings is the whole difference.
+- Independent stages **multiply**; mutually exclusive cases **add**.
+- Repeated identical items → divide by each repeat's factorial; round tables → \((n-1)!\).
+- For "at least one," count the complement and subtract.
diff --git a/content/lessons/quant/quant-number-properties-odds-evens-signs.md b/content/lessons/quant/quant-number-properties-odds-evens-signs.md
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+---
+id: quant-number-properties-odds-evens-signs
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+title: "Odds, Evens & Signs"
+tags: [parity, odd-even, signs, number-properties]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Parity (whether a number is odd or even) and sign rules are the quiet workhorses of GMAT Focus
+number-properties questions. On their own they look trivial, but the exam hides them inside
+"must be true," "could be true," and Data-Insights logic problems where you never learn the actual
+numbers. Knowing the rules cold lets you answer those without any arithmetic at all.
+
+## Core concepts
+
+**Parity under addition and subtraction.** The result is even exactly when the two numbers share
+the same parity:
+
+- even ± even = **even**
+- odd ± odd = **even**
+- even ± odd = **odd**
+
+A useful shortcut: adding or subtracting an **even** number never changes parity; adding or
+subtracting an **odd** number always flips it.
+
+**Parity under multiplication.** A product is even as soon as **one** factor is even:
+
+- even × (anything) = **even**
+- odd × odd = **odd**
+
+So a product of integers is odd only if *every* factor is odd.
+
+**Sums of several odd numbers.** Adding \(k\) odd numbers gives an **odd** total when \(k\) is odd,
+and an **even** total when \(k\) is even. (Each odd number flips the parity once.) Even numbers in
+the sum never affect the parity.
+
+**Consecutive integers.** Among any two consecutive integers \(n\) and \(n+1\), exactly one is
+even, so the product \(n(n+1)\) is **always even**. This is why expressions like \(n^2 + n\) are
+guaranteed even.
+
+**Sign rules for products.** The sign of a product depends only on **how many negative factors**
+there are:
+
+- an **even** number of negative factors → the product is **positive**
+- an **odd** number of negative factors → the product is **negative**
+- any factor of \(0\) → the product is \(0\)
+
+**Sign logic with two variables.** Two facts appear constantly:
+
+- \(xy > 0\) means \(x\) and \(y\) have the **same sign** (both positive or both negative).
+- \(xy < 0\) means they have **opposite signs**.
+
+Combine that with a fact about \(x + y\): if \(xy > 0\) and \(x + y < 0\), the shared sign must be
+negative, so **both are negative**.
+
+## Worked examples
+
+**"Must be odd."** If \(n\) is even, is \(n + 1\) odd? Adding the odd number \(1\) flips even to
+odd, so yes — always. By contrast \(2n\), \(3n\), \(n^2\), and \(n + 2\) are all still even.
+
+**Counting signs.** Evaluate \((-3)(-4)(-2)\). The magnitude is \(3 \cdot 4 \cdot 2 = 24\); there
+are three negative factors, an odd count, so the product is negative: \(-24\).
+
+**Parity from a product.** If \(mn\) is odd, what do you know? A product is odd only when every
+factor is odd, so both \(m\) and \(n\) are odd — and therefore \(m + n\) is even.
+
+## Common traps
+
+- **Confusing "adds up" with "multiplies."** Four copies of an odd number add to an even total
+  (even count of odds), even though each is odd.
+- **Losing track of signs.** With three negative factors the product is negative, not positive —
+  count the negatives, don't just take the magnitude.
+- **Assuming \(n^2\) is even.** \(n^2\) has the same parity as \(n\); it is odd whenever \(n\) is
+  odd.
+- **Forgetting zero.** "\(xy > 0\)" excludes \(x = 0\) or \(y = 0\); "same sign" quietly rules zero
+  out.
+
+## Key takeaways
+
+- Add/subtract: same parity → even, different parity → odd. Adding an even number never changes
+  parity.
+- A product is even if any factor is even; odd only if every factor is odd.
+- Product sign: even number of negatives → positive, odd number → negative, any zero → zero.
+- \(n(n+1)\) is always even — consecutive integers guarantee an even factor.
+- \(xy > 0\) means same sign; pair it with the sign of \(x + y\) to pin down both.
diff --git a/content/lessons/quant/quant-number-properties-remainders.md b/content/lessons/quant/quant-number-properties-remainders.md
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+---
+id: quant-number-properties-remainders
+section: quant
+topic: number-properties
+subtopic: remainders
+title: "Remainders"
+tags: [remainders, divisibility, modular-arithmetic, number-properties]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Remainder questions look like they need long division, but almost none of them do. The whole
+topic rests on one identity and a few rules for how remainders behave when you **add**,
+**multiply**, and take **powers**. Master those and you can answer most remainder questions in
+under a minute — often just by picking the smallest number that fits and testing it.
+
+## Core concepts
+
+**The division identity.** When a positive integer \(n\) is divided by a positive divisor \(d\),
+there is a unique quotient \(q\) and remainder \(r\) with
+
+\[n = dq + r, \qquad 0 \le r < d.\]
+
+The remainder is always **less than the divisor** (a remainder of 7 is impossible when dividing
+by 5). Reading a word problem, "quotient 7, remainder 4, divisor 6" becomes
+\(n = 6\cdot 7 + 4 = 46\) — nothing more.
+
+**Remainders add and multiply, then reduce.** If you only care about the remainder mod \(d\), you
+may replace any number by its remainder at every step, as long as you **reduce again at the end**:
+
+\[\text{rem}(a+b) = \text{rem}\big(\text{rem}(a) + \text{rem}(b)\big), \qquad
+\text{rem}(ab) = \text{rem}\big(\text{rem}(a)\cdot \text{rem}(b)\big).\]
+
+So if \(a\) leaves remainder 5 and \(b\) leaves remainder 8 when divided by 12, then \(ab\) leaves
+the remainder of \(5\times 8 = 40\), which is \(40 - 36 = 4\). The single most common mistake is
+stopping at 40 (or at \(5+8=13\)) and forgetting the final reduction.
+
+**Powers cycle.** Remainders of \(a^1, a^2, a^3, \dots\) repeat in a short cycle. Dividing by 10
+is the same as asking for the **units digit**, and units digits cycle with period at most 4:
+
+\[7^1 \to 7,\quad 7^2 \to 9,\quad 7^3 \to 3,\quad 7^4 \to 1,\quad 7^5 \to 7,\ \dots\]
+
+To find \(7^{20} \bmod 10\), note the cycle length is 4 and \(20\) is a multiple of 4, so \(7^{20}\)
+lands on the **last** entry of the cycle: remainder **1**. Match the exponent to its position with
+"exponent mod cycle-length," and remember that a remainder of 0 means the *end* of the cycle, not
+the start.
+
+**Pick the smallest number that fits.** "\(n\) leaves remainder 2 when divided by 5" just means
+\(n\) is one of \(2, 7, 12, 17, \dots\). For two conditions at once, list each and find the first
+overlap; the pattern then repeats every \(\text{lcm}\) of the divisors.
+
+## Worked examples
+
+**Translate and compute.** A positive integer divided by 6 gives quotient 7 and remainder 4.
+The integer is \(6\cdot 7 + 4 = 46\). (Swapping quotient and remainder — \(6\cdot 4 + 7 = 31\) — is
+the classic trap.)
+
+**Combine two conditions.** Find the smallest integer above 10 that leaves remainder 2 mod 5 and
+remainder 1 mod 3. Numbers \(\equiv 2 \pmod 5\): \(2, 7, 12, 17, 22, \dots\); of these, the ones
+\(\equiv 1 \pmod 3\) are \(7, 22, 37, \dots\) (they repeat every \(\text{lcm}(5,3)=15\)). The first
+one above 10 is **22**.
+
+**Reason with squares.** If \(n^2\) leaves remainder 4 when divided by 7, what can \(n\) leave?
+Test each possible remainder \(r = 0,1,\dots,6\): \(r^2 \bmod 7\) gives \(0,1,4,2,2,4,1\). The value
+4 appears at \(r = 2\) and \(r = 5\), so \(n\) leaves remainder **2 or 5** — *not* 4. The remainder of
+\(n\) is generally **not** the same as the remainder of \(n^2\).
+
+## Common traps
+
+- **Forgetting to reduce at the end.** \(5\times 8 = 40\) is not the answer when dividing by 12; the
+  remainder is \(40 - 36 = 4\).
+- **Adding remainders when the numbers are multiplied** (or vice versa). Combine remainders the
+  same way you combine the numbers, then reduce.
+- **Swapping quotient and remainder** in \(n = dq + r\).
+- **A remainder as large as the divisor.** Dividing by \(d\), the remainder must be \(0\) to \(d-1\).
+- **Assuming \(n\) and \(n^2\) (or \(n\) and \(2n\)) share a remainder.** They usually don't — test it.
+
+## Key takeaways
+
+- Everything flows from \(n = dq + r\) with \(0 \le r < d\).
+- To combine numbers, combine their remainders **and reduce again**: works for sums and products.
+- Powers of a number cycle; use "exponent mod cycle-length," and treat a result of 0 as the end of
+  the cycle.
+- For "which could be" and two-condition problems, list the smallest values that fit and look for
+  the overlap; it repeats every lcm of the divisors.
diff --git a/content/lessons/quant/quant-word-problems-interest.md b/content/lessons/quant/quant-word-problems-interest.md
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+---
+id: quant-word-problems-interest
+section: quant
+topic: word-problems
+subtopic: interest
+title: "Simple and Compound Interest"
+tags: [interest, simple-interest, compound-interest, growth, percents]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Interest problems ask what a sum of money becomes after it earns a percentage return over time.
+There are only two engines you need: **simple interest**, where the interest is a fixed slice of
+the original principal every period, and **compound interest**, where each period's interest is
+itself added to the balance and earns interest afterward. Almost every GMAT interest question is a
+matter of picking the right engine and being careful about what the question asks for — the
+**interest earned** or the **final balance**. Confusing those two is the single most common mistake.
+
+## Core concepts
+
+**Notation.** Let \(P\) be the principal (the starting amount), \(r\) the interest rate per period
+written as a decimal, and \(t\) the number of periods.
+
+**Simple interest** pays \(r\) of the *original* principal each period. The interest is
+
+\[I = P \cdot r \cdot t\]
+
+and the final balance (or "amount") is \(A = P + I = P(1 + rt)\). Because only the principal ever
+earns, simple interest grows in a straight line.
+
+**Compound interest** adds each period's interest to the balance before the next period, so the
+balance grows geometrically:
+
+\[A = P(1 + r)^{t}\]
+
+The **interest earned** is \(A - P = P\big[(1 + r)^{t} - 1\big]\). Notice \(A\) is the *balance*,
+not the interest — subtract \(P\) whenever the question asks how much was *earned*.
+
+**Compounding more than once a year.** If a stated annual rate \(R\) is compounded \(n\) times per
+year for \(y\) years, split the rate and multiply the periods:
+
+\[A = P\left(1 + \frac{R}{n}\right)^{n y}\]
+
+So 8% compounded semiannually for one year is two periods at 4%: \(P(1.04)^2\), **not** \(P(1.08)\).
+
+**Simple vs. compound over exactly 2 years — a shortcut.** For the same \(P\) and \(r\), compound
+interest beats simple interest after 2 years by exactly
+
+\[P r^{2}\]
+
+(the interest that the first year's interest earns in the second year). It's a fast way to compute
+the gap without expanding both formulas.
+
+## Worked examples
+
+**Simple interest.** $2,000 at 5% simple annual interest for 3 years earns
+\(I = 2000 \times 0.05 \times 3 = 300\), for a balance of \(2000 + 300 = 2300\).
+
+**Compound interest.** $5,000 at 10% compounded annually for 2 years grows to
+\(5000(1.1)^2 = 5000 \times 1.21 = 6050\). The interest *earned* is \(6050 - 5000 = 1050\) — more
+than the \(1000\) simple interest would give, because the first year's $500 itself earns 10%.
+
+**The 2-year gap.** With \(P = 4000\) and \(r = 0.10\), the difference between compound and simple
+interest over 2 years is \(P r^2 = 4000 \times 0.01 = 40\). Check: compound earns
+\(4000(1.21) - 4000 = 840\); simple earns \(4000 \times 0.1 \times 2 = 800\); the gap is \(40\). ✓
+
+## Common traps
+
+- **Balance vs. interest earned.** \(P(1+r)^t\) and \(P(1+rt)\) are *balances*. If the question
+  asks for interest, subtract the principal. Reading "how much interest" and answering with the
+  total is the classic trap.
+- **Compounding periods.** A rate compounded semiannually or quarterly must be divided by the
+  number of periods per year, and the exponent multiplied by it. Applying the full annual rate each
+  sub-period massively overstates the result.
+- **Simple where compound is meant (and vice versa).** Over one period they agree; after that they
+  diverge, and the difference is exactly what many questions test.
+- **Decimal slips.** 6% is \(0.06\), not \(0.6\) or \(0.006\). A misplaced decimal changes the
+  answer by a factor of ten.
+- **Reversing growth by subtraction.** To find the principal behind a compounded balance you must
+  *divide* by \((1+r)^t\); taking a flat percentage off the final amount is not the inverse.
+
+## Key takeaways
+
+- Simple interest: \(I = Prt\), balance \(= P(1 + rt)\) — linear, only the principal earns.
+- Compound interest: balance \(= P(1+r)^t\), interest earned \(= P[(1+r)^t - 1]\) — geometric.
+- Sub-annual compounding: rate \(\to R/n\), periods \(\to ny\).
+- The 2-year compound-minus-simple gap is \(P r^2\).
+- Always confirm whether the question wants the **interest** or the **final balance**.
diff --git a/content/lessons/quant/quant-word-problems-mixtures.md b/content/lessons/quant/quant-word-problems-mixtures.md
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+---
+id: quant-word-problems-mixtures
+section: quant
+topic: word-problems
+subtopic: mixtures
+title: "Mixtures"
+tags: [mixtures, concentration, alligation, weighted-average]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Mixture problems ask you to combine things of different "strengths" — salt in water, alcohol in a
+solution, cheap coffee with expensive coffee — and reason about the result. Almost every one of
+them yields to a single idea: **track the pure quantity of the thing you care about, not the
+percentages.** Percentages don't add; amounts do.
+
+## Core concepts
+
+**Concentration is a part-over-whole.** For a solution,
+
+\[\text{concentration} = \frac{\text{amount of substance}}{\text{total amount}}\]
+
+So 40 liters of a 15% salt solution contains \(0.15 \times 40 = 6\) liters of salt. When you mix or
+dilute, the **amount of substance** is what you carry from line to line.
+
+**Mixing is a weighted average.** Combine \(a\) units at concentration \(p\) with \(b\) units at
+concentration \(q\). Add the *amounts*, then divide by the *total*:
+
+\[\text{result} = \frac{p\,a + q\,b}{a + b}\]
+
+The answer always lands **between** \(p\) and \(q\), pulled toward whichever quantity is larger.
+A simple average \(\tfrac{p+q}{2}\) is only right when \(a = b\).
+
+**Diluting or concentrating.** Adding pure water adds to the total but **not** to the substance.
+Adding the pure substance adds to both. Set up an equation on the quantity that stays fixed:
+
+> Add \(w\) liters of water to 10 liters of 40% salt to reach 25%. The salt is fixed at
+> \(0.40 \times 10 = 4\) liters, so \(\dfrac{4}{10 + w} = 0.25 \Rightarrow 10 + w = 16 \Rightarrow w = 6\).
+
+**Alligation (a shortcut).** The ratio of the two quantities in a mix is the *inverse* of their
+distances from the blend value:
+
+\[\frac{a}{b} = \frac{q - m}{m - p}\]
+
+where \(m\) is the mixture's concentration. It's the weighted average solved for the ratio — handy
+when a question asks "in what ratio?"
+
+## Worked examples
+
+**Weighted average.** Mix 20 L of 30% acid with 30 L of 80% acid. Acid \(= 0.30(20) + 0.80(30) =
+6 + 24 = 30\) L in \(50\) L total, so \(\tfrac{30}{50} = 60\%\). (Note it's above the simple average
+of 55%, because more of the stronger solution is present.)
+
+**Alligation.** In what ratio should $8/kg and $12/kg coffee be mixed to sell at $9/kg?
+\(\dfrac{\text{amt of \$8}}{\text{amt of \$12}} = \dfrac{12 - 9}{9 - 8} = \dfrac{3}{1}\) — three parts
+cheap to one part expensive.
+
+## Common traps
+
+- **Averaging the percentages.** \(\tfrac{30\% + 80\%}{2} = 55\%\) is wrong unless the volumes are
+  equal. Weight by the amounts.
+- **Adding water to the substance too.** Diluting with water raises the total but leaves the amount
+  of substance unchanged — that fixed amount is your equation.
+- **Answering the wrong part.** If you solve for one component, re-read to check whether the
+  question wanted that one or the other (or the total).
+- **Inverting the alligation ratio.** The ratio of amounts is inverse to the distances — the
+  quantity *farther* from the blend value is the *smaller* share.
+
+## Key takeaways
+
+- Track the **amount** of the substance (part = concentration × total), not the percentages.
+- A mixture is a weighted average; the result sits between the inputs, nearer the larger quantity.
+- Dilution/concentration: write an equation on the quantity that stays fixed.
+- Alligation gives the ratio directly: distances from the blend value, inverted.
diff --git a/content/lessons/quant/quant-word-problems-translation.md b/content/lessons/quant/quant-word-problems-translation.md
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+---
+id: quant-word-problems-translation
+section: quant
+topic: word-problems
+subtopic: translation
+title: "Translating Words into Equations"
+tags: [word-problem, translation, algebra, setup]
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Overview
+
+Most GMAT Focus word problems are not hard *algebra* — they are hard *reading*. The whole game is
+turning an English sentence into a correct equation. Once the equation is right, the solving is
+usually routine. This lesson is a phrasebook: it maps the words you actually see on the exam onto
+the symbols they stand for, and it flags the phrasings that are engineered to be misread.
+
+## Core concepts
+
+**Name the unknown first.** Before writing any equation, decide what the variable is and write it
+down: "let \(n\) = the number," "let \(a\) = adult tickets." Nearly every wrong answer on a word
+problem is the value of *some* quantity in the problem — just not the one that was asked. Naming the
+unknown keeps you honest about which number to report at the end.
+
+**A translation dictionary.** These map almost every arithmetic phrase you will meet:
+
+| English | Symbol |
+|---------|--------|
+| is, was, will be, gives, results in | \(=\) |
+| sum, more than, increased by, added to, exceeds by | \(+\) |
+| difference, less than, decreased by, fewer than | \(-\) |
+| of, times, product, twice (\(\times 2\)), triple (\(\times 3\)) | \(\times\) |
+| per, out of, ratio, quotient | \(\div\) |
+| a number, an unknown | a variable, e.g. \(n\) |
+
+**Order matters for subtraction and division.** "5 **less than** \(x\)" is \(x - 5\), **not**
+\(5 - x\) — the thing after "less than" is what you subtract *from*. Likewise "8 less than 3 times a
+number" is \(3n - 8\). Read these right-to-left.
+
+**Grouping: watch what the multiplier attaches to.** "Twice a number increased by 6" is
+\(2n + 6\). "Twice the sum of a number and 6" is \(2(n + 6)\). The words *the sum of* signal a
+group — a set of parentheses.
+
+**Consecutive integers.** Consecutive integers are \(n,\ n+1,\ n+2,\dots\); consecutive **even** or
+**odd** integers are \(n,\ n+2,\ n+4,\dots\). Their sum is not \(3n\) — the \(+1,+2\) corrections
+matter.
+
+**Two unknowns, one relationship.** When a problem describes one quantity *in terms of* another
+("the larger is 8 less than 3 times the smaller"), write both in terms of a single variable, then
+use the remaining fact (usually a sum or total) to get one equation in one unknown.
+
+## Worked examples
+
+**Basic.** "Twice a number, increased by 6, is 20." Let \(n\) be the number:
+\(2n + 6 = 20 \Rightarrow 2n = 14 \Rightarrow n = 7\).
+
+**Two people, one variable.** "Maria is 4 years older than twice her brother's age; their ages sum
+to 34." Let \(b\) = brother's age, so Maria \(= 2b + 4\). Then
+\(b + (2b + 4) = 34 \Rightarrow 3b = 30 \Rightarrow b = 10\), and Maria \(= 2(10) + 4 = 24\). The
+question asks for **Maria**, so the answer is 24 — not the 10 you solved for.
+
+**"Less than" order.** "The larger of two numbers is 8 less than 3 times the smaller; the two sum to
+84." Smaller \(= s\), larger \(= 3s - 8\). Then \(s + (3s - 8) = 84 \Rightarrow 4s = 92
+\Rightarrow s = 23\), and the larger is \(3(23) - 8 = 61\).
+
+## Common traps
+
+- **Answering the wrong quantity.** You solved for the smaller number, the brother's age, or the
+  number of adults — but the question asked for the larger, Maria, or the students. Re-read the last
+  line before you bubble.
+- **Reversing subtraction/division.** "\(x\) less than \(y\)" is \(y - x\). "The quotient of \(x\)
+  and \(y\)" is \(x / y\). Getting the order backward flips the sign or inverts the fraction.
+- **Misplacing the multiplier.** \(2n + 6\) vs. \(2(n + 6)\) — "the sum of" is your cue to group.
+- **Forgetting the consecutive-integer correction.** Three consecutive integers summing to 72 give
+  \(3n + 3 = 72\), so \(n = 23\) — not \(72 / 3 = 24\).
+
+## Key takeaways
+
+- Name the unknown in words first; report the quantity the question actually asks for.
+- Use the phrase-to-symbol dictionary, and read "less than" / "fewer than" right-to-left.
+- "The sum of …" means parentheses — group before you multiply.
+- Express everything in one variable, then spend the last fact (a total or sum) to close the system.
diff --git a/content/questions/quant/quant-algebra-sequences-functions-0052.md b/content/questions/quant/quant-algebra-sequences-functions-0052.md
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+---
+id: quant-algebra-sequences-functions-0052
+section: quant
+topic: algebra
+subtopic: sequences-functions
+type: problem-solving
+difficulty: easy
+tags: [arithmetic-sequence, nth-term]
+choices:
+  A: "76"
+  B: "80"
+  C: "83"
+  D: "87"
+  E: "91"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+An arithmetic sequence has first term 7 and common difference 4. What is its 20th term?
+
+## Explanation
+
+For an arithmetic sequence, the \(n\)th term is
+
+\[a_n = a_1 + (n-1)\,d.\]
+
+The key is the multiplier \((n-1)\): to reach the 20th term you add the common difference **19**
+times, not 20, because the first term itself takes zero steps.
+
+\[a_{20} = 7 + (20-1)\cdot 4 = 7 + 19\cdot 4 = 7 + 76 = 83.\]
+
+So the 20th term is **83**.
+
+## Hints
+
+- The \(n\)th term is \(a_1 + (n-1)d\) — count the *gaps* between terms, not the terms.
+- From the 1st term to the 20th term there are 19 steps of size 4.
diff --git a/content/questions/quant/quant-algebra-sequences-functions-0053.md b/content/questions/quant/quant-algebra-sequences-functions-0053.md
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+---
+id: quant-algebra-sequences-functions-0053
+section: quant
+topic: algebra
+subtopic: sequences-functions
+type: problem-solving
+difficulty: easy
+tags: [geometric-sequence, nth-term]
+choices:
+  A: "32"
+  B: "48"
+  C: "96"
+  D: "160"
+  E: "192"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A geometric sequence has first term 3 and common ratio 2. What is its 6th term?
+
+## Explanation
+
+For a geometric sequence, the \(n\)th term is
+
+\[a_n = a_1 \, r^{\,n-1}.\]
+
+As with arithmetic sequences, the exponent is \((n-1)\): reaching the 6th term takes **5**
+multiplications by the ratio, not 6.
+
+\[a_6 = 3 \cdot 2^{6-1} = 3 \cdot 2^{5} = 3 \cdot 32 = 96.\]
+
+So the 6th term is **96**. (Writing the terms out confirms it: 3, 6, 12, 24, 48, **96**.)
+
+## Hints
+
+- The \(n\)th term of a geometric sequence is \(a_1 r^{\,n-1}\).
+- To land on the 6th term you multiply by the ratio 5 times, then don't forget the factor of 3.
diff --git a/content/questions/quant/quant-algebra-sequences-functions-0054.md b/content/questions/quant/quant-algebra-sequences-functions-0054.md
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+---
+id: quant-algebra-sequences-functions-0054
+section: quant
+topic: algebra
+subtopic: sequences-functions
+type: problem-solving
+difficulty: medium
+tags: [arithmetic-sequence, series-sum]
+choices:
+  A: "855"
+  B: "900"
+  C: "950"
+  D: "1000"
+  E: "1900"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+An arithmetic sequence has first term 5 and last term 95, and it contains 19 terms in all. What is
+the sum of all 19 terms?
+
+## Explanation
+
+The sum of an arithmetic sequence is the number of terms times the **average** of the first and last
+term:
+
+\[S_n = \frac{n\,(a_1 + a_n)}{2}.\]
+
+You don't need the common difference — the first and last terms and the count are enough. Here
+\(n = 19\), \(a_1 = 5\), \(a_{n} = 95\):
+
+\[S_{19} = \frac{19\,(5 + 95)}{2} = \frac{19 \cdot 100}{2} = 19 \cdot 50 = 950.\]
+
+So the sum is **950**.
+
+(As a check, the common difference is \(d = \frac{95 - 5}{19 - 1} = \frac{90}{18} = 5\), so the terms
+are \(5, 10, 15, \dots, 95\) — exactly 19 of them.)
+
+## Hints
+
+- Sum \(= \dfrac{n(a_1 + a_n)}{2}\): count of terms times the average of the ends.
+- The average of the first and last term is \(\frac{5+95}{2} = 50\); you have 19 of them.
+
diff --git a/content/questions/quant/quant-algebra-sequences-functions-0055.md b/content/questions/quant/quant-algebra-sequences-functions-0055.md
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+---
+id: quant-algebra-sequences-functions-0055
+section: quant
+topic: algebra
+subtopic: sequences-functions
+type: problem-solving
+difficulty: medium
+tags: [functions, function-notation, composition]
+choices:
+  A: "5"
+  B: "7"
+  C: "10"
+  D: "11"
+  E: "13"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+The function \(f\) is defined by \(f(x) = 2x - 3\). What is the value of \(f(f(4))\)?
+
+## Explanation
+
+A composition like \(f(f(4))\) is evaluated **inside-out**: compute the inner function first, then feed
+that result back into \(f\).
+
+**Inner:** \(f(4) = 2\cdot 4 - 3 = 8 - 3 = 5.\)
+
+**Outer:** \(f(5) = 2\cdot 5 - 3 = 10 - 3 = 7.\)
+
+So \(f(f(4)) = 7\).
+
+Two things to keep straight: don't stop after the inner step (that gives 5, not the final answer), and
+don't drop the \(-3\) — it applies **every** time you use the rule.
+
+## Hints
+
+- \(f(f(4))\) means: find \(f(4)\) first, then apply \(f\) to that answer.
+- Substitute into \(2x - 3\) each time; the \(-3\) doesn't disappear on the second application.
diff --git a/content/questions/quant/quant-algebra-sequences-functions-0056.md b/content/questions/quant/quant-algebra-sequences-functions-0056.md
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+---
+id: quant-algebra-sequences-functions-0056
+section: quant
+topic: algebra
+subtopic: sequences-functions
+type: problem-solving
+difficulty: medium
+tags: [recursive-sequence, recursion]
+choices:
+  A: "9"
+  B: "16"
+  C: "17"
+  D: "31"
+  E: "33"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A sequence is defined by \(a_1 = 2\) and \(a_{n+1} = 2a_n - 1\) for every \(n \ge 1\). What is the
+value of \(a_5\)?
+
+## Explanation
+
+There's no shortcut needed — apply the rule one step at a time, labeling each term so you stop at the
+right one:
+
+\[a_1 = 2\]
+\[a_2 = 2(2) - 1 = 3\]
+\[a_3 = 2(3) - 1 = 5\]
+\[a_4 = 2(5) - 1 = 9\]
+\[a_5 = 2(9) - 1 = 17\]
+
+So \(a_5 = 17\).
+
+The most common slip is stopping one term early (\(a_4 = 9\)) or going one too far (\(a_6 = 33\)).
+Labeling each line prevents the off-by-one. (If you spotted the pattern, the closed form is
+\(a_n = 2^{\,n-1} + 1\), which also gives \(a_5 = 16 + 1 = 17\).)
+
+## Hints
+
+- Compute term by term: \(a_2, a_3, a_4, a_5\) — and write the index next to each so you stop at \(a_5\).
+- Each step is "double the previous term, then subtract 1."
diff --git a/content/questions/quant/quant-algebra-sequences-functions-0057.md b/content/questions/quant/quant-algebra-sequences-functions-0057.md
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+---
+id: quant-algebra-sequences-functions-0057
+section: quant
+topic: algebra
+subtopic: sequences-functions
+type: problem-solving
+difficulty: hard
+tags: [arithmetic-sequence, system-of-equations]
+choices:
+  A: "-3"
+  B: "2"
+  C: "5"
+  D: "7"
+  E: "12"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+In an arithmetic sequence, the 4th term is 17 and the 9th term is 42. What is the first term?
+
+## Explanation
+
+Write each given term with the formula \(a_n = a_1 + (n-1)d\):
+
+\[a_4 = a_1 + 3d = 17\]
+\[a_9 = a_1 + 8d = 42\]
+
+Subtract the first equation from the second to eliminate \(a_1\):
+
+\[(a_1 + 8d) - (a_1 + 3d) = 42 - 17 \;\Rightarrow\; 5d = 25 \;\Rightarrow\; d = 5.\]
+
+Now back-substitute into \(a_1 + 3d = 17\):
+
+\[a_1 + 3(5) = 17 \;\Rightarrow\; a_1 = 17 - 15 = 2.\]
+
+So the first term is **2**.
+
+Watch two traps: the question asks for the first term, not the common difference \(d = 5\); and the
+4th term is \(a_1 + 3d\) (three steps), not \(a_1 + 4d\).
+
+## Hints
+
+- Set up \(a_1 + 3d = 17\) and \(a_1 + 8d = 42\), then subtract to find \(d\) first.
+- There are 5 steps between the 4th and 9th terms, and the values differ by \(42 - 17 = 25\).
diff --git a/content/questions/quant/quant-algebra-sequences-functions-0058.md b/content/questions/quant/quant-algebra-sequences-functions-0058.md
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+---
+id: quant-algebra-sequences-functions-0058
+section: quant
+topic: algebra
+subtopic: sequences-functions
+type: problem-solving
+difficulty: hard
+tags: [geometric-sequence, threshold, exponents]
+choices:
+  A: "405"
+  B: "729"
+  C: "1215"
+  D: "2025"
+  E: "3645"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A geometric sequence has first term 5 and common ratio 3. What is the value of the first term in the
+sequence that is greater than 1000?
+
+## Explanation
+
+The terms are \(a_n = 5 \cdot 3^{\,n-1}\). Because each term triples, they climb quickly — just list
+them until one clears 1000:
+
+\[5,\; 15,\; 45,\; 135,\; 405,\; 1215,\; \dots\]
+
+The 5th term is \(5 \cdot 3^4 = 405\), which is still below 1000. The 6th term is
+
+\[a_6 = 5 \cdot 3^{5} = 5 \cdot 243 = 1215,\]
+
+which is the first term to exceed 1000. So the answer is **1215**.
+
+Common misses: stopping at the last term *below* the threshold (405), overshooting to the next term
+(3645), or dropping the leading factor of 5 and using a bare power of 3 (\(3^6 = 729\)).
+
+## Hints
+
+- List the terms — tripling each time, they pass 1000 fast, so you only need a few.
+- Find the last term below 1000, then take one more step (×3) and check that it's the first one over.
diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0017.md b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md
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+---
+id: quant-arithmetic-exponents-roots-0017
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [exponents, power-rules]
+choices:
+  A: "4"
+  B: "8"
+  C: "16"
+  D: "64"
+  E: "512"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(\dfrac{2^{5} \cdot 2^{3}}{2^{6}}\)?
+
+## Explanation
+
+Keep everything in base 2 and just track exponents. When you multiply like bases you **add**
+exponents; when you divide, you **subtract**:
+
+\[\frac{2^{5} \cdot 2^{3}}{2^{6}} = 2^{\,5 + 3 - 6} = 2^{2} = 4\]
+
+**The trap:** answer **E (512)** comes from *multiplying* the exponents in the numerator
+(\(5 \times 3 = 15\), then \(15 - 6 = 9\), giving \(2^{9} = 512\)). You add exponents when the bases
+are multiplied, not multiply them. Answer **D (64)** is \(2^{6}\), what you get if you drop one of
+the numerator terms.
+
+## Hints
+
+- Rewrite the whole expression as a single power of 2.
+- Multiply like bases → add exponents; divide → subtract exponents.
diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0018.md b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md
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+---
+id: quant-arithmetic-exponents-roots-0018
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: easy
+tags: [exponents, negative-exponent, zero-exponent]
+choices:
+  A: "0"
+  B: "1/9"
+  C: "10/9"
+  D: "2"
+  E: "9"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(3^{-2} + 3^{0}\)?
+
+## Explanation
+
+Two rules do all the work here. A **negative exponent** means take the reciprocal, and **anything
+nonzero raised to the 0 power is 1**:
+
+\[3^{-2} = \frac{1}{3^{2}} = \frac{1}{9}, \qquad 3^{0} = 1\]
+
+So the sum is
+
+\[\frac{1}{9} + 1 = \frac{1}{9} + \frac{9}{9} = \frac{10}{9}\]
+
+**The traps:** answer **B (1/9)** drops the \(3^{0} = 1\) term (a studier who thinks \(3^0 = 0\)).
+Answer **D (2)** comes from mistakenly treating \(3^{-2}\) as \(1\) as well. A negative exponent is
+never a negative number, so **A (0)** and **E (9)** are out.
+
+## Hints
+
+- A negative exponent means "reciprocal," not "negative": \(3^{-2} = \tfrac{1}{3^2}\).
+- Remember that \(3^{0} = 1\), not \(0\).
diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0019.md b/content/questions/quant/quant-arithmetic-exponents-roots-0019.md
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+---
+id: quant-arithmetic-exponents-roots-0019
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [exponents, common-base, equations]
+choices:
+  A: "1"
+  B: "2"
+  C: "3"
+  D: "4"
+  E: "5"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(2^{\,x+1} = 8^{\,x-1}\), what is the value of \(x\)?
+
+## Explanation
+
+The two sides have different bases, so first rewrite them with the **same base**. Since
+\(8 = 2^{3}\):
+
+\[8^{\,x-1} = \left(2^{3}\right)^{x-1} = 2^{\,3(x-1)} = 2^{\,3x-3}\]
+
+Now both sides are powers of 2, so the exponents must be equal:
+
+\[x + 1 = 3x - 3 \;\Rightarrow\; 4 = 2x \;\Rightarrow\; x = 2\]
+
+Check: \(2^{2+1} = 2^{3} = 8\) and \(8^{2-1} = 8^{1} = 8\). ✓
+
+**The trap:** answer **A (1)** comes from forgetting to distribute the 3, i.e. writing the exponent
+as \(3x - 1\) instead of \(3(x-1) = 3x - 3\). Distributing carefully is the whole question.
+
+## Hints
+
+- Rewrite \(8\) as a power of \(2\) so both sides share a base.
+- When two equal powers have the same base, set the exponents equal — and distribute the exponent
+  fully.
diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0020.md b/content/questions/quant/quant-arithmetic-exponents-roots-0020.md
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+---
+id: quant-arithmetic-exponents-roots-0020
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [exponents, fractional-exponents, roots]
+choices:
+  A: "6"
+  B: "9"
+  C: "18"
+  D: "54"
+  E: "729"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \(27^{2/3}\)?
+
+## Explanation
+
+A fractional exponent \(m/n\) means "take the \(n\)th root, then raise to the \(m\)th power." The
+denominator is the root; the numerator is the power. Take the root first to keep the numbers small:
+
+\[27^{2/3} = \left(27^{1/3}\right)^{2} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9\]
+
+**The traps:**
+
+- **E (729)** ignores the cube-root denominator and just squares: \(27^{2} = 729\).
+- **A (6)** multiplies the root by the numerator instead of raising to it: \(3 \times 2 = 6\).
+- **C (18)** and **D (54)** treat the exponent as ordinary multiplication (\(27 \times \tfrac{2}{3} = 18\),
+  or \(27 \times 2 = 54\)) rather than as a root-and-power.
+
+## Hints
+
+- The denominator of the exponent is a root; the numerator is a power.
+- \(27^{2/3} = (\sqrt[3]{27})^{2}\) — take the cube root first.
diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0021.md b/content/questions/quant/quant-arithmetic-exponents-roots-0021.md
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+---
+id: quant-arithmetic-exponents-roots-0021
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: medium
+tags: [roots, radicals, simplifying]
+choices:
+  A: "\(2\sqrt{17}\)"
+  B: "\(15\sqrt{2}\)"
+  C: "\(8\sqrt{2}\)"
+  D: "\(4\sqrt{2}\)"
+  E: "\(34\sqrt{2}\)"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is \(\sqrt{50} + \sqrt{18}\) in simplest form?
+
+## Explanation
+
+You can only add radicals once the parts **under** the root match, so simplify each one by pulling
+out perfect squares:
+
+\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \qquad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\]
+
+Now both terms are multiples of \(\sqrt{2}\), so add the coefficients:
+
+\[5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\]
+
+**The traps:**
+
+- **A (\(2\sqrt{17}\))** comes from illegally merging the roots: \(\sqrt{50} + \sqrt{18} \neq \sqrt{68}\).
+  Square roots do not add that way.
+- **B (\(15\sqrt{2}\))** multiplies the coefficients (\(5 \times 3\)) instead of adding them.
+
+## Hints
+
+- Simplify each radical first by factoring out the largest perfect square.
+- You can add radicals only when the number under the root is the same — then add the coefficients.
diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0022.md b/content/questions/quant/quant-arithmetic-exponents-roots-0022.md
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+---
+id: quant-arithmetic-exponents-roots-0022
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: hard
+tags: [exponents, factoring-powers, common-base]
+choices:
+  A: "3"
+  B: "6"
+  C: "10"
+  D: "12"
+  E: "48"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(2^{n} + 2^{n} + 2^{n} + 2^{n} = 2^{12}\), what is the value of \(n\)?
+
+## Explanation
+
+The left side is **four copies of the same term**, so factor — do not treat it as a new base.
+Adding four equal terms is multiplying by 4:
+
+\[2^{n} + 2^{n} + 2^{n} + 2^{n} = 4 \cdot 2^{n}\]
+
+Now write the coefficient as a power of 2 so both sides share a base. Since \(4 = 2^{2}\):
+
+\[4 \cdot 2^{n} = 2^{2} \cdot 2^{n} = 2^{\,n+2}\]
+
+Setting \(2^{\,n+2} = 2^{12}\) gives \(n + 2 = 12\), so \(n = 10\).
+
+**The traps:**
+
+- **A (3)** treats the sum as a *product* of the four terms, \((2^{n})^{4} = 2^{4n} = 2^{12}\),
+  giving \(4n = 12\). Adding equal powers is not the same as multiplying them.
+- **B (6)** correctly writes \(4 \cdot 2^{n}\) but then *multiplies* the exponents (\(2 \cdot n = 12\))
+  instead of adding: \(2^{2} \cdot 2^{n} = 2^{n+2}\), not \(2^{2n}\).
+- **D (12)** simply ignores the coefficient of 4.
+
+## Hints
+
+- Four copies of \(2^{n}\) is \(4 \cdot 2^{n}\), not \((2^{n})^{4}\).
+- Rewrite \(4\) as \(2^{2}\), then combine with \(2^{n}\) by **adding** exponents.
diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0023.md b/content/questions/quant/quant-arithmetic-exponents-roots-0023.md
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+---
+id: quant-arithmetic-exponents-roots-0023
+section: quant
+topic: arithmetic
+subtopic: exponents-roots
+type: problem-solving
+difficulty: hard
+tags: [exponents, common-base, factoring-powers]
+choices:
+  A: "1/9"
+  B: "2"
+  C: "8"
+  D: "9"
+  E: "\(3^{38}\)"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(x = 3^{20}\) and \(y = 3^{18}\), what is the value of \(\dfrac{x - y}{y}\)?
+
+## Explanation
+
+Do not try to evaluate \(3^{20}\) or \(3^{18}\) — they are enormous. Split the fraction so the huge
+powers turn into a ratio you can simplify:
+
+\[\frac{x - y}{y} = \frac{x}{y} - \frac{y}{y} = \frac{3^{20}}{3^{18}} - 1\]
+
+Dividing like bases means **subtracting** exponents:
+
+\[\frac{3^{20}}{3^{18}} = 3^{\,20 - 18} = 3^{2} = 9\]
+
+So the value is \(9 - 1 = 8\).
+
+**The traps:**
+
+- **D (9)** is \(\dfrac{x}{y} = 3^{2}\) — correct so far, but it forgets to subtract the \(\dfrac{y}{y} = 1\)
+  term.
+- **B (2)** stops at the exponent difference \(20 - 18 = 2\), as if the whole expression equaled
+  \(3^{20-18}\) without the "\(-1\)".
+- **E (\(3^{38}\))** comes from *multiplying* the powers (adding exponents \(20 + 18\)) instead of
+  dividing.
+
+## Hints
+
+- Break \(\dfrac{x - y}{y}\) into \(\dfrac{x}{y} - 1\) before touching the exponents.
+- Dividing like bases subtracts exponents: \(\dfrac{3^{20}}{3^{18}} = 3^{2}\).
diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0031.md b/content/questions/quant/quant-counting-probability-combinatorics-0031.md
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+---
+id: quant-counting-probability-combinatorics-0031
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: easy
+tags: [combinatorics, combinations, committee]
+choices:
+  A: "21"
+  B: "24"
+  C: "56"
+  D: "336"
+  E: "512"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A book club will choose a 3-person panel from its 8 members. In how many different ways can the
+panel be chosen?
+
+## Explanation
+
+A panel is an **unordered** selection — being picked "first" versus "third" makes no difference,
+so this is a combination:
+
+\[C(8,3) = \frac{8 \cdot 7 \cdot 6}{3!} = \frac{336}{6} = 56\]
+
+**The trap:** answer **D (336)** is \(P(8,3) = 8\cdot7\cdot6\), the *ordered* count. That treats
+"Ann, then Bob, then Cara" as different from "Cara, then Bob, then Ann," but a panel is the same
+group either way — so you must divide by \(3! = 6\). Answer **B (24)** comes from \(8 \times 3\)
+and **A (21)** from \(8 + 7 + 6\); neither reflects how counting works. Answer **E (512)** is
+\(8^3\), which would allow the same person to be picked three times.
+
+## Hints
+
+- Does the order in which the three people are chosen change who is on the panel?
+- If order does not matter, use \(C(n,k) = \dfrac{n!}{k!\,(n-k)!}\).
diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0032.md b/content/questions/quant/quant-counting-probability-combinatorics-0032.md
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+---
+id: quant-counting-probability-combinatorics-0032
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: easy
+tags: [combinatorics, permutations, factorial, arrangement]
+choices:
+  A: "20"
+  B: "25"
+  C: "60"
+  D: "120"
+  E: "3125"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+In how many different orders can 5 distinct books be arranged on a shelf?
+
+## Explanation
+
+Every book is distinct and every position is distinct, so this is a straight arrangement of all
+5 objects:
+
+\[5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\]
+
+**The trap:** answer **E (3125)** is \(5^5\), which would count arrangements as if each of the 5
+slots could independently hold any of the 5 books — but once a book is placed it cannot be reused,
+so the choices shrink \(5 \to 4 \to 3 \to \dots\). Answer **A (20)** is \(5 \times 4\) (arranging
+only 2 of the books), and **C (60)** is \(\tfrac{5!}{2}\). Answer **B (25)** is \(5^2\).
+
+## Hints
+
+- Placing the first book leaves 4 choices for the next slot, then 3, and so on.
+- Multiply the shrinking number of choices: that product is \(5!\).
diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0033.md b/content/questions/quant/quant-counting-probability-combinatorics-0033.md
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+---
+id: quant-counting-probability-combinatorics-0033
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: medium
+tags: [combinatorics, combinations, multiplication-principle]
+choices:
+  A: "16"
+  B: "60"
+  C: "40"
+  D: "126"
+  E: "3600"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A committee of 5 is to be formed from a group of 5 seniors and 4 juniors. If the committee must
+contain exactly 3 seniors and exactly 2 juniors, how many different committees are possible?
+
+## Explanation
+
+Choosing the seniors and choosing the juniors are two independent stages, and within each stage
+order does not matter — so use combinations and **multiply** (the "AND" principle):
+
+\[C(5,3) \times C(4,2) = 10 \times 6 = 60\]
+
+Here \(C(5,3) = \dfrac{5\cdot4\cdot3}{3!} = 10\) and \(C(4,2) = \dfrac{4\cdot3}{2!} = 6\).
+
+**The trap:** answer **A (16)** *adds* the two counts (\(10 + 6\)) instead of multiplying — but you
+need a group of seniors **and** a group of juniors together, so the counts multiply. Answer
+**C (40)** swaps the roles, computing \(C(5,2)\times C(4,3) = 10 \times 4\) (2 seniors, 3 juniors).
+Answer **D (126)** is \(C(9,5)\), which ignores the required senior/junior split. Answer
+**E (3600)** treats each selection as ordered (\(P(5,3)\times P(4,2)\)).
+
+## Hints
+
+- Count the ways to pick the seniors and the ways to pick the juniors separately.
+- "3 seniors AND 2 juniors" — do the two counts add or multiply?
+- Within each group, does the order you pick people in matter?
diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0034.md b/content/questions/quant/quant-counting-probability-combinatorics-0034.md
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+---
+id: quant-counting-probability-combinatorics-0034
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: medium
+tags: [combinatorics, arrangements, identical-items, permutations]
+choices:
+  A: "120"
+  B: "60"
+  C: "30"
+  D: "20"
+  E: "10"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+How many distinct arrangements can be made using all the letters of the word **LEVEL**?
+
+## Explanation
+
+LEVEL has 5 letters, but they are not all distinct: **L** appears twice, **E** appears twice, and
+**V** appears once. When items repeat, divide \(n!\) by the factorial of each repeat count so that
+swapping identical letters is not counted as a new arrangement:
+
+\[\frac{5!}{2!\,\cdot\,2!\,\cdot\,1!} = \frac{120}{2 \cdot 2} = \frac{120}{4} = 30\]
+
+**The trap:** answer **A (120)** is \(5!\), which counts the two L's (and the two E's) as though
+they were distinguishable — but "L\(_1\)EVEL\(_2\)" looks identical to "L\(_2\)EVEL\(_1\)," so those
+duplicates must be divided out. Answer **B (60)** is \(\tfrac{5!}{2!}\), correcting for only *one*
+of the two repeated letters. Answers **D** and **E** under-count further.
+
+## Hints
+
+- Count how many times each letter appears: which letters repeat?
+- Divide \(5!\) by the factorial of *each* repeated letter's count.
diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0035.md b/content/questions/quant/quant-counting-probability-combinatorics-0035.md
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@@ -0,0 +1,44 @@
+---
+id: quant-counting-probability-combinatorics-0035
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: medium
+tags: [combinatorics, circular-arrangement, permutations]
+choices:
+  A: "24"
+  B: "60"
+  C: "360"
+  D: "720"
+  E: "120"
+answer: E
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+In how many distinct ways can 6 people be seated around a circular table, if two seatings are
+considered the same when one can be obtained from the other by rotating the whole table?
+
+## Explanation
+
+Around a circle there is no fixed "first" seat: rotating everyone one seat to the left produces the
+same arrangement of neighbors. Fix one person's seat as a reference point and arrange the remaining
+\(6 - 1 = 5\) people in the other seats:
+
+\[(6-1)! = 5! = 120\]
+
+**The trap:** answer **D (720)** is \(6!\), the count for a *straight row* of 6 seats. In a row all
+6 positions are distinct, but a circle's rotations collapse every arrangement into a group of 6
+equivalent rotations, so you divide \(6!\) by 6 — equivalently, use \((n-1)!\). Answer **C (360)**
+is \(\tfrac{6!}{2}\), an incorrect partial correction.
+
+## Hints
+
+- Around a round table, is there a fixed "first" chair, or do rotations look identical?
+- Pin one person down as a reference, then arrange the rest: that gives \((n-1)!\).
diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0036.md b/content/questions/quant/quant-counting-probability-combinatorics-0036.md
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@@ -0,0 +1,48 @@
+---
+id: quant-counting-probability-combinatorics-0036
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: hard
+tags: [combinatorics, combinations, complement, at-least]
+choices:
+  A: "31"
+  B: "4"
+  C: "18"
+  D: "35"
+  E: "45"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A 3-person team is to be selected from a pool of 4 men and 3 women. How many different teams
+include **at least one woman**?
+
+## Explanation
+
+"At least one" is fastest to count with the **complement**: take every possible team and subtract
+the teams that violate the condition (the all-men teams).
+
+\[\underbrace{C(7,3)}_{\text{all teams}} - \underbrace{C(4,3)}_{\text{all men}} = 35 - 4 = 31\]
+
+**The trap:** answer **E (45)** comes from "pick 1 woman, then any 2 of the remaining 6 people":
+\(C(3,1)\times C(6,2) = 3 \times 15 = 45\). That double-counts — a team with two women gets counted
+once for each woman chosen as "the guaranteed one." Answer **C (18)** counts only *exactly one*
+woman, \(C(3,1)\times C(4,2) = 3 \times 6 = 18\), missing the two-woman and three-woman teams.
+Answer **D (35)** forgets to remove the all-men teams, and **B (4)** is just the all-men count you
+were supposed to subtract.
+
+You can confirm 31 the direct way: exactly 1 woman \(= 18\), exactly 2 \(= C(3,2)C(4,1) = 12\),
+exactly 3 \(= C(3,3)C(4,0) = 1\); \(18 + 12 + 1 = 31\).
+
+## Hints
+
+- Counting "at least one" directly is error-prone — try counting the opposite.
+- Total teams minus all-men teams leaves exactly the teams with at least one woman.
diff --git a/content/questions/quant/quant-counting-probability-combinatorics-0037.md b/content/questions/quant/quant-counting-probability-combinatorics-0037.md
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+---
+id: quant-counting-probability-combinatorics-0037
+section: quant
+topic: counting-probability
+subtopic: combinatorics
+type: problem-solving
+difficulty: hard
+tags: [combinatorics, permutations, arrangements, restrictions, block-method]
+choices:
+  A: "24"
+  B: "48"
+  C: "72"
+  D: "120"
+  E: "240"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Five students are to be arranged in a row for a photograph. Two of them, Priya and Quinn, insist on
+standing next to each other. In how many different arrangements can the five students stand?
+
+## Explanation
+
+Use the **block method** for "must be together." Glue Priya and Quinn into a single unit. That
+leaves 4 items to arrange in the row — the block plus the other 3 students:
+
+\[4! = 24 \text{ ways to order the units.}\]
+
+Within the block, Priya and Quinn can stand in 2 orders (Priya–Quinn or Quinn–Priya), so multiply:
+
+\[4! \times 2! = 24 \times 2 = 48\]
+
+**The trap:** answer **C (72)** is the count for the *opposite* condition — arrangements where the
+two are **not** adjacent: \(5! - 48 = 120 - 48 = 72\). If you solve for the wrong condition (or
+subtract when you should not), you land on 72. Answer **A (24)** forgets the internal \(2!\) for the
+block's two orders, and **D (120)** is \(5!\), ignoring the restriction entirely. Answer
+**E (240)** is \(5! \times 2\), which double-applies the arrangement.
+
+## Hints
+
+- Treat the two who must be adjacent as one combined block, then count the arrangements of the blocks.
+- Don't forget the two people inside the block can switch places.
diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0059.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0059.md
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+---
+id: quant-number-properties-odds-evens-signs-0059
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: easy
+tags: [parity, odd-even, must-be-true]
+choices:
+  A: "n + 1"
+  B: "n + 2"
+  C: "2n"
+  D: "3n"
+  E: "n^2"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(n\) is an even integer, which of the following must be **odd**?
+
+## Explanation
+
+Start from "\(n\) is even" and apply the parity rules to each option:
+
+- **A. \(n + 1\)** — adding the odd number \(1\) flips even to **odd**. ✓
+- **B. \(n + 2\)** — adding an even number keeps parity, so this stays **even**.
+- **C. \(2n\)** — any integer times \(2\) is **even**.
+- **D. \(3n\)** — odd × even is **even** (the even factor \(n\) wins).
+- **E. \(n^2\)** — \(n \cdot n\) is even × even = **even**.
+
+Only \(n + 1\) is guaranteed odd, so the answer is **A**.
+
+**The traps:** **D (\(3n\))** tempts people who see the odd coefficient \(3\) and forget that
+multiplying by the even \(n\) makes the product even. **E (\(n^2\))** catches the idea that
+squaring changes parity — it does not; \(n^2\) has the same parity as \(n\).
+
+## Hints
+
+- Adding an even number leaves parity unchanged; adding an odd number flips it.
+- A product is even the moment any single factor is even.
diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0060.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0060.md
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+---
+id: quant-number-properties-odds-evens-signs-0060
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: easy
+tags: [signs, negative-numbers, products]
+choices:
+  A: "-24"
+  B: "-9"
+  C: "9"
+  D: "12"
+  E: "24"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the value of \((-3)(-4)(-2)\)?
+
+## Explanation
+
+First find the magnitude, then decide the sign separately.
+
+Magnitude: \(3 \cdot 4 \cdot 2 = 24\).
+
+Sign: count the negative factors. There are **three** of them — an **odd** number — so the product
+is **negative**. Therefore the value is \(-24\).
+
+You can also multiply step by step: \((-3)(-4) = +12\) (two negatives make a positive), and then
+\((+12)(-2) = -24\).
+
+**The traps:**
+
+- **E (24)** has the right magnitude but the wrong sign — it treats three negatives as if they
+  cancel to positive. An odd number of negatives stays negative.
+- **D (12)** stops after multiplying the first two factors, \((-3)(-4) = 12\), and forgets the
+  third factor.
+- **C (9)** adds the magnitudes (\(3 + 4 + 2\)) instead of multiplying.
+- **B (-9)** adds the numbers keeping their signs (\(-3 - 4 - 2\)) instead of multiplying.
+
+## Hints
+
+- Handle magnitude and sign as two separate questions.
+- Count the negative factors: an odd count gives a negative product, an even count a positive one.
diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0061.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0061.md
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+---
+id: quant-number-properties-odds-evens-signs-0061
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: medium
+tags: [parity, products, must-be-true]
+choices:
+  A: "m and n are both odd"
+  B: "m and n are both even"
+  C: "m + n is odd"
+  D: "at least one of m and n is even"
+  E: "m - n is odd"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(m\) and \(n\) are integers and the product \(mn\) is odd, which of the following **must** be
+true?
+
+## Explanation
+
+A product of integers is odd only when **every** factor is odd — a single even factor would make
+the product even. Since \(mn\) is odd, both \(m\) and \(n\) must be odd. That makes **A** correct.
+
+Check the rest against "both \(m\) and \(n\) are odd":
+
+- **B** (both even) — impossible; even factors give an even product.
+- **D** (at least one even) — same contradiction.
+- **C** (\(m + n\) odd) — odd + odd = **even**, so \(m + n\) is even, not odd.
+- **E** (\(m - n\) odd) — odd − odd = **even**, so this is false too.
+
+The answer is **A**.
+
+**The traps:** **C** and **E** bait the reflex that "odd things make odd results." Two odd numbers
+*add and subtract* to an even number — parity flips only when the two numbers differ in parity.
+
+## Hints
+
+- A product is odd only if none of its factors is even.
+- Once you know both numbers are odd, test what odd ± odd gives.
diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0062.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0062.md
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+---
+id: quant-number-properties-odds-evens-signs-0062
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: medium
+tags: [signs, inequalities, must-be-true]
+choices:
+  A: "Both x and y are negative."
+  B: "Both x and y are positive."
+  C: "x and y have opposite signs."
+  D: "The product xy is negative."
+  E: "Exactly one of x and y is zero."
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(xy > 0\) and \(x + y < 0\), which of the following must be true?
+
+## Explanation
+
+Read the two conditions one at a time.
+
+\(xy > 0\) tells you \(x\) and \(y\) have the **same sign** — either both positive or both negative.
+(It also rules out either being \(0\), since that would make the product \(0\), not positive.)
+
+\(x + y < 0\) tells you their sum is negative. Two positive numbers can never sum to something
+negative, so the "both positive" case is out. That leaves only **both negative**, which is
+answer **A**.
+
+**The traps:**
+
+- **B** (both positive) contradicts \(x + y < 0\).
+- **C** (opposite signs) and **D** (\(xy < 0\)) both describe \(xy < 0\), the opposite of what is
+  given.
+- **E** (one is zero) is impossible: a zero factor makes \(xy = 0\), not \(xy > 0\).
+
+## Hints
+
+- \(xy > 0\) means the two numbers share a sign; \(xy < 0\) means opposite signs.
+- If they share a sign but add to a negative number, which sign must it be?
diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0063.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0063.md
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+---
+id: quant-number-properties-odds-evens-signs-0063
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: medium
+tags: [parity, consecutive-integers, must-be-true]
+choices:
+  A: "n^2 + n"
+  B: "n^2 + 1"
+  C: "2n + 1"
+  D: "n^2"
+  E: "n^2 - 1"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(n\) is a positive integer, which of the following must be **even**?
+
+## Explanation
+
+Factor where you can, and test each expression against both parities of \(n\).
+
+- **A. \(n^2 + n = n(n + 1)\)** — this is the product of two **consecutive** integers, and one of
+  any two consecutive integers is always even. So \(n(n+1)\) is **always even**. ✓
+- **B. \(n^2 + 1\)** — if \(n\) is even, \(n^2 + 1\) is odd. Not always even.
+- **C. \(2n + 1\)** — always **odd** (an even number plus 1).
+- **D. \(n^2\)** — has the same parity as \(n\), so it is odd whenever \(n\) is odd.
+- **E. \(n^2 - 1 = (n-1)(n+1)\)** — if \(n\) is even, both factors are odd, giving an odd result.
+
+Only **A** is guaranteed even, so the answer is **A**.
+
+**The traps:** **D (\(n^2\))** relies on the false belief that squaring always yields an even
+number. **B** and **E** look symmetric to A but flip to odd when \(n\) is even — the deciding test
+is trying an even value of \(n\).
+
+## Hints
+
+- \(n^2 + n\) factors into \(n(n+1)\) — what is special about two consecutive integers?
+- For each option, just try \(n = 2\) and \(n = 3\); the "must be even" answer survives both.
diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0064.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0064.md
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@@ -0,0 +1,53 @@
+---
+id: quant-number-properties-odds-evens-signs-0064
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: hard
+tags: [consecutive-integers, divisibility, cannot-be-true]
+choices:
+  A: "18"
+  B: "33"
+  C: "44"
+  D: "57"
+  E: "72"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Which of the following **cannot** be the sum of three consecutive integers?
+
+## Explanation
+
+Write three consecutive integers as \(n-1\), \(n\), and \(n+1\). Their sum is
+
+\[(n-1) + n + (n+1) = 3n,\]
+
+which is exactly \(3\) times the middle integer. So the sum of any three consecutive integers is
+**always a multiple of 3**. A number that is *not* divisible by 3 cannot be such a sum.
+
+Check each option's divisibility by 3 (add the digits):
+
+- **A. 18** → \(1+8 = 9\), divisible by 3. Possible: \(5 + 6 + 7\).
+- **B. 33** → \(3+3 = 6\), divisible by 3. Possible: \(10 + 11 + 12\).
+- **C. 44** → \(4+4 = 8\), **not** divisible by 3. **Impossible.** ✓
+- **D. 57** → \(5+7 = 12\), divisible by 3. Possible: \(18 + 19 + 20\).
+- **E. 72** → \(7+2 = 9\), divisible by 3. Possible: \(23 + 24 + 25\).
+
+Only \(44\) fails, so the answer is **C**.
+
+**The trap:** without the "sum \(= 3n\)" insight, every option looks equally plausible and you are
+left guessing. The four wrong answers are all deliberately multiples of 3 so that only the
+divisibility test separates them.
+
+## Hints
+
+- Represent the three integers as \(n-1,\ n,\ n+1\) and add them.
+- The sum simplifies to \(3n\), so it must be divisible by 3 — test each choice.
diff --git a/content/questions/quant/quant-number-properties-odds-evens-signs-0065.md b/content/questions/quant/quant-number-properties-odds-evens-signs-0065.md
new file mode 100644
index 0000000..bbe25a3
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@@ -0,0 +1,51 @@
+---
+id: quant-number-properties-odds-evens-signs-0065
+section: quant
+topic: number-properties
+subtopic: odds-evens-signs
+type: problem-solving
+difficulty: hard
+tags: [parity, consecutive-integers, always-odd]
+choices:
+  A: "n^2 + n + 1"
+  B: "n^2 + n"
+  C: "2n^2 + 3n"
+  D: "n^3 + n"
+  E: "3n + 3"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+If \(n\) is an integer, which of the following is **always odd**?
+
+## Explanation
+
+The key building block is \(n^2 + n = n(n+1)\), the product of two consecutive integers, which is
+**always even**. Use that to read off each option's parity.
+
+- **A. \(n^2 + n + 1\)** — this is (always even) \(+\,1\), so it is **always odd**. ✓
+- **B. \(n^2 + n\)** — always **even** (the consecutive-integer product itself).
+- **C. \(2n^2 + 3n\)** — \(2n^2\) is even, and \(3n\) has the same parity as \(n\); the total has
+  the parity of \(n\), so it is even when \(n\) is even. Not always odd.
+- **D. \(n^3 + n = n(n^2 + 1)\)** — if \(n\) is even the whole thing is even; if \(n\) is odd then
+  \(n^2 + 1\) is even, so it is even again. Always **even**.
+- **E. \(3n + 3 = 3(n + 1)\)** — its parity matches \(n + 1\), so it is even when \(n\) is odd.
+  Not always odd.
+
+Only **A** is odd for every integer \(n\), so the answer is **A**.
+
+**The traps:** **C** and **E** are odd for some values of \(n\) (try \(n = 1\)) but even for
+others (try \(n = 2\)) — "sometimes odd" is not "always odd." **D** looks like it should alternate,
+but factoring shows it is always even.
+
+## Hints
+
+- Anchor on the fact that \(n^2 + n = n(n+1)\) is always even, then adjust by the constant.
+- "Always odd" must survive both an even and an odd test value — check \(n = 2\) and \(n = 3\).
+
diff --git a/content/questions/quant/quant-number-properties-remainders-0024.md b/content/questions/quant/quant-number-properties-remainders-0024.md
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@@ -0,0 +1,46 @@
+---
+id: quant-number-properties-remainders-0024
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: easy
+tags: [remainders, sums]
+choices:
+  A: "2"
+  B: "3"
+  C: "5"
+  D: "7"
+  E: "12"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When the positive integer \(n\) is divided by 9, the remainder is 5. What is the remainder when
+\(n + 7\) is divided by 9?
+
+## Explanation
+
+You don't need to know \(n\) itself. Adding 7 to \(n\) adds 7 to the remainder, and then you reduce
+mod 9:
+
+\[\text{remainder of } (n+7) = \text{remainder of } (5 + 7) = \text{remainder of } 12 = 12 - 9 = 3.\]
+
+You can confirm with the smallest such \(n\): \(n = 5\) gives \(n + 7 = 12\), and \(12 \div 9\) leaves
+remainder **3**.
+
+**Traps:** answer **E (12)** is \(5 + 7\) with the final reduction forgotten. Answer **D (7)** is
+just the number being added, as if the original remainder didn't matter. Answer **C (5)** assumes
+adding 7 leaves the remainder unchanged. Answer **A (2)** comes from *subtracting* the remainders
+(\(7 - 5\)) instead of adding.
+
+## Hints
+
+- You never need the actual value of \(n\) — work with its remainder.
+- Add the remainders, then reduce mod 9 one more time.
diff --git a/content/questions/quant/quant-number-properties-remainders-0025.md b/content/questions/quant/quant-number-properties-remainders-0025.md
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@@ -0,0 +1,47 @@
+---
+id: quant-number-properties-remainders-0025
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: easy
+tags: [remainders, division-identity]
+choices:
+  A: "31"
+  B: "42"
+  C: "46"
+  D: "52"
+  E: "66"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When a positive integer is divided by 6, the quotient is 7 and the remainder is 4. What is the
+integer?
+
+## Explanation
+
+Every division fits the identity
+
+\[n = (\text{divisor})(\text{quotient}) + (\text{remainder}).\]
+
+Here the divisor is 6, the quotient is 7, and the remainder is 4, so
+
+\[n = 6 \times 7 + 4 = 42 + 4 = 46.\]
+
+Check: \(46 \div 6 = 7\) with \(46 - 42 = 4\) left over. ✓
+
+**Traps:** answer **A (31)** swaps the quotient and remainder (\(6 \times 4 + 7\)). Answer
+**B (42)** is \(6 \times 7\) with the remainder dropped. Answer **D (52)** multiplies by one too many
+(\(6 \times 8 + 4\)). Answer **E (66)** treats the problem as \(6 \times (7 + 4)\).
+
+## Hints
+
+- Use \(n = \text{divisor} \times \text{quotient} + \text{remainder}\).
+- Keep the quotient and the remainder in their correct roles — one gets multiplied, one gets added.
diff --git a/content/questions/quant/quant-number-properties-remainders-0026.md b/content/questions/quant/quant-number-properties-remainders-0026.md
new file mode 100644
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@@ -0,0 +1,47 @@
+---
+id: quant-number-properties-remainders-0026
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: medium
+tags: [remainders, products]
+choices:
+  A: "1"
+  B: "2"
+  C: "4"
+  D: "8"
+  E: "11"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When the positive integers \(a\) and \(b\) are each divided by 12, the remainders are 5 and 8
+respectively. What is the remainder when the product \(ab\) is divided by 12?
+
+## Explanation
+
+To get the remainder of a product, multiply the remainders and then reduce mod 12:
+
+\[\text{remainder of } ab = \text{remainder of } (5 \times 8) = \text{remainder of } 40.\]
+
+Since \(40 = 12 \times 3 + 4\), the remainder is **4**.
+
+Confirm with the smallest values: \(a = 5\), \(b = 8\), so \(ab = 40\), and \(40 \div 12\) leaves
+remainder **4**. ✓
+
+**Traps:** answer **A (1)** *adds* the remainders (\(5 + 8 = 13\), then \(13 - 12 = 1\)) instead of
+multiplying them. Answer **D (8)** and answer **E (11)** come from botching the final reduction of
+40 (subtracting 32 or 29 rather than a multiple of 12). The key discipline: combine the remainders
+the same way the numbers are combined — here, multiply — then reduce one last time.
+
+## Hints
+
+- The remainder of a product is the remainder of (remainder \(\times\) remainder).
+- \(5 \times 8 = 40\) is not the final answer — reduce 40 modulo 12.
diff --git a/content/questions/quant/quant-number-properties-remainders-0027.md b/content/questions/quant/quant-number-properties-remainders-0027.md
new file mode 100644
index 0000000..f792de4
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@@ -0,0 +1,47 @@
+---
+id: quant-number-properties-remainders-0027
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: medium
+tags: [remainders, powers, cyclicity, units-digit]
+choices:
+  A: "1"
+  B: "3"
+  C: "7"
+  D: "9"
+  E: "0"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+What is the remainder when \(7^{20}\) is divided by 10?
+
+## Explanation
+
+Dividing by 10 asks for the **units digit**. The units digit of powers of 7 repeats in a cycle of
+length 4:
+
+\[7^1 \to 7, \quad 7^2 \to 9, \quad 7^3 \to 3, \quad 7^4 \to 1, \quad 7^5 \to 7, \ \dots\]
+
+To find where \(7^{20}\) lands, divide the exponent by the cycle length: \(20 = 4 \times 5\), so 20 is
+a multiple of 4 and \(7^{20}\) sits at the **end** of a cycle — the same spot as \(7^4\), whose units
+digit is **1**. The remainder when \(7^{20}\) is divided by 10 is **1**.
+
+**Traps:** answer **C (7)** treats "\(20 \div 4\) has remainder 0" as if it pointed to the *first*
+entry (7) — but a remainder of 0 means the *last* entry of the cycle, not the first. Answers
+**D (9)** and **B (3)** are the units digits of \(7^2\) and \(7^3\), chosen by miscounting the
+position. Answer **E (0)** wrongly assumes any large power is divisible by 10.
+
+## Hints
+
+- Remainder mod 10 is just the units digit.
+- Units digits of \(7^n\) cycle 7, 9, 3, 1; use "exponent mod 4," and treat a result of 0 as the end
+  of the cycle.
diff --git a/content/questions/quant/quant-number-properties-remainders-0028.md b/content/questions/quant/quant-number-properties-remainders-0028.md
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+---
+id: quant-number-properties-remainders-0028
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: medium
+tags: [remainders, simultaneous-conditions]
+choices:
+  A: "12"
+  B: "16"
+  C: "17"
+  D: "22"
+  E: "31"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+The positive integer \(n\) leaves a remainder of 2 when divided by 5 and a remainder of 1 when
+divided by 3. What is the least possible value of \(n\) that is greater than 10?
+
+## Explanation
+
+List the numbers that satisfy each condition and find the overlap.
+
+- Remainder 2 when divided by 5: \(2, 7, 12, 17, 22, 27, \dots\)
+- Remainder 1 when divided by 3: \(1, 4, 7, 10, 13, 16, 19, 22, 25, \dots\)
+
+The values in **both** lists are \(7, 22, 37, \dots\) — they repeat every \(\text{lcm}(5, 3) = 15\).
+The smallest one greater than 10 is **22**.
+
+Check: \(22 = 5 \times 4 + 2\) (remainder 2) and \(22 = 3 \times 7 + 1\) (remainder 1). ✓
+
+**Traps:** answer **A (12)** satisfies only the mod-5 condition (\(12 \div 3\) leaves remainder 0).
+Answer **B (16)** satisfies only the mod-3 condition (\(16 \div 5\) leaves remainder 1). Answer
+**C (17)** leaves remainder 2 mod 5 but remainder 2 mod 3, so it fails the second condition. Answer
+**E (31)** leaves remainder 1 mod 5, satisfying neither cleanly. Only 22 meets both.
+
+## Hints
+
+- Write out the numbers that fit each condition separately, then look for the first common value.
+- Above the first overlap, solutions repeat every lcm of the two divisors.
diff --git a/content/questions/quant/quant-number-properties-remainders-0029.md b/content/questions/quant/quant-number-properties-remainders-0029.md
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+---
+id: quant-number-properties-remainders-0029
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: hard
+tags: [remainders, squares, could-be]
+choices:
+  A: "2"
+  B: "3"
+  C: "4"
+  D: "6"
+  E: "0"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+The positive integer \(n\) is such that \(n^2\) leaves a remainder of 4 when divided by 7. Which of
+the following could be the remainder when \(n\) is divided by 7?
+
+## Explanation
+
+The remainder of \(n^2\) depends only on the remainder of \(n\), so test every possible remainder
+\(r = 0, 1, 2, \dots, 6\) and see which give \(r^2 \equiv 4 \pmod 7\):
+
+\[0^2 \to 0,\quad 1^2 \to 1,\quad 2^2 \to 4,\quad 3^2 \to 2,\quad 4^2 \to 2,\quad 5^2 \to 4,\quad 6^2 \to 1.\]
+
+(For example, \(5^2 = 25 = 7 \times 3 + 4\).) The value 4 appears when \(r = 2\) or \(r = 5\). Of the
+answer choices, only **2** is possible.
+
+Check: if \(n = 2\), then \(n^2 = 4\), and \(4 \div 7\) leaves remainder 4. ✓
+
+**Traps:** answer **C (4)** assumes \(n\) leaves the *same* remainder as \(n^2\) — but a number and
+its square rarely share a remainder (here \(4^2 = 16\) leaves remainder 2, not 4). Answers **B (3)**,
+**D (6)**, and **E (0)** all fail: their squares leave remainders 2, 1, and 0 respectively.
+
+## Hints
+
+- The remainder of \(n^2\) depends only on the remainder of \(n\), so just test \(r = 0\) through 6.
+- Don't assume \(n\) and \(n^2\) leave the same remainder — square each candidate and check.
diff --git a/content/questions/quant/quant-number-properties-remainders-0030.md b/content/questions/quant/quant-number-properties-remainders-0030.md
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+---
+id: quant-number-properties-remainders-0030
+section: quant
+topic: number-properties
+subtopic: remainders
+type: problem-solving
+difficulty: hard
+tags: [remainders, simultaneous-conditions, lcm]
+choices:
+  A: "1"
+  B: "5"
+  C: "7"
+  D: "9"
+  E: "11"
+answer: E
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+When the positive integer \(n\) is divided by 4, the remainder is 3, and when \(n\) is divided by 6,
+the remainder is 5. What is the remainder when \(n\) is divided by 12?
+
+## Explanation
+
+Find a number that satisfies both conditions, then reduce mod 12. List each:
+
+- Remainder 3 when divided by 4: \(3, 7, 11, 15, 19, 23, \dots\)
+- Remainder 5 when divided by 6: \(5, 11, 17, 23, 29, \dots\)
+
+The first common value is **11**. Both divisors go into \(\text{lcm}(4, 6) = 12\), so the solutions
+repeat every 12: \(11, 23, 35, \dots\), all of which leave remainder **11** when divided by 12.
+
+Check: \(11 = 4 \times 2 + 3\) (remainder 3) and \(11 = 6 \times 1 + 5\) (remainder 5). ✓
+
+**Traps:** answers **B (5)** and **C (7)** grab one of the given remainders (or their sum minus one)
+without combining the conditions. Answer **D (9)** and answer **A (1)** come from guessing a value
+that fits only one of the two divisions. A shortcut worth noticing: each condition says \(n\) is one
+short of a multiple (\(n + 1\) is divisible by 4 **and** by 6, hence by 12), so \(n\) is one less than
+a multiple of 12 — remainder **11**.
+
+## Hints
+
+- Find the smallest \(n\) satisfying both conditions, then take its remainder mod 12.
+- Notice that \(n + 1\) is divisible by both 4 and 6 — what does that say about \(n + 1\) and 12?
diff --git a/content/questions/quant/quant-word-problems-interest-0045.md b/content/questions/quant/quant-word-problems-interest-0045.md
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+---
+id: quant-word-problems-interest-0045
+section: quant
+topic: word-problems
+subtopic: interest
+type: problem-solving
+difficulty: easy
+tags: [interest, simple-interest]
+choices:
+  A: "$24"
+  B: "$120"
+  C: "$240"
+  D: "$249.60"
+  E: "$1,740"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Priya deposits $1,500 in an account paying 8% simple annual interest. How much interest does the deposit earn in 2 years?
+
+## Explanation
+
+Simple interest is a fixed slice of the *original* principal each year:
+
+\[I = P \cdot r \cdot t = 1500 \times 0.08 \times 2 = 240\]
+
+So the deposit earns **$240**.
+
+**The traps:**
+- **B ($120)** computes the interest for a single year (\(1500 \times 0.08\)) and forgets to multiply by the 2 years.
+- **E ($1,740)** is the final *balance* \(P + I = 1500 + 240\), not the interest the question asked for.
+- **D ($249.60)** compounds the interest annually (\(1500 \times 1.08^2 - 1500\)); the account pays simple interest, so nothing compounds.
+- **A ($24)** slips the decimal, using \(0.008\) instead of \(0.08\) for the rate.
+
+## Hints
+
+- Simple interest uses the original principal every year: \(I = P \cdot r \cdot t\).
+- The question asks for the interest earned, not the total in the account.
diff --git a/content/questions/quant/quant-word-problems-interest-0046.md b/content/questions/quant/quant-word-problems-interest-0046.md
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+---
+id: quant-word-problems-interest-0046
+section: quant
+topic: word-problems
+subtopic: interest
+type: problem-solving
+difficulty: easy
+tags: [interest, simple-interest, balance]
+choices:
+  A: "$192"
+  B: "$848"
+  C: "$992"
+  D: "$1,010"
+  E: "$2,720"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+An account holds $800 and earns 6% simple annual interest. What is the balance in the account after 4 years?
+
+## Explanation
+
+First find the simple interest, then add it back to the principal:
+
+\[I = P \cdot r \cdot t = 800 \times 0.06 \times 4 = 192\]
+\[A = P + I = 800 + 192 = 992\]
+
+The balance is **$992**.
+
+**The traps:**
+- **A ($192)** is the interest alone — the question asks for the balance, so the $800 principal must be added back.
+- **B ($848)** adds only one year of interest (\(800 + 48\)) instead of four.
+- **D ($1,010)** compounds annually (\(800 \times 1.06^4 \approx 1{,}010\)); simple interest does not compound.
+- **E ($2,720)** slips the decimal, treating 6% as \(0.6\): \(800 + 800(0.6)(4)\).
+
+## Hints
+
+- Compute the interest with \(I = P \cdot r \cdot t\), then add the principal for the balance.
+- Simple interest is the same amount each year — no compounding.
diff --git a/content/questions/quant/quant-word-problems-interest-0047.md b/content/questions/quant/quant-word-problems-interest-0047.md
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+---
+id: quant-word-problems-interest-0047
+section: quant
+topic: word-problems
+subtopic: interest
+type: problem-solving
+difficulty: medium
+tags: [interest, compound-interest]
+choices:
+  A: "$500"
+  B: "$1,000"
+  C: "$1,050"
+  D: "$1,500"
+  E: "$6,050"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+$5,000 is invested at 10% compounded annually. How much interest does the investment earn over 2 years?
+
+## Explanation
+
+Compound interest adds each year's interest to the balance before the next year. The balance after 2 years is
+
+\[A = P(1 + r)^{t} = 5000(1.10)^2 = 5000 \times 1.21 = 6050\]
+
+The interest *earned* is the balance minus the principal:
+
+\[A - P = 6050 - 5000 = 1050\]
+
+So the investment earns **$1,050**. (The first year earns $500; the second year earns 10% of the new $5,500 balance, or $550.)
+
+**The traps:**
+- **B ($1,000)** uses simple interest, \(5000 \times 0.10 \times 2\), ignoring that the first year's interest also earns interest.
+- **E ($6,050)** is the final *balance*, not the interest earned — the $5,000 principal must be subtracted.
+- **A ($500)** counts only the first year's interest.
+- **D ($1,500)** applies simple interest over 3 years instead of 2.
+
+## Hints
+
+- Compound balance is \(P(1+r)^t\); the interest earned is that balance minus \(P\).
+- Over the second year, the interest is charged on the *new* balance, not the original principal.
diff --git a/content/questions/quant/quant-word-problems-interest-0048.md b/content/questions/quant/quant-word-problems-interest-0048.md
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+---
+id: quant-word-problems-interest-0048
+section: quant
+topic: word-problems
+subtopic: interest
+type: problem-solving
+difficulty: medium
+tags: [interest, simple-interest, rate]
+choices:
+  A: "6%"
+  B: "9%"
+  C: "18%"
+  D: "54%"
+  E: "0.6%"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+An investment of $1,500 earns $270 in simple interest over 3 years. What is the annual interest rate?
+
+## Explanation
+
+Start from \(I = P \cdot r \cdot t\) and solve for the rate:
+
+\[r = \frac{I}{P \cdot t} = \frac{270}{1500 \times 3} = \frac{270}{4500} = 0.06\]
+
+So the annual rate is **6%**.
+
+**The traps:**
+- **C (18%)** divides the interest by the principal but forgets the 3 years: \(270 / 1500 = 0.18\).
+- **B (9%)** divides by only 2 years instead of 3: \(270 / (1500 \times 2)\).
+- **D (54%)** multiplies by the time instead of dividing: \(270 \times 3 / 1500\).
+- **E (0.6%)** slips a decimal place in the final rate.
+
+## Hints
+
+- Rearrange \(I = P \cdot r \cdot t\) to \(r = \dfrac{I}{P \cdot t}\).
+- Divide by *both* the principal and the number of years.
diff --git a/content/questions/quant/quant-word-problems-interest-0049.md b/content/questions/quant/quant-word-problems-interest-0049.md
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+---
+id: quant-word-problems-interest-0049
+section: quant
+topic: word-problems
+subtopic: interest
+type: problem-solving
+difficulty: medium
+tags: [interest, compound-interest, simple-interest, comparison]
+choices:
+  A: "$4"
+  B: "$40"
+  C: "$80"
+  D: "$400"
+  E: "$840"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+$4,000 is invested for 2 years at 10% per year. How much more interest does the money earn if the interest is compounded annually rather than paid as simple interest?
+
+## Explanation
+
+Compute both and subtract.
+
+Simple interest: \(I = 4000 \times 0.10 \times 2 = 800\).
+
+Compound interest: \(A = 4000(1.10)^2 = 4000 \times 1.21 = 4840\), so the interest is \(4840 - 4000 = 840\).
+
+The difference is \(840 - 800 = 40\).
+
+A faster route: over exactly 2 years the compound-minus-simple gap is \(P r^2 = 4000 \times (0.10)^2 = 4000 \times 0.01 = 40\) — the interest that the first year's $400 earns in the second year.
+
+So compounding earns **$40** more.
+
+**The traps:**
+- **E ($840)** is the total compound interest, not the *difference* from simple interest.
+- **D ($400)** is one year's interest (\(4000 \times 0.10\)), a common stand-in for the gap.
+- **C ($80)** doubles the correct gap, as if the $40 recurred in both years.
+- **A ($4)** slips a decimal, using \(r^2 = 0.001\) in the shortcut.
+
+## Hints
+
+- Find the simple interest and the compound interest separately, then subtract.
+- Over 2 years the gap equals \(P r^2\) — the interest earned on the first year's interest.
diff --git a/content/questions/quant/quant-word-problems-interest-0050.md b/content/questions/quant/quant-word-problems-interest-0050.md
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+---
+id: quant-word-problems-interest-0050
+section: quant
+topic: word-problems
+subtopic: interest
+type: problem-solving
+difficulty: hard
+tags: [interest, compound-interest, semiannual-compounding]
+choices:
+  A: "$400"
+  B: "$800"
+  C: "$816"
+  D: "$1,600"
+  E: "$1,664"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+$10,000 is invested at a nominal annual rate of 8%, compounded semiannually. How much interest does the investment earn in one year?
+
+## Explanation
+
+Semiannual compounding means two periods per year, each at half the annual rate: \(8\% / 2 = 4\%\) per period, for \(2\) periods.
+
+\[A = P\left(1 + \frac{R}{n}\right)^{ny} = 10000(1.04)^2 = 10000 \times 1.0816 = 10816\]
+
+The interest earned is \(10816 - 10000 = 816\).
+
+So the investment earns **$816**. (First half-year: $400 of interest; second half-year: 4% of the new $10,400 balance, or $416.)
+
+**The traps:**
+- **B ($800)** ignores compounding and applies a flat 8% for the year (\(10000 \times 0.08\)).
+- **A ($400)** counts only one semiannual period (\(10000 \times 0.04\)).
+- **D ($1,600)** applies the full 8% to *each* half-year as simple interest (\(10000 \times 0.08 \times 2\)).
+- **E ($1,664)** compounds but forgets to halve the rate, using 8% per period: \(10000(1.08)^2 - 10000\).
+
+## Hints
+
+- With semiannual compounding, halve the rate and double the number of periods.
+- Two periods at 4% is \(P(1.04)^2\) — then subtract the principal for the interest.
diff --git a/content/questions/quant/quant-word-problems-interest-0051.md b/content/questions/quant/quant-word-problems-interest-0051.md
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+---
+id: quant-word-problems-interest-0051
+section: quant
+topic: word-problems
+subtopic: interest
+type: problem-solving
+difficulty: hard
+tags: [interest, compound-interest, reverse, principal]
+choices:
+  A: "$7,744"
+  B: "$8,000"
+  C: "$8,067"
+  D: "$8,800"
+  E: "$10,648"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A sum of money invested at 10% compounded annually grows to $9,680 after 2 years. What was the amount originally invested?
+
+## Explanation
+
+Compound growth multiplies the principal by \((1+r)^t\), so to recover the principal you *divide* by that factor:
+
+\[P = \frac{A}{(1 + r)^t} = \frac{9680}{(1.10)^2} = \frac{9680}{1.21} = 8000\]
+
+Check: \(8000 \times 1.10 = 8800\), and \(8800 \times 1.10 = 9680\). ✓
+
+So the original investment was **$8,000**.
+
+**The traps:**
+- **D ($8,800)** reverses only one of the two years, dividing by 1.10 once.
+- **A ($7,744)** takes a flat 20% off the final amount (\(9680 \times 0.8\)) — but subtracting a percentage is not the inverse of compound growth.
+- **C ($8,067)** divides by 1.20, treating the growth as simple interest over 2 years.
+- **E ($10,648)** grows the amount by another year (\(9680 \times 1.10\)) instead of working backward.
+
+## Hints
+
+- To undo compound growth, divide the final amount by \((1+r)^t\), not subtract a percentage.
+- Over 2 years at 10%, the factor is \(1.10^2 = 1.21\).
+
diff --git a/content/questions/quant/quant-word-problems-mixtures-0038.md b/content/questions/quant/quant-word-problems-mixtures-0038.md
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+---
+id: quant-word-problems-mixtures-0038
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: easy
+tags: [mixtures, concentration, percents]
+choices:
+  A: "2.7"
+  B: "6"
+  C: "15"
+  D: "34"
+  E: "60"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A 40-liter salt solution is 15% salt by volume. How many liters of salt does the solution contain?
+
+## Explanation
+
+The amount of a substance is its concentration times the total: multiply the percent (as a decimal)
+by the total volume.
+
+\[0.15 \times 40 = 6 \text{ liters}\]
+
+So the solution contains **6 liters** of salt.
+
+**The traps:**
+- **C (15)** just repeats the percent number — but 15% of 40 is not 15.
+- **D (34)** is the amount of *water* (\(40 - 6\)), not the salt.
+- **A (2.7)** divides instead of multiplying (\(40 \div 15\)).
+- **E (60)** reads "15%" as "times 1.5" (\(40 \times 1.5\)).
+
+## Hints
+
+- Amount of substance = concentration (as a decimal) × total volume.
+- 15% means \(0.15\), so compute \(0.15 \times 40\).
diff --git a/content/questions/quant/quant-word-problems-mixtures-0039.md b/content/questions/quant/quant-word-problems-mixtures-0039.md
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index 0000000..c50325e
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@@ -0,0 +1,46 @@
+---
+id: quant-word-problems-mixtures-0039
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: easy
+tags: [mixtures, part-whole, percents]
+choices:
+  A: "20%"
+  B: "25%"
+  C: "33 1/3 %"
+  D: "50%"
+  E: "75%"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A trail mix is made from 3 kilograms of peanuts and 1 kilogram of raisins. What percent of the
+mixture, by weight, is raisins?
+
+## Explanation
+
+Concentration is part over **whole**. The whole mixture weighs \(3 + 1 = 4\) kilograms, and the
+raisins are 1 of those kilograms:
+
+\[\frac{1}{4} = 0.25 = 25\%\]
+
+So raisins make up **25%** of the mixture.
+
+**The traps:**
+- **C (33⅓%)** divides raisins by *peanuts* (\(\tfrac{1}{3}\)) instead of by the whole (\(\tfrac{1}{4}\)).
+- **E (75%)** is the *peanut* share, not the raisin share.
+- **D (50%)** assumes an even split, ignoring the 3-to-1 amounts.
+- **A (20%)** uses a total of 5 (\(\tfrac{1}{5}\)) — miscounting the whole.
+
+## Hints
+
+- Percent of the mixture = (that ingredient) ÷ (total of all ingredients).
+- The denominator is the *whole* mixture: \(3 + 1\), not just the peanuts.
diff --git a/content/questions/quant/quant-word-problems-mixtures-0040.md b/content/questions/quant/quant-word-problems-mixtures-0040.md
new file mode 100644
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@@ -0,0 +1,53 @@
+---
+id: quant-word-problems-mixtures-0040
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: medium
+tags: [mixtures, weighted-average, concentration]
+choices:
+  A: "48%"
+  B: "50%"
+  C: "55%"
+  D: "60%"
+  E: "110%"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+20 liters of a 30% acid solution is mixed with 30 liters of an 80% acid solution. What is the acid
+concentration of the resulting mixture?
+
+## Explanation
+
+Add up the actual **amount of acid**, then divide by the **total volume** — never average the
+percents directly.
+
+Acid from each solution:
+
+\[0.30 \times 20 = 6 \text{ L} \qquad 0.80 \times 30 = 24 \text{ L}\]
+
+Total acid \(= 6 + 24 = 30\) L in a total volume of \(20 + 30 = 50\) L:
+
+\[\frac{30}{50} = 60\%\]
+
+So the mixture is **60%** acid. (It's above the halfway point because there's more of the strong
+solution.)
+
+**The traps:**
+- **C (55%)** is the unweighted average \(\tfrac{30 + 80}{2}\) — wrong because the volumes differ.
+- **E (110%)** adds the two percentages, which can't exceed 100%.
+- **B (50%)** swaps the volumes and concentrations (\(0.30 \times 30 + 0.80 \times 20 = 25\), then \(\tfrac{25}{50}\)).
+- **A (48%)** counts only the strong solution's acid (\(\tfrac{24}{50}\)), forgetting the 6 L from the weak one.
+
+## Hints
+
+- Compute the liters of acid in each solution separately, then add.
+- Concentration = total acid ÷ total volume; weight by the amounts, don't average the percents.
diff --git a/content/questions/quant/quant-word-problems-mixtures-0041.md b/content/questions/quant/quant-word-problems-mixtures-0041.md
new file mode 100644
index 0000000..9ebfba5
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-mixtures-0041.md
@@ -0,0 +1,50 @@
+---
+id: quant-word-problems-mixtures-0041
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: medium
+tags: [mixtures, dilution, concentration]
+choices:
+  A: "4"
+  B: "6"
+  C: "10"
+  D: "15"
+  E: "16"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+How many liters of water must be added to 10 liters of a 40% salt solution to dilute it to a 25%
+salt solution?
+
+## Explanation
+
+Adding water changes the total volume but not the amount of salt, so the salt is the quantity that
+stays fixed. Start by finding it:
+
+\[\text{salt} = 0.40 \times 10 = 4 \text{ liters}\]
+
+After adding \(w\) liters of water, the salt is 25% of the new total \((10 + w)\):
+
+\[\frac{4}{10 + w} = 0.25 \;\Rightarrow\; 10 + w = \frac{4}{0.25} = 16 \;\Rightarrow\; w = 6\]
+
+So you must add **6 liters** of water.
+
+**The traps:**
+- **E (16)** is the new *total* volume — you still have to subtract the original 10 liters.
+- **A (4)** is the amount of salt, not the water added.
+- **D (15)** treats the 15-percentage-point drop as \(1.5 \times 10\) liters.
+- **C (10)** guesses that you simply double the volume.
+
+## Hints
+
+- The salt stays constant while you add water — find that fixed amount first.
+- Set (salt) ÷ (new total) = 0.25, solve for the new total, then subtract the original 10 liters.
diff --git a/content/questions/quant/quant-word-problems-mixtures-0042.md b/content/questions/quant/quant-word-problems-mixtures-0042.md
new file mode 100644
index 0000000..ec4060e
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+++ b/content/questions/quant/quant-word-problems-mixtures-0042.md
@@ -0,0 +1,48 @@
+---
+id: quant-word-problems-mixtures-0042
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: medium
+tags: [mixtures, alligation, price-mixture]
+choices:
+  A: "5"
+  B: "8"
+  C: "10"
+  D: "12"
+  E: "15"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A shopkeeper blends coffee costing $8 per kilogram with coffee costing $12 per kilogram to make
+20 kilograms of a blend that costs $9 per kilogram. How many kilograms of the $12 coffee are in the
+blend?
+
+## Explanation
+
+Let \(x\) be the kilograms of $12 coffee; then \(20 - x\) kilograms are the $8 coffee. The total cost
+of the blend is $9 × 20 = $180, so match the costs:
+
+\[8(20 - x) + 12x = 180\]
+\[160 - 8x + 12x = 180 \;\Rightarrow\; 4x = 20 \;\Rightarrow\; x = 5\]
+
+So the blend contains **5 kilograms** of the $12 coffee. (Alligation agrees: the ratio of $12 to $8
+coffee is \((9-8):(12-9) = 1:3\), so the $12 coffee is \(\tfrac14 \times 20 = 5\) kg.)
+
+**The traps:**
+- **E (15)** is the amount of the *$8* coffee (the other part), or the alligation ratio read upside down.
+- **C (10)** assumes a 50-50 split of the 20 kg.
+- **B (8)** and **D (12)** are the per-kilogram *prices*, not weights — the question asks for kilograms.
+
+## Hints
+
+- Let \(x\) be the $12 coffee and \(20 - x\) the $8 coffee; total cost is \(9 \times 20 = 180\).
+- Alligation: the ratio of the two coffees is the inverse of their distances from $9.
diff --git a/content/questions/quant/quant-word-problems-mixtures-0043.md b/content/questions/quant/quant-word-problems-mixtures-0043.md
new file mode 100644
index 0000000..988ccd1
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-mixtures-0043.md
@@ -0,0 +1,51 @@
+---
+id: quant-word-problems-mixtures-0043
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: hard
+tags: [mixtures, replacement, successive-dilution]
+choices:
+  A: "3"
+  B: "6"
+  C: "6.75"
+  D: "7.5"
+  E: "9"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A 12-liter tank is full of pure acid. 3 liters are drained off and replaced with water; the tank is
+stirred, then 3 liters of the mixture are drained off and replaced with water again. How many liters
+of pure acid remain in the tank?
+
+## Explanation
+
+Each step removes the same **fraction** of whatever acid is present, then tops up with water. Draining
+3 of 12 liters removes \(\tfrac{3}{12} = \tfrac14\) of the contents, leaving \(\tfrac34\) of the acid.
+Because the tank is refilled to 12 liters each time, the *second* draining also removes \(\tfrac14\)
+of the acid then present.
+
+So the acid is multiplied by \(\tfrac34\) twice:
+
+\[12 \times \left(\frac{3}{4}\right)^2 = 12 \times \frac{9}{16} = \frac{108}{16} = 6.75 \text{ liters}\]
+
+So **6.75 liters** of pure acid remain.
+
+**The traps:**
+- **E (9)** stops after the first replacement (\(12 \times \tfrac34 = 9\)).
+- **D (7.5)** subtracts the fractions instead of multiplying: \(12\left(1 - \tfrac14 - \tfrac14\right) = 12 \times \tfrac12\).
+- **B (6)** subtracts 3 liters of acid twice (\(12 - 3 - 3\)), as if each draining removed pure acid.
+- **A (3)** is the amount of water added, not the acid remaining.
+
+## Hints
+
+- After the first swap the liquid is a mixture, so the second draining removes acid *and* water — think in fractions, not fixed liters.
+- Each replacement multiplies the remaining acid by \(\tfrac34\); apply it twice.
diff --git a/content/questions/quant/quant-word-problems-mixtures-0044.md b/content/questions/quant/quant-word-problems-mixtures-0044.md
new file mode 100644
index 0000000..2ce42f8
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-mixtures-0044.md
@@ -0,0 +1,51 @@
+---
+id: quant-word-problems-mixtures-0044
+section: quant
+topic: word-problems
+subtopic: mixtures
+type: problem-solving
+difficulty: hard
+tags: [mixtures, weighted-average, solving]
+choices:
+  A: "4"
+  B: "5"
+  C: "10"
+  D: "25"
+  E: "80"
+answer: B
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A chemist has a 60% alcohol solution and a 10% alcohol solution. How many liters of the 60% solution
+must be mixed with 20 liters of the 10% solution to produce a 20% alcohol solution?
+
+## Explanation
+
+Let \(x\) be the liters of 60% solution added. Track the actual alcohol on each side. The alcohol
+before mixing must equal the alcohol after:
+
+\[0.60x + 0.10(20) = 0.20(x + 20)\]
+\[0.60x + 2 = 0.20x + 4\]
+\[0.40x = 2 \;\Rightarrow\; x = 5\]
+
+So you need **5 liters** of the 60% solution.
+
+Check: alcohol \(= 0.60(5) + 0.10(20) = 3 + 2 = 5\) L in \(5 + 20 = 25\) L, and \(\tfrac{5}{25} = 20\%\). ✓
+
+**The traps:**
+- **E (80)** inverts the alligation ratio. The correct ratio of 60% to 10% solution is
+  \((20-10):(60-20) = 10:40 = 1:4\); flipping it to 4:1 gives \(4 \times 20 = 80\).
+- **A (4)** uses the ratio's "1 part" as if it were already the number of liters.
+- **D (25)** is the *total* volume of the final mixture, not the amount added.
+- **C (10)** guesses half of the 20 liters.
+
+## Hints
+
+- Write "alcohol in = alcohol out": \(0.60x + 0.10(20) = 0.20(x + 20)\).
+- Or use alligation: the amounts are in the ratio \((20-10):(60-20)\), with the 10% solution being the larger share.
diff --git a/content/questions/quant/quant-word-problems-translation-0066.md b/content/questions/quant/quant-word-problems-translation-0066.md
new file mode 100644
index 0000000..2a2a615
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-translation-0066.md
@@ -0,0 +1,56 @@
+---
+id: quant-word-problems-translation-0066
+section: quant
+topic: word-problems
+subtopic: translation
+type: problem-solving
+difficulty: easy
+tags: [translation, linear-equations, setup]
+choices:
+  A: "7"
+  B: "10"
+  C: "13"
+  D: "14"
+  E: "26"
+answer: A
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Twice a number, increased by 6, is equal to 20. What is the number?
+
+## Explanation
+
+Translate the sentence piece by piece. Let \(n\) be the number.
+
+- "Twice a number" → \(2n\)
+- "increased by 6" → \(+\,6\)
+- "is equal to 20" → \(=\,20\)
+
+So the equation is:
+
+\[2n + 6 = 20\]
+
+Solve it in the usual order — undo the addition first, then the multiplication:
+
+\[2n = 20 - 6 = 14, \qquad n = \frac{14}{2} = 7\]
+
+The number is **7**. (Check: \(2(7) + 6 = 20\). ✓)
+
+**The traps:**
+- **D (14)** stops one step early: it's \(20 - 6\), forgetting to divide by 2.
+- **B (10)** ignores the "\(+6\)" entirely and just computes \(20 \div 2\).
+- **C (13)** *adds* 6 instead of subtracting: \((20 + 6)/2\). Undo operations in reverse — the last
+  thing done to \(n\) was "add 6," so subtract 6 first.
+- **E (26)** is simply \(20 + 6\), with no equation solved at all.
+
+## Hints
+
+- Let \(n\) be the number and turn each phrase into a symbol: "twice" is \(\times 2\), "increased by"
+  is \(+\).
+- To undo \(2n + 6 = 20\), subtract 6 first, then divide by 2.
diff --git a/content/questions/quant/quant-word-problems-translation-0067.md b/content/questions/quant/quant-word-problems-translation-0067.md
new file mode 100644
index 0000000..55c2121
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-translation-0067.md
@@ -0,0 +1,54 @@
+---
+id: quant-word-problems-translation-0067
+section: quant
+topic: word-problems
+subtopic: translation
+type: problem-solving
+difficulty: easy
+tags: [translation, consecutive-integers, setup]
+choices:
+  A: "23"
+  B: "24"
+  C: "25"
+  D: "26"
+  E: "72"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+The sum of three consecutive integers is 72. What is the largest of the three integers?
+
+## Explanation
+
+Represent consecutive integers with one variable. Let the smallest be \(n\); the next two are
+\(n + 1\) and \(n + 2\). "The sum … is 72" translates to:
+
+\[n + (n + 1) + (n + 2) = 72\]
+
+Combine like terms — note the constants add up to \(+3\), not \(0\):
+
+\[3n + 3 = 72 \quad\Rightarrow\quad 3n = 69 \quad\Rightarrow\quad n = 23\]
+
+So the three integers are \(23,\ 24,\ 25\), and the **largest** is **25**.
+(Check: \(23 + 24 + 25 = 72\). ✓)
+
+**The traps:**
+- **D (26)** comes from forgetting the "\(+3\)" correction: solving \(3n = 72\) gives \(n = 24\),
+  and calling that the smallest makes the largest \(26\). But the three constants sum to \(3\), so
+  \(3n + 3 = 72\).
+- **B (24)** is \(72 \div 3\), the *middle* integer (the average), not the largest.
+- **A (23)** is the smallest integer — the value of \(n\) — not what the question asked for.
+- **E (72)** is the given total, untouched.
+
+## Hints
+
+- Call the smallest integer \(n\); the others are \(n+1\) and \(n+2\). Their sum is \(3n + 3\), not
+  \(3n\).
+- After you find \(n\), the question wants the *largest* value, \(n + 2\) — re-read which one it asks
+  for.
diff --git a/content/questions/quant/quant-word-problems-translation-0068.md b/content/questions/quant/quant-word-problems-translation-0068.md
new file mode 100644
index 0000000..ec02031
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-translation-0068.md
@@ -0,0 +1,61 @@
+---
+id: quant-word-problems-translation-0068
+section: quant
+topic: word-problems
+subtopic: translation
+type: problem-solving
+difficulty: medium
+tags: [translation, ages, setup]
+choices:
+  A: "10"
+  B: "20"
+  C: "22"
+  D: "24"
+  E: "28"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Maria is 4 years older than twice her brother's age. If the sum of their ages is 34, how old is
+Maria?
+
+## Explanation
+
+Express both ages with a single variable. Let \(b\) be the brother's age. Then:
+
+- "twice her brother's age" → \(2b\)
+- "4 years older than …" → \(2b + 4\), so Maria \(= 2b + 4\)
+
+"The sum of their ages is 34" translates to:
+
+\[b + (2b + 4) = 34\]
+
+Solve for \(b\):
+
+\[3b + 4 = 34 \quad\Rightarrow\quad 3b = 30 \quad\Rightarrow\quad b = 10\]
+
+The brother is 10, so Maria is \(2(10) + 4 = 24\).
+(Check: \(10 + 24 = 34\). ✓)
+
+The answer is **24**.
+
+**The traps:**
+- **A (10)** is the brother's age — the variable you solved for, but not the person the question
+  asks about. Always report the requested quantity.
+- **B (20)** is \(2b\), forgetting the "\(+4\)."
+- **E (28)** solves the sum incorrectly as \(3b = 36 \Rightarrow b = 12\) and then computes
+  \(2b + 4\) — a slip in isolating \(b\).
+- **C (22)** splits 34 loosely (near half) without setting up the relationship at all.
+
+## Hints
+
+- Let \(b\) be the brother's age and write Maria's age in terms of \(b\): "4 more than twice \(b\)"
+  is \(2b + 4\).
+- Add the two expressions, set the sum to 34, solve for \(b\) — then compute *Maria's* age, not the
+  brother's.
diff --git a/content/questions/quant/quant-word-problems-translation-0069.md b/content/questions/quant/quant-word-problems-translation-0069.md
new file mode 100644
index 0000000..a0ba480
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-translation-0069.md
@@ -0,0 +1,55 @@
+---
+id: quant-word-problems-translation-0069
+section: quant
+topic: word-problems
+subtopic: translation
+type: problem-solving
+difficulty: medium
+tags: [translation, coins, systems, setup]
+choices:
+  A: "8"
+  B: "14"
+  C: "15"
+  D: "16"
+  E: "20"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A jar contains only nickels (5 cents each) and dimes (10 cents each). There are 30 coins in the jar
+with a total value of $2.30. How many dimes are in the jar?
+
+## Explanation
+
+Two facts give two equations. Let \(n\) be the number of nickels and \(d\) the number of dimes, and
+work in cents ($2.30 = 230 cents).
+
+- "30 coins" → \(n + d = 30\)
+- "total value is 230 cents" → \(5n + 10d = 230\)
+
+From the first equation, \(n = 30 - d\). Substitute into the value equation:
+
+\[5(30 - d) + 10d = 230\]
+\[150 - 5d + 10d = 230\]
+\[150 + 5d = 230 \quad\Rightarrow\quad 5d = 80 \quad\Rightarrow\quad d = 16\]
+
+So there are **16 dimes** (and \(30 - 16 = 14\) nickels).
+(Check: \(14 \times 5 + 16 \times 10 = 70 + 160 = 230\) cents. ✓)
+
+**The traps:**
+- **B (14)** is the number of *nickels* — the other unknown. The question asks for dimes.
+- **C (15)** assumes the coins split evenly, \(30 \div 2\), ignoring the value equation.
+- **E (20)** and **A (8)** come from mixing up the coin values (e.g. treating nickels as 10 cents
+  and dimes as 5) or a sign slip in the substitution. Keep 5-cent and 10-cent coins straight.
+
+## Hints
+
+- Set up two equations: one counting coins (\(n + d = 30\)) and one totaling value in cents
+  (\(5n + 10d = 230\)).
+- Substitute \(n = 30 - d\) into the value equation, then make sure you report *dimes*, not nickels.
diff --git a/content/questions/quant/quant-word-problems-translation-0070.md b/content/questions/quant/quant-word-problems-translation-0070.md
new file mode 100644
index 0000000..941d321
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-translation-0070.md
@@ -0,0 +1,56 @@
+---
+id: quant-word-problems-translation-0070
+section: quant
+topic: word-problems
+subtopic: translation
+type: problem-solving
+difficulty: medium
+tags: [translation, ratio, revenue, setup]
+choices:
+  A: "20"
+  B: "28"
+  C: "40"
+  D: "60"
+  E: "84"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+A theater sold three times as many student tickets as adult tickets. Student tickets cost $6 each
+and adult tickets cost $10 each. If total ticket revenue was $560, how many student tickets were
+sold?
+
+## Explanation
+
+Tie the two quantities to one variable using the ratio. Let \(a\) be the number of adult tickets.
+"Three times as many student tickets as adult tickets" means student tickets \(= 3a\).
+
+Now translate the revenue:
+
+\[\underbrace{6(3a)}_{\text{student \$}} + \underbrace{10a}_{\text{adult \$}} = 560\]
+
+\[18a + 10a = 560 \quad\Rightarrow\quad 28a = 560 \quad\Rightarrow\quad a = 20\]
+
+So there were 20 adult tickets and \(3a = 3(20) = 60\) student tickets.
+(Check: \(60 \times \$6 + 20 \times \$10 = \$360 + \$200 = \$560\). ✓)
+
+The answer is **60**.
+
+**The traps:**
+- **A (20)** is the number of *adult* tickets, \(a\). The question asks for student tickets, which
+  is \(3a\) — don't stop at the variable you solved for.
+- **C (40)** doubles \(a\) instead of tripling it (\(2a\) rather than \(3a\)).
+- **B (28)** is the coefficient \(28\) from \(28a = 560\), reported by mistake.
+- **E (84)** ignores the prices and just splits the ratio across a wrong total.
+
+## Hints
+
+- Let \(a\) = adult tickets, so student tickets \(= 3a\). "Three times as many students" multiplies
+  the *student* count.
+- Write revenue as \(6(3a) + 10a = 560\); after solving for \(a\), the question wants \(3a\).
diff --git a/content/questions/quant/quant-word-problems-translation-0071.md b/content/questions/quant/quant-word-problems-translation-0071.md
new file mode 100644
index 0000000..f93d345
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-translation-0071.md
@@ -0,0 +1,64 @@
+---
+id: quant-word-problems-translation-0071
+section: quant
+topic: word-problems
+subtopic: translation
+type: problem-solving
+difficulty: hard
+tags: [translation, fractions, before-after, setup]
+choices:
+  A: "24"
+  B: "36"
+  C: "40"
+  D: "60"
+  E: "72"
+answer: D
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+Two-fifths of the marbles in a bag are blue. If 12 more blue marbles were added to the bag, then
+blue marbles would make up one-half of all the marbles in the bag. How many marbles are in the bag
+originally?
+
+## Explanation
+
+The key is that adding blue marbles raises **both** the blue count and the total. Let \(t\) be the
+original total number of marbles. Then the original number of blue marbles is \(\tfrac{2}{5}t\).
+
+After adding 12 blue marbles:
+
+- new blue count \(= \tfrac{2}{5}t + 12\)
+- new total \(= t + 12\)
+
+"Blue would make up one-half of all the marbles" translates to:
+
+\[\frac{2}{5}t + 12 = \frac{1}{2}\,(t + 12)\]
+
+Clear the fractions by multiplying every term by 10:
+
+\[4t + 120 = 5(t + 12) = 5t + 60\]
+\[120 - 60 = 5t - 4t \quad\Rightarrow\quad t = 60\]
+
+So the bag originally holds **60** marbles.
+(Check: originally \(\tfrac{2}{5}(60) = 24\) blue out of 60. Add 12 blue: \(36\) blue out of
+\(72\), and \(36/72 = \tfrac{1}{2}\). ✓)
+
+**The traps:**
+- **A (24)** is the original number of *blue* marbles, \(\tfrac{2}{5}t\), not the total.
+- **E (72)** is the *new* total after adding 12 — a "which total?" mix-up.
+- **B (36)** is the new blue count after adding 12.
+- **C (40)** results from adding 12 only to the blue side while forgetting to add it to the total,
+  i.e. solving \(\tfrac{2}{5}t + 12 = \tfrac{1}{2}t\).
+
+## Hints
+
+- Let \(t\) be the original total; blue \(= \tfrac{2}{5}t\). Adding 12 blue marbles changes *both*
+  the numerator and the denominator.
+- Set \(\tfrac{2}{5}t + 12 = \tfrac{1}{2}(t + 12)\), then clear fractions by multiplying through by
+  10.
diff --git a/content/questions/quant/quant-word-problems-translation-0072.md b/content/questions/quant/quant-word-problems-translation-0072.md
new file mode 100644
index 0000000..f6448dc
--- /dev/null
+++ b/content/questions/quant/quant-word-problems-translation-0072.md
@@ -0,0 +1,63 @@
+---
+id: quant-word-problems-translation-0072
+section: quant
+topic: word-problems
+subtopic: translation
+type: problem-solving
+difficulty: hard
+tags: [translation, systems, less-than, setup]
+choices:
+  A: "23"
+  B: "57"
+  C: "61"
+  D: "65"
+  E: "69"
+answer: C
+author: openmat
+reviewers: []
+status: in-review
+original: true
+license: CC-BY-SA-4.0
+---
+
+## Question
+
+The sum of two numbers is 84. The larger number is 8 less than 3 times the smaller number. What is
+the larger number?
+
+## Explanation
+
+Two relationships, so express both numbers with one variable. Let \(s\) be the smaller number and
+\(L\) the larger.
+
+The phrase "8 less than 3 times the smaller number" must be read right-to-left: take \(3s\), then
+subtract 8. So:
+
+\[L = 3s - 8\]
+
+"The sum of the two numbers is 84" gives \(s + L = 84\). Substitute:
+
+\[s + (3s - 8) = 84\]
+\[4s - 8 = 84 \quad\Rightarrow\quad 4s = 92 \quad\Rightarrow\quad s = 23\]
+
+Then the larger number is:
+
+\[L = 3(23) - 8 = 69 - 8 = 61\]
+
+(Check: \(23 + 61 = 84\), and \(61\) is \(8\) less than \(3 \times 23 = 69\). ✓)
+
+The answer is **61**.
+
+**The traps:**
+- **A (23)** is the *smaller* number, \(s\). The question asks for the larger.
+- **E (69)** is \(3s\) — forgetting to subtract the 8 at the end.
+- **D (65)** reverses the sign, reading "8 *more* than 3 times the smaller" as \(L = 3s + 8\), which
+  gives \(s = 19\) and \(L = 65\). "Less than" subtracts.
+- **B (57)** mis-groups the phrase as \(3(s - 8)\) instead of \(3s - 8\), giving \(s = 27\) and
+  \(L = 57\). The "3 times" multiplies only the smaller number, not the whole "\(s - 8\)."
+
+## Hints
+
+- Let \(s\) be the smaller number. "8 less than 3 times \(s\)" is \(3s - 8\) (subtract *after*
+  multiplying) — not \(8 - 3s\) and not \(3(s - 8)\).
+- Substitute into \(s + L = 84\), solve for \(s\), then compute the *larger* value \(L\).
diff --git a/docs/data/content.json b/docs/data/content.json
index 9ef981c..48d5094 100644
--- a/docs/data/content.json
+++ b/docs/data/content.json
@@ -7,8 +7,8 @@
   "stats": {
     "quant": {
       "label": "Quantitative",
-      "questions": 16,
-      "lessons": 5
+      "questions": 72,
+      "lessons": 13
     },
     "verbal": {
       "label": "Verbal",
@@ -22,8 +22,8 @@
     }
   },
   "counts": {
-    "questions": 32,
-    "lessons": 8
+    "questions": 88,
+    "lessons": 16
   },
   "questions": [
     {
@@ -301,6 +301,420 @@
         "Or use the coefficient shortcut: sum of roots \\(= -b\\), product \\(= c\\)."
       ]
     },
+    {
+      "id": "quant-algebra-sequences-functions-0052",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "arithmetic-sequence",
+        "nth-term"
+      ],
+      "choices": {
+        "A": "76",
+        "B": "80",
+        "C": "83",
+        "D": "87",
+        "E": "91"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "An arithmetic sequence has first term 7 and common difference 4. What is its 20th term?",
+      "explanation": "For an arithmetic sequence, the \\(n\\)th term is\n\n\\[a_n = a_1 + (n-1)\\,d.\\]\n\nThe key is the multiplier \\((n-1)\\): to reach the 20th term you add the common difference **19**\ntimes, not 20, because the first term itself takes zero steps.\n\n\\[a_{20} = 7 + (20-1)\\cdot 4 = 7 + 19\\cdot 4 = 7 + 76 = 83.\\]\n\nSo the 20th term is **83**.",
+      "hints": [
+        "The \\(n\\)th term is \\(a_1 + (n-1)d\\) — count the *gaps* between terms, not the terms.",
+        "From the 1st term to the 20th term there are 19 steps of size 4."
+      ]
+    },
+    {
+      "id": "quant-algebra-sequences-functions-0053",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "geometric-sequence",
+        "nth-term"
+      ],
+      "choices": {
+        "A": "32",
+        "B": "48",
+        "C": "96",
+        "D": "160",
+        "E": "192"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A geometric sequence has first term 3 and common ratio 2. What is its 6th term?",
+      "explanation": "For a geometric sequence, the \\(n\\)th term is\n\n\\[a_n = a_1 \\, r^{\\,n-1}.\\]\n\nAs with arithmetic sequences, the exponent is \\((n-1)\\): reaching the 6th term takes **5**\nmultiplications by the ratio, not 6.\n\n\\[a_6 = 3 \\cdot 2^{6-1} = 3 \\cdot 2^{5} = 3 \\cdot 32 = 96.\\]\n\nSo the 6th term is **96**. (Writing the terms out confirms it: 3, 6, 12, 24, 48, **96**.)",
+      "hints": [
+        "The \\(n\\)th term of a geometric sequence is \\(a_1 r^{\\,n-1}\\).",
+        "To land on the 6th term you multiply by the ratio 5 times, then don't forget the factor of 3."
+      ]
+    },
+    {
+      "id": "quant-algebra-sequences-functions-0054",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "arithmetic-sequence",
+        "series-sum"
+      ],
+      "choices": {
+        "A": "855",
+        "B": "900",
+        "C": "950",
+        "D": "1000",
+        "E": "1900"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "An arithmetic sequence has first term 5 and last term 95, and it contains 19 terms in all. What is\nthe sum of all 19 terms?",
+      "explanation": "The sum of an arithmetic sequence is the number of terms times the **average** of the first and last\nterm:\n\n\\[S_n = \\frac{n\\,(a_1 + a_n)}{2}.\\]\n\nYou don't need the common difference — the first and last terms and the count are enough. Here\n\\(n = 19\\), \\(a_1 = 5\\), \\(a_{n} = 95\\):\n\n\\[S_{19} = \\frac{19\\,(5 + 95)}{2} = \\frac{19 \\cdot 100}{2} = 19 \\cdot 50 = 950.\\]\n\nSo the sum is **950**.\n\n(As a check, the common difference is \\(d = \\frac{95 - 5}{19 - 1} = \\frac{90}{18} = 5\\), so the terms\nare \\(5, 10, 15, \\dots, 95\\) — exactly 19 of them.)",
+      "hints": [
+        "Sum \\(= \\dfrac{n(a_1 + a_n)}{2}\\): count of terms times the average of the ends.",
+        "The average of the first and last term is \\(\\frac{5+95}{2} = 50\\); you have 19 of them."
+      ]
+    },
+    {
+      "id": "quant-algebra-sequences-functions-0055",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "functions",
+        "function-notation",
+        "composition"
+      ],
+      "choices": {
+        "A": "5",
+        "B": "7",
+        "C": "10",
+        "D": "11",
+        "E": "13"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "The function \\(f\\) is defined by \\(f(x) = 2x - 3\\). What is the value of \\(f(f(4))\\)?",
+      "explanation": "A composition like \\(f(f(4))\\) is evaluated **inside-out**: compute the inner function first, then feed\nthat result back into \\(f\\).\n\n**Inner:** \\(f(4) = 2\\cdot 4 - 3 = 8 - 3 = 5.\\)\n\n**Outer:** \\(f(5) = 2\\cdot 5 - 3 = 10 - 3 = 7.\\)\n\nSo \\(f(f(4)) = 7\\).\n\nTwo things to keep straight: don't stop after the inner step (that gives 5, not the final answer), and\ndon't drop the \\(-3\\) — it applies **every** time you use the rule.",
+      "hints": [
+        "\\(f(f(4))\\) means: find \\(f(4)\\) first, then apply \\(f\\) to that answer.",
+        "Substitute into \\(2x - 3\\) each time; the \\(-3\\) doesn't disappear on the second application."
+      ]
+    },
+    {
+      "id": "quant-algebra-sequences-functions-0056",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "recursive-sequence",
+        "recursion"
+      ],
+      "choices": {
+        "A": "9",
+        "B": "16",
+        "C": "17",
+        "D": "31",
+        "E": "33"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A sequence is defined by \\(a_1 = 2\\) and \\(a_{n+1} = 2a_n - 1\\) for every \\(n \\ge 1\\). What is the\nvalue of \\(a_5\\)?",
+      "explanation": "There's no shortcut needed — apply the rule one step at a time, labeling each term so you stop at the\nright one:\n\n\\[a_1 = 2\\]\n\\[a_2 = 2(2) - 1 = 3\\]\n\\[a_3 = 2(3) - 1 = 5\\]\n\\[a_4 = 2(5) - 1 = 9\\]\n\\[a_5 = 2(9) - 1 = 17\\]\n\nSo \\(a_5 = 17\\).\n\nThe most common slip is stopping one term early (\\(a_4 = 9\\)) or going one too far (\\(a_6 = 33\\)).\nLabeling each line prevents the off-by-one. (If you spotted the pattern, the closed form is\n\\(a_n = 2^{\\,n-1} + 1\\), which also gives \\(a_5 = 16 + 1 = 17\\).)",
+      "hints": [
+        "Compute term by term: \\(a_2, a_3, a_4, a_5\\) — and write the index next to each so you stop at \\(a_5\\).",
+        "Each step is \"double the previous term, then subtract 1.\""
+      ]
+    },
+    {
+      "id": "quant-algebra-sequences-functions-0057",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "arithmetic-sequence",
+        "system-of-equations"
+      ],
+      "choices": {
+        "A": "-3",
+        "B": "2",
+        "C": "5",
+        "D": "7",
+        "E": "12"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "In an arithmetic sequence, the 4th term is 17 and the 9th term is 42. What is the first term?",
+      "explanation": "Write each given term with the formula \\(a_n = a_1 + (n-1)d\\):\n\n\\[a_4 = a_1 + 3d = 17\\]\n\\[a_9 = a_1 + 8d = 42\\]\n\nSubtract the first equation from the second to eliminate \\(a_1\\):\n\n\\[(a_1 + 8d) - (a_1 + 3d) = 42 - 17 \\;\\Rightarrow\\; 5d = 25 \\;\\Rightarrow\\; d = 5.\\]\n\nNow back-substitute into \\(a_1 + 3d = 17\\):\n\n\\[a_1 + 3(5) = 17 \\;\\Rightarrow\\; a_1 = 17 - 15 = 2.\\]\n\nSo the first term is **2**.\n\nWatch two traps: the question asks for the first term, not the common difference \\(d = 5\\); and the\n4th term is \\(a_1 + 3d\\) (three steps), not \\(a_1 + 4d\\).",
+      "hints": [
+        "Set up \\(a_1 + 3d = 17\\) and \\(a_1 + 8d = 42\\), then subtract to find \\(d\\) first.",
+        "There are 5 steps between the 4th and 9th terms, and the values differ by \\(42 - 17 = 25\\)."
+      ]
+    },
+    {
+      "id": "quant-algebra-sequences-functions-0058",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "geometric-sequence",
+        "threshold",
+        "exponents"
+      ],
+      "choices": {
+        "A": "405",
+        "B": "729",
+        "C": "1215",
+        "D": "2025",
+        "E": "3645"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A geometric sequence has first term 5 and common ratio 3. What is the value of the first term in the\nsequence that is greater than 1000?",
+      "explanation": "The terms are \\(a_n = 5 \\cdot 3^{\\,n-1}\\). Because each term triples, they climb quickly — just list\nthem until one clears 1000:\n\n\\[5,\\; 15,\\; 45,\\; 135,\\; 405,\\; 1215,\\; \\dots\\]\n\nThe 5th term is \\(5 \\cdot 3^4 = 405\\), which is still below 1000. The 6th term is\n\n\\[a_6 = 5 \\cdot 3^{5} = 5 \\cdot 243 = 1215,\\]\n\nwhich is the first term to exceed 1000. So the answer is **1215**.\n\nCommon misses: stopping at the last term *below* the threshold (405), overshooting to the next term\n(3645), or dropping the leading factor of 5 and using a bare power of 3 (\\(3^6 = 729\\)).",
+      "hints": [
+        "List the terms — tripling each time, they pass 1000 fast, so you only need a few.",
+        "Find the last term below 1000, then take one more step (×3) and check that it's the first one over."
+      ]
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots-0017",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "exponents",
+        "power-rules"
+      ],
+      "choices": {
+        "A": "4",
+        "B": "8",
+        "C": "16",
+        "D": "64",
+        "E": "512"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "What is the value of \\(\\dfrac{2^{5} \\cdot 2^{3}}{2^{6}}\\)?",
+      "explanation": "Keep everything in base 2 and just track exponents. When you multiply like bases you **add**\nexponents; when you divide, you **subtract**:\n\n\\[\\frac{2^{5} \\cdot 2^{3}}{2^{6}} = 2^{\\,5 + 3 - 6} = 2^{2} = 4\\]\n\n**The trap:** answer **E (512)** comes from *multiplying* the exponents in the numerator\n(\\(5 \\times 3 = 15\\), then \\(15 - 6 = 9\\), giving \\(2^{9} = 512\\)). You add exponents when the bases\nare multiplied, not multiply them. Answer **D (64)** is \\(2^{6}\\), what you get if you drop one of\nthe numerator terms.",
+      "hints": [
+        "Rewrite the whole expression as a single power of 2.",
+        "Multiply like bases → add exponents; divide → subtract exponents."
+      ]
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots-0018",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "exponents",
+        "negative-exponent",
+        "zero-exponent"
+      ],
+      "choices": {
+        "A": "0",
+        "B": "1/9",
+        "C": "10/9",
+        "D": "2",
+        "E": "9"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "What is the value of \\(3^{-2} + 3^{0}\\)?",
+      "explanation": "Two rules do all the work here. A **negative exponent** means take the reciprocal, and **anything\nnonzero raised to the 0 power is 1**:\n\n\\[3^{-2} = \\frac{1}{3^{2}} = \\frac{1}{9}, \\qquad 3^{0} = 1\\]\n\nSo the sum is\n\n\\[\\frac{1}{9} + 1 = \\frac{1}{9} + \\frac{9}{9} = \\frac{10}{9}\\]\n\n**The traps:** answer **B (1/9)** drops the \\(3^{0} = 1\\) term (a studier who thinks \\(3^0 = 0\\)).\nAnswer **D (2)** comes from mistakenly treating \\(3^{-2}\\) as \\(1\\) as well. A negative exponent is\nnever a negative number, so **A (0)** and **E (9)** are out.",
+      "hints": [
+        "A negative exponent means \"reciprocal,\" not \"negative\": \\(3^{-2} = \\tfrac{1}{3^2}\\).",
+        "Remember that \\(3^{0} = 1\\), not \\(0\\)."
+      ]
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots-0019",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "exponents",
+        "common-base",
+        "equations"
+      ],
+      "choices": {
+        "A": "1",
+        "B": "2",
+        "C": "3",
+        "D": "4",
+        "E": "5"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(2^{\\,x+1} = 8^{\\,x-1}\\), what is the value of \\(x\\)?",
+      "explanation": "The two sides have different bases, so first rewrite them with the **same base**. Since\n\\(8 = 2^{3}\\):\n\n\\[8^{\\,x-1} = \\left(2^{3}\\right)^{x-1} = 2^{\\,3(x-1)} = 2^{\\,3x-3}\\]\n\nNow both sides are powers of 2, so the exponents must be equal:\n\n\\[x + 1 = 3x - 3 \\;\\Rightarrow\\; 4 = 2x \\;\\Rightarrow\\; x = 2\\]\n\nCheck: \\(2^{2+1} = 2^{3} = 8\\) and \\(8^{2-1} = 8^{1} = 8\\). ✓\n\n**The trap:** answer **A (1)** comes from forgetting to distribute the 3, i.e. writing the exponent\nas \\(3x - 1\\) instead of \\(3(x-1) = 3x - 3\\). Distributing carefully is the whole question.",
+      "hints": [
+        "Rewrite \\(8\\) as a power of \\(2\\) so both sides share a base.",
+        "When two equal powers have the same base, set the exponents equal — and distribute the exponent"
+      ]
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots-0020",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "exponents",
+        "fractional-exponents",
+        "roots"
+      ],
+      "choices": {
+        "A": "6",
+        "B": "9",
+        "C": "18",
+        "D": "54",
+        "E": "729"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "What is the value of \\(27^{2/3}\\)?",
+      "explanation": "A fractional exponent \\(m/n\\) means \"take the \\(n\\)th root, then raise to the \\(m\\)th power.\" The\ndenominator is the root; the numerator is the power. Take the root first to keep the numbers small:\n\n\\[27^{2/3} = \\left(27^{1/3}\\right)^{2} = \\left(\\sqrt[3]{27}\\right)^{2} = 3^{2} = 9\\]\n\n**The traps:**\n\n- **E (729)** ignores the cube-root denominator and just squares: \\(27^{2} = 729\\).\n- **A (6)** multiplies the root by the numerator instead of raising to it: \\(3 \\times 2 = 6\\).\n- **C (18)** and **D (54)** treat the exponent as ordinary multiplication (\\(27 \\times \\tfrac{2}{3} = 18\\),\n  or \\(27 \\times 2 = 54\\)) rather than as a root-and-power.",
+      "hints": [
+        "The denominator of the exponent is a root; the numerator is a power.",
+        "\\(27^{2/3} = (\\sqrt[3]{27})^{2}\\) — take the cube root first."
+      ]
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots-0021",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "roots",
+        "radicals",
+        "simplifying"
+      ],
+      "choices": {
+        "A": "\\(2\\sqrt{17}\\)",
+        "B": "\\(15\\sqrt{2}\\)",
+        "C": "\\(8\\sqrt{2}\\)",
+        "D": "\\(4\\sqrt{2}\\)",
+        "E": "\\(34\\sqrt{2}\\)"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "What is \\(\\sqrt{50} + \\sqrt{18}\\) in simplest form?",
+      "explanation": "You can only add radicals once the parts **under** the root match, so simplify each one by pulling\nout perfect squares:\n\n\\[\\sqrt{50} = \\sqrt{25 \\cdot 2} = 5\\sqrt{2}, \\qquad \\sqrt{18} = \\sqrt{9 \\cdot 2} = 3\\sqrt{2}\\]\n\nNow both terms are multiples of \\(\\sqrt{2}\\), so add the coefficients:\n\n\\[5\\sqrt{2} + 3\\sqrt{2} = 8\\sqrt{2}\\]\n\n**The traps:**\n\n- **A (\\(2\\sqrt{17}\\))** comes from illegally merging the roots: \\(\\sqrt{50} + \\sqrt{18} \\neq \\sqrt{68}\\).\n  Square roots do not add that way.\n- **B (\\(15\\sqrt{2}\\))** multiplies the coefficients (\\(5 \\times 3\\)) instead of adding them.",
+      "hints": [
+        "Simplify each radical first by factoring out the largest perfect square.",
+        "You can add radicals only when the number under the root is the same — then add the coefficients."
+      ]
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots-0022",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "exponents",
+        "factoring-powers",
+        "common-base"
+      ],
+      "choices": {
+        "A": "3",
+        "B": "6",
+        "C": "10",
+        "D": "12",
+        "E": "48"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(2^{n} + 2^{n} + 2^{n} + 2^{n} = 2^{12}\\), what is the value of \\(n\\)?",
+      "explanation": "The left side is **four copies of the same term**, so factor — do not treat it as a new base.\nAdding four equal terms is multiplying by 4:\n\n\\[2^{n} + 2^{n} + 2^{n} + 2^{n} = 4 \\cdot 2^{n}\\]\n\nNow write the coefficient as a power of 2 so both sides share a base. Since \\(4 = 2^{2}\\):\n\n\\[4 \\cdot 2^{n} = 2^{2} \\cdot 2^{n} = 2^{\\,n+2}\\]\n\nSetting \\(2^{\\,n+2} = 2^{12}\\) gives \\(n + 2 = 12\\), so \\(n = 10\\).\n\n**The traps:**\n\n- **A (3)** treats the sum as a *product* of the four terms, \\((2^{n})^{4} = 2^{4n} = 2^{12}\\),\n  giving \\(4n = 12\\). Adding equal powers is not the same as multiplying them.\n- **B (6)** correctly writes \\(4 \\cdot 2^{n}\\) but then *multiplies* the exponents (\\(2 \\cdot n = 12\\))\n  instead of adding: \\(2^{2} \\cdot 2^{n} = 2^{n+2}\\), not \\(2^{2n}\\).\n- **D (12)** simply ignores the coefficient of 4.",
+      "hints": [
+        "Four copies of \\(2^{n}\\) is \\(4 \\cdot 2^{n}\\), not \\((2^{n})^{4}\\).",
+        "Rewrite \\(4\\) as \\(2^{2}\\), then combine with \\(2^{n}\\) by **adding** exponents."
+      ]
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots-0023",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "exponents",
+        "common-base",
+        "factoring-powers"
+      ],
+      "choices": {
+        "A": "1/9",
+        "B": "2",
+        "C": "8",
+        "D": "9",
+        "E": "\\(3^{38}\\)"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(x = 3^{20}\\) and \\(y = 3^{18}\\), what is the value of \\(\\dfrac{x - y}{y}\\)?",
+      "explanation": "Do not try to evaluate \\(3^{20}\\) or \\(3^{18}\\) — they are enormous. Split the fraction so the huge\npowers turn into a ratio you can simplify:\n\n\\[\\frac{x - y}{y} = \\frac{x}{y} - \\frac{y}{y} = \\frac{3^{20}}{3^{18}} - 1\\]\n\nDividing like bases means **subtracting** exponents:\n\n\\[\\frac{3^{20}}{3^{18}} = 3^{\\,20 - 18} = 3^{2} = 9\\]\n\nSo the value is \\(9 - 1 = 8\\).\n\n**The traps:**\n\n- **D (9)** is \\(\\dfrac{x}{y} = 3^{2}\\) — correct so far, but it forgets to subtract the \\(\\dfrac{y}{y} = 1\\)\n  term.\n- **B (2)** stops at the exponent difference \\(20 - 18 = 2\\), as if the whole expression equaled\n  \\(3^{20-18}\\) without the \"\\(-1\\)\".\n- **E (\\(3^{38}\\))** comes from *multiplying* the powers (adding exponents \\(20 + 18\\)) instead of\n  dividing.",
+      "hints": [
+        "Break \\(\\dfrac{x - y}{y}\\) into \\(\\dfrac{x}{y} - 1\\) before touching the exponents.",
+        "Dividing like bases subtracts exponents: \\(\\dfrac{3^{20}}{3^{18}} = 3^{2}\\)."
+      ]
+    },
     {
       "id": "quant-arithmetic-fractions-decimals-0015",
       "section": "quant",
@@ -360,29 +774,245 @@
       ]
     },
     {
-      "id": "quant-counting-probability-probability-0007",
+      "id": "quant-counting-probability-combinatorics-0031",
       "section": "quant",
       "topic": "counting-probability",
-      "subtopic": "probability",
+      "subtopic": "combinatorics",
       "type": "problem-solving",
-      "difficulty": "hard",
+      "difficulty": "easy",
       "tags": [
-        "probability",
-        "without-replacement",
-        "dependent-events"
+        "combinatorics",
+        "combinations",
+        "committee"
       ],
       "choices": {
-        "A": "3/10",
-        "B": "9/25",
-        "C": "2/5",
-        "D": "1/2",
-        "E": "3/5"
+        "A": "21",
+        "B": "24",
+        "C": "56",
+        "D": "336",
+        "E": "512"
       },
-      "answer": "A",
+      "answer": "C",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "A bag contains 3 red marbles and 2 blue marbles. Two marbles are drawn at random, one after\nthe other, **without replacement**. What is the probability that both marbles are red?",
+      "prompt": "A book club will choose a 3-person panel from its 8 members. In how many different ways can the\npanel be chosen?",
+      "explanation": "A panel is an **unordered** selection — being picked \"first\" versus \"third\" makes no difference,\nso this is a combination:\n\n\\[C(8,3) = \\frac{8 \\cdot 7 \\cdot 6}{3!} = \\frac{336}{6} = 56\\]\n\n**The trap:** answer **D (336)** is \\(P(8,3) = 8\\cdot7\\cdot6\\), the *ordered* count. That treats\n\"Ann, then Bob, then Cara\" as different from \"Cara, then Bob, then Ann,\" but a panel is the same\ngroup either way — so you must divide by \\(3! = 6\\). Answer **B (24)** comes from \\(8 \\times 3\\)\nand **A (21)** from \\(8 + 7 + 6\\); neither reflects how counting works. Answer **E (512)** is\n\\(8^3\\), which would allow the same person to be picked three times.",
+      "hints": [
+        "Does the order in which the three people are chosen change who is on the panel?",
+        "If order does not matter, use \\(C(n,k) = \\dfrac{n!}{k!\\,(n-k)!}\\)."
+      ]
+    },
+    {
+      "id": "quant-counting-probability-combinatorics-0032",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "combinatorics",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "combinatorics",
+        "permutations",
+        "factorial",
+        "arrangement"
+      ],
+      "choices": {
+        "A": "20",
+        "B": "25",
+        "C": "60",
+        "D": "120",
+        "E": "3125"
+      },
+      "answer": "D",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "In how many different orders can 5 distinct books be arranged on a shelf?",
+      "explanation": "Every book is distinct and every position is distinct, so this is a straight arrangement of all\n5 objects:\n\n\\[5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120\\]\n\n**The trap:** answer **E (3125)** is \\(5^5\\), which would count arrangements as if each of the 5\nslots could independently hold any of the 5 books — but once a book is placed it cannot be reused,\nso the choices shrink \\(5 \\to 4 \\to 3 \\to \\dots\\). Answer **A (20)** is \\(5 \\times 4\\) (arranging\nonly 2 of the books), and **C (60)** is \\(\\tfrac{5!}{2}\\). Answer **B (25)** is \\(5^2\\).",
+      "hints": [
+        "Placing the first book leaves 4 choices for the next slot, then 3, and so on.",
+        "Multiply the shrinking number of choices: that product is \\(5!\\)."
+      ]
+    },
+    {
+      "id": "quant-counting-probability-combinatorics-0033",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "combinatorics",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "combinatorics",
+        "combinations",
+        "multiplication-principle"
+      ],
+      "choices": {
+        "A": "16",
+        "B": "60",
+        "C": "40",
+        "D": "126",
+        "E": "3600"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A committee of 5 is to be formed from a group of 5 seniors and 4 juniors. If the committee must\ncontain exactly 3 seniors and exactly 2 juniors, how many different committees are possible?",
+      "explanation": "Choosing the seniors and choosing the juniors are two independent stages, and within each stage\norder does not matter — so use combinations and **multiply** (the \"AND\" principle):\n\n\\[C(5,3) \\times C(4,2) = 10 \\times 6 = 60\\]\n\nHere \\(C(5,3) = \\dfrac{5\\cdot4\\cdot3}{3!} = 10\\) and \\(C(4,2) = \\dfrac{4\\cdot3}{2!} = 6\\).\n\n**The trap:** answer **A (16)** *adds* the two counts (\\(10 + 6\\)) instead of multiplying — but you\nneed a group of seniors **and** a group of juniors together, so the counts multiply. Answer\n**C (40)** swaps the roles, computing \\(C(5,2)\\times C(4,3) = 10 \\times 4\\) (2 seniors, 3 juniors).\nAnswer **D (126)** is \\(C(9,5)\\), which ignores the required senior/junior split. Answer\n**E (3600)** treats each selection as ordered (\\(P(5,3)\\times P(4,2)\\)).",
+      "hints": [
+        "Count the ways to pick the seniors and the ways to pick the juniors separately.",
+        "\"3 seniors AND 2 juniors\" — do the two counts add or multiply?",
+        "Within each group, does the order you pick people in matter?"
+      ]
+    },
+    {
+      "id": "quant-counting-probability-combinatorics-0034",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "combinatorics",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "combinatorics",
+        "arrangements",
+        "identical-items",
+        "permutations"
+      ],
+      "choices": {
+        "A": "120",
+        "B": "60",
+        "C": "30",
+        "D": "20",
+        "E": "10"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "How many distinct arrangements can be made using all the letters of the word **LEVEL**?",
+      "explanation": "LEVEL has 5 letters, but they are not all distinct: **L** appears twice, **E** appears twice, and\n**V** appears once. When items repeat, divide \\(n!\\) by the factorial of each repeat count so that\nswapping identical letters is not counted as a new arrangement:\n\n\\[\\frac{5!}{2!\\,\\cdot\\,2!\\,\\cdot\\,1!} = \\frac{120}{2 \\cdot 2} = \\frac{120}{4} = 30\\]\n\n**The trap:** answer **A (120)** is \\(5!\\), which counts the two L's (and the two E's) as though\nthey were distinguishable — but \"L\\(_1\\)EVEL\\(_2\\)\" looks identical to \"L\\(_2\\)EVEL\\(_1\\),\" so those\nduplicates must be divided out. Answer **B (60)** is \\(\\tfrac{5!}{2!}\\), correcting for only *one*\nof the two repeated letters. Answers **D** and **E** under-count further.",
+      "hints": [
+        "Count how many times each letter appears: which letters repeat?",
+        "Divide \\(5!\\) by the factorial of *each* repeated letter's count."
+      ]
+    },
+    {
+      "id": "quant-counting-probability-combinatorics-0035",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "combinatorics",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "combinatorics",
+        "circular-arrangement",
+        "permutations"
+      ],
+      "choices": {
+        "A": "24",
+        "B": "60",
+        "C": "360",
+        "D": "720",
+        "E": "120"
+      },
+      "answer": "E",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "In how many distinct ways can 6 people be seated around a circular table, if two seatings are\nconsidered the same when one can be obtained from the other by rotating the whole table?",
+      "explanation": "Around a circle there is no fixed \"first\" seat: rotating everyone one seat to the left produces the\nsame arrangement of neighbors. Fix one person's seat as a reference point and arrange the remaining\n\\(6 - 1 = 5\\) people in the other seats:\n\n\\[(6-1)! = 5! = 120\\]\n\n**The trap:** answer **D (720)** is \\(6!\\), the count for a *straight row* of 6 seats. In a row all\n6 positions are distinct, but a circle's rotations collapse every arrangement into a group of 6\nequivalent rotations, so you divide \\(6!\\) by 6 — equivalently, use \\((n-1)!\\). Answer **C (360)**\nis \\(\\tfrac{6!}{2}\\), an incorrect partial correction.",
+      "hints": [
+        "Around a round table, is there a fixed \"first\" chair, or do rotations look identical?",
+        "Pin one person down as a reference, then arrange the rest: that gives \\((n-1)!\\)."
+      ]
+    },
+    {
+      "id": "quant-counting-probability-combinatorics-0036",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "combinatorics",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "combinatorics",
+        "combinations",
+        "complement",
+        "at-least"
+      ],
+      "choices": {
+        "A": "31",
+        "B": "4",
+        "C": "18",
+        "D": "35",
+        "E": "45"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A 3-person team is to be selected from a pool of 4 men and 3 women. How many different teams\ninclude **at least one woman**?",
+      "explanation": "\"At least one\" is fastest to count with the **complement**: take every possible team and subtract\nthe teams that violate the condition (the all-men teams).\n\n\\[\\underbrace{C(7,3)}_{\\text{all teams}} - \\underbrace{C(4,3)}_{\\text{all men}} = 35 - 4 = 31\\]\n\n**The trap:** answer **E (45)** comes from \"pick 1 woman, then any 2 of the remaining 6 people\":\n\\(C(3,1)\\times C(6,2) = 3 \\times 15 = 45\\). That double-counts — a team with two women gets counted\nonce for each woman chosen as \"the guaranteed one.\" Answer **C (18)** counts only *exactly one*\nwoman, \\(C(3,1)\\times C(4,2) = 3 \\times 6 = 18\\), missing the two-woman and three-woman teams.\nAnswer **D (35)** forgets to remove the all-men teams, and **B (4)** is just the all-men count you\nwere supposed to subtract.\n\nYou can confirm 31 the direct way: exactly 1 woman \\(= 18\\), exactly 2 \\(= C(3,2)C(4,1) = 12\\),\nexactly 3 \\(= C(3,3)C(4,0) = 1\\); \\(18 + 12 + 1 = 31\\).",
+      "hints": [
+        "Counting \"at least one\" directly is error-prone — try counting the opposite.",
+        "Total teams minus all-men teams leaves exactly the teams with at least one woman."
+      ]
+    },
+    {
+      "id": "quant-counting-probability-combinatorics-0037",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "combinatorics",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "combinatorics",
+        "permutations",
+        "arrangements",
+        "restrictions",
+        "block-method"
+      ],
+      "choices": {
+        "A": "24",
+        "B": "48",
+        "C": "72",
+        "D": "120",
+        "E": "240"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "Five students are to be arranged in a row for a photograph. Two of them, Priya and Quinn, insist on\nstanding next to each other. In how many different arrangements can the five students stand?",
+      "explanation": "Use the **block method** for \"must be together.\" Glue Priya and Quinn into a single unit. That\nleaves 4 items to arrange in the row — the block plus the other 3 students:\n\n\\[4! = 24 \\text{ ways to order the units.}\\]\n\nWithin the block, Priya and Quinn can stand in 2 orders (Priya–Quinn or Quinn–Priya), so multiply:\n\n\\[4! \\times 2! = 24 \\times 2 = 48\\]\n\n**The trap:** answer **C (72)** is the count for the *opposite* condition — arrangements where the\ntwo are **not** adjacent: \\(5! - 48 = 120 - 48 = 72\\). If you solve for the wrong condition (or\nsubtract when you should not), you land on 72. Answer **A (24)** forgets the internal \\(2!\\) for the\nblock's two orders, and **D (120)** is \\(5!\\), ignoring the restriction entirely. Answer\n**E (240)** is \\(5! \\times 2\\), which double-applies the arrangement.",
+      "hints": [
+        "Treat the two who must be adjacent as one combined block, then count the arrangements of the blocks.",
+        "Don't forget the two people inside the block can switch places."
+      ]
+    },
+    {
+      "id": "quant-counting-probability-probability-0007",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "probability",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "probability",
+        "without-replacement",
+        "dependent-events"
+      ],
+      "choices": {
+        "A": "3/10",
+        "B": "9/25",
+        "C": "2/5",
+        "D": "1/2",
+        "E": "3/5"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A bag contains 3 red marbles and 2 blue marbles. Two marbles are drawn at random, one after\nthe other, **without replacement**. What is the probability that both marbles are red?",
       "explanation": "For \"both\" events happening in sequence, multiply the probability of the first by the\nprobability of the second *given the first has happened*.\n\n- First marble red: \\(\\tfrac{3}{5}\\) (3 red out of 5 total).\n- Second marble red, given one red is already gone: now 2 red out of 4 remaining, \\(\\tfrac{2}{4} = \\tfrac{1}{2}\\).\n\n\\[P(\\text{both red}) = \\frac{3}{5} \\times \\frac{1}{2} = \\frac{3}{10}\\]\n\n**The trap:** answer **B (9/25)** treats the draws as *with replacement* \\(\\left(\\tfrac{3}{5}\\times\\tfrac{3}{5}\\right)\\).\nWithout replacement, the totals shrink between draws, so the second fraction has denominator 4,\nnot 5.",
       "hints": [
         "\"Without replacement\" means the second draw's totals change — the pool shrinks by one.",
@@ -411,247 +1041,1298 @@
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "How many positive divisors does 72 have?",
-      "explanation": "Listing all divisors by hand is slow and error-prone. Use the **prime factorization**.\n\n\\[72 = 8 \\times 9 = 2^3 \\times 3^2\\]\n\nFor a number written as \\(p^a \\times q^b \\times \\dots\\), the number of positive divisors is\n\\((a+1)(b+1)\\dots\\) — because each divisor independently uses the prime \\(p\\) to a power from\n\\(0\\) to \\(a\\), and \\(q\\) to a power from \\(0\\) to \\(b\\), and so on.\n\nHere that gives:\n\n\\[(3+1)(2+1) = 4 \\times 3 = 12\\]\n\nSo 72 has **12** positive divisors. (Check by listing: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 — twelve of them.)",
+      "prompt": "How many positive divisors does 72 have?",
+      "explanation": "Listing all divisors by hand is slow and error-prone. Use the **prime factorization**.\n\n\\[72 = 8 \\times 9 = 2^3 \\times 3^2\\]\n\nFor a number written as \\(p^a \\times q^b \\times \\dots\\), the number of positive divisors is\n\\((a+1)(b+1)\\dots\\) — because each divisor independently uses the prime \\(p\\) to a power from\n\\(0\\) to \\(a\\), and \\(q\\) to a power from \\(0\\) to \\(b\\), and so on.\n\nHere that gives:\n\n\\[(3+1)(2+1) = 4 \\times 3 = 12\\]\n\nSo 72 has **12** positive divisors. (Check by listing: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 — twelve of them.)",
+      "hints": [
+        "Don't list divisors — factor the number into primes first.",
+        "If \\(N = p^a q^b\\), the divisor count is \\((a+1)(b+1)\\). Remember the \"+1\" accounts for using the prime zero times."
+      ]
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs-0059",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "parity",
+        "odd-even",
+        "must-be-true"
+      ],
+      "choices": {
+        "A": "n + 1",
+        "B": "n + 2",
+        "C": "2n",
+        "D": "3n",
+        "E": "n^2"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(n\\) is an even integer, which of the following must be **odd**?",
+      "explanation": "Start from \"\\(n\\) is even\" and apply the parity rules to each option:\n\n- **A. \\(n + 1\\)** — adding the odd number \\(1\\) flips even to **odd**. ✓\n- **B. \\(n + 2\\)** — adding an even number keeps parity, so this stays **even**.\n- **C. \\(2n\\)** — any integer times \\(2\\) is **even**.\n- **D. \\(3n\\)** — odd × even is **even** (the even factor \\(n\\) wins).\n- **E. \\(n^2\\)** — \\(n \\cdot n\\) is even × even = **even**.\n\nOnly \\(n + 1\\) is guaranteed odd, so the answer is **A**.\n\n**The traps:** **D (\\(3n\\))** tempts people who see the odd coefficient \\(3\\) and forget that\nmultiplying by the even \\(n\\) makes the product even. **E (\\(n^2\\))** catches the idea that\nsquaring changes parity — it does not; \\(n^2\\) has the same parity as \\(n\\).",
+      "hints": [
+        "Adding an even number leaves parity unchanged; adding an odd number flips it.",
+        "A product is even the moment any single factor is even."
+      ]
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs-0060",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "signs",
+        "negative-numbers",
+        "products"
+      ],
+      "choices": {
+        "A": "-24",
+        "B": "-9",
+        "C": "9",
+        "D": "12",
+        "E": "24"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "What is the value of \\((-3)(-4)(-2)\\)?",
+      "explanation": "First find the magnitude, then decide the sign separately.\n\nMagnitude: \\(3 \\cdot 4 \\cdot 2 = 24\\).\n\nSign: count the negative factors. There are **three** of them — an **odd** number — so the product\nis **negative**. Therefore the value is \\(-24\\).\n\nYou can also multiply step by step: \\((-3)(-4) = +12\\) (two negatives make a positive), and then\n\\((+12)(-2) = -24\\).\n\n**The traps:**\n\n- **E (24)** has the right magnitude but the wrong sign — it treats three negatives as if they\n  cancel to positive. An odd number of negatives stays negative.\n- **D (12)** stops after multiplying the first two factors, \\((-3)(-4) = 12\\), and forgets the\n  third factor.\n- **C (9)** adds the magnitudes (\\(3 + 4 + 2\\)) instead of multiplying.\n- **B (-9)** adds the numbers keeping their signs (\\(-3 - 4 - 2\\)) instead of multiplying.",
+      "hints": [
+        "Handle magnitude and sign as two separate questions.",
+        "Count the negative factors: an odd count gives a negative product, an even count a positive one."
+      ]
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs-0061",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "parity",
+        "products",
+        "must-be-true"
+      ],
+      "choices": {
+        "A": "m and n are both odd",
+        "B": "m and n are both even",
+        "C": "m + n is odd",
+        "D": "at least one of m and n is even",
+        "E": "m - n is odd"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(m\\) and \\(n\\) are integers and the product \\(mn\\) is odd, which of the following **must** be\ntrue?",
+      "explanation": "A product of integers is odd only when **every** factor is odd — a single even factor would make\nthe product even. Since \\(mn\\) is odd, both \\(m\\) and \\(n\\) must be odd. That makes **A** correct.\n\nCheck the rest against \"both \\(m\\) and \\(n\\) are odd\":\n\n- **B** (both even) — impossible; even factors give an even product.\n- **D** (at least one even) — same contradiction.\n- **C** (\\(m + n\\) odd) — odd + odd = **even**, so \\(m + n\\) is even, not odd.\n- **E** (\\(m - n\\) odd) — odd − odd = **even**, so this is false too.\n\nThe answer is **A**.\n\n**The traps:** **C** and **E** bait the reflex that \"odd things make odd results.\" Two odd numbers\n*add and subtract* to an even number — parity flips only when the two numbers differ in parity.",
+      "hints": [
+        "A product is odd only if none of its factors is even.",
+        "Once you know both numbers are odd, test what odd ± odd gives."
+      ]
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs-0062",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "signs",
+        "inequalities",
+        "must-be-true"
+      ],
+      "choices": {
+        "A": "Both x and y are negative.",
+        "B": "Both x and y are positive.",
+        "C": "x and y have opposite signs.",
+        "D": "The product xy is negative.",
+        "E": "Exactly one of x and y is zero."
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(xy > 0\\) and \\(x + y < 0\\), which of the following must be true?",
+      "explanation": "Read the two conditions one at a time.\n\n\\(xy > 0\\) tells you \\(x\\) and \\(y\\) have the **same sign** — either both positive or both negative.\n(It also rules out either being \\(0\\), since that would make the product \\(0\\), not positive.)\n\n\\(x + y < 0\\) tells you their sum is negative. Two positive numbers can never sum to something\nnegative, so the \"both positive\" case is out. That leaves only **both negative**, which is\nanswer **A**.\n\n**The traps:**\n\n- **B** (both positive) contradicts \\(x + y < 0\\).\n- **C** (opposite signs) and **D** (\\(xy < 0\\)) both describe \\(xy < 0\\), the opposite of what is\n  given.\n- **E** (one is zero) is impossible: a zero factor makes \\(xy = 0\\), not \\(xy > 0\\).",
+      "hints": [
+        "\\(xy > 0\\) means the two numbers share a sign; \\(xy < 0\\) means opposite signs.",
+        "If they share a sign but add to a negative number, which sign must it be?"
+      ]
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs-0063",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "parity",
+        "consecutive-integers",
+        "must-be-true"
+      ],
+      "choices": {
+        "A": "n^2 + n",
+        "B": "n^2 + 1",
+        "C": "2n + 1",
+        "D": "n^2",
+        "E": "n^2 - 1"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(n\\) is a positive integer, which of the following must be **even**?",
+      "explanation": "Factor where you can, and test each expression against both parities of \\(n\\).\n\n- **A. \\(n^2 + n = n(n + 1)\\)** — this is the product of two **consecutive** integers, and one of\n  any two consecutive integers is always even. So \\(n(n+1)\\) is **always even**. ✓\n- **B. \\(n^2 + 1\\)** — if \\(n\\) is even, \\(n^2 + 1\\) is odd. Not always even.\n- **C. \\(2n + 1\\)** — always **odd** (an even number plus 1).\n- **D. \\(n^2\\)** — has the same parity as \\(n\\), so it is odd whenever \\(n\\) is odd.\n- **E. \\(n^2 - 1 = (n-1)(n+1)\\)** — if \\(n\\) is even, both factors are odd, giving an odd result.\n\nOnly **A** is guaranteed even, so the answer is **A**.\n\n**The traps:** **D (\\(n^2\\))** relies on the false belief that squaring always yields an even\nnumber. **B** and **E** look symmetric to A but flip to odd when \\(n\\) is even — the deciding test\nis trying an even value of \\(n\\).",
+      "hints": [
+        "\\(n^2 + n\\) factors into \\(n(n+1)\\) — what is special about two consecutive integers?",
+        "For each option, just try \\(n = 2\\) and \\(n = 3\\); the \"must be even\" answer survives both."
+      ]
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs-0064",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "consecutive-integers",
+        "divisibility",
+        "cannot-be-true"
+      ],
+      "choices": {
+        "A": "18",
+        "B": "33",
+        "C": "44",
+        "D": "57",
+        "E": "72"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "Which of the following **cannot** be the sum of three consecutive integers?",
+      "explanation": "Write three consecutive integers as \\(n-1\\), \\(n\\), and \\(n+1\\). Their sum is\n\n\\[(n-1) + n + (n+1) = 3n,\\]\n\nwhich is exactly \\(3\\) times the middle integer. So the sum of any three consecutive integers is\n**always a multiple of 3**. A number that is *not* divisible by 3 cannot be such a sum.\n\nCheck each option's divisibility by 3 (add the digits):\n\n- **A. 18** → \\(1+8 = 9\\), divisible by 3. Possible: \\(5 + 6 + 7\\).\n- **B. 33** → \\(3+3 = 6\\), divisible by 3. Possible: \\(10 + 11 + 12\\).\n- **C. 44** → \\(4+4 = 8\\), **not** divisible by 3. **Impossible.** ✓\n- **D. 57** → \\(5+7 = 12\\), divisible by 3. Possible: \\(18 + 19 + 20\\).\n- **E. 72** → \\(7+2 = 9\\), divisible by 3. Possible: \\(23 + 24 + 25\\).\n\nOnly \\(44\\) fails, so the answer is **C**.\n\n**The trap:** without the \"sum \\(= 3n\\)\" insight, every option looks equally plausible and you are\nleft guessing. The four wrong answers are all deliberately multiples of 3 so that only the\ndivisibility test separates them.",
+      "hints": [
+        "Represent the three integers as \\(n-1,\\ n,\\ n+1\\) and add them.",
+        "The sum simplifies to \\(3n\\), so it must be divisible by 3 — test each choice."
+      ]
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs-0065",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "parity",
+        "consecutive-integers",
+        "always-odd"
+      ],
+      "choices": {
+        "A": "n^2 + n + 1",
+        "B": "n^2 + n",
+        "C": "2n^2 + 3n",
+        "D": "n^3 + n",
+        "E": "3n + 3"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "If \\(n\\) is an integer, which of the following is **always odd**?",
+      "explanation": "The key building block is \\(n^2 + n = n(n+1)\\), the product of two consecutive integers, which is\n**always even**. Use that to read off each option's parity.\n\n- **A. \\(n^2 + n + 1\\)** — this is (always even) \\(+\\,1\\), so it is **always odd**. ✓\n- **B. \\(n^2 + n\\)** — always **even** (the consecutive-integer product itself).\n- **C. \\(2n^2 + 3n\\)** — \\(2n^2\\) is even, and \\(3n\\) has the same parity as \\(n\\); the total has\n  the parity of \\(n\\), so it is even when \\(n\\) is even. Not always odd.\n- **D. \\(n^3 + n = n(n^2 + 1)\\)** — if \\(n\\) is even the whole thing is even; if \\(n\\) is odd then\n  \\(n^2 + 1\\) is even, so it is even again. Always **even**.\n- **E. \\(3n + 3 = 3(n + 1)\\)** — its parity matches \\(n + 1\\), so it is even when \\(n\\) is odd.\n  Not always odd.\n\nOnly **A** is odd for every integer \\(n\\), so the answer is **A**.\n\n**The traps:** **C** and **E** are odd for some values of \\(n\\) (try \\(n = 1\\)) but even for\nothers (try \\(n = 2\\)) — \"sometimes odd\" is not \"always odd.\" **D** looks like it should alternate,\nbut factoring shows it is always even.",
+      "hints": [
+        "Anchor on the fact that \\(n^2 + n = n(n+1)\\) is always even, then adjust by the constant.",
+        "\"Always odd\" must survive both an even and an odd test value — check \\(n = 2\\) and \\(n = 3\\)."
+      ]
+    },
+    {
+      "id": "quant-number-properties-remainders-0024",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "remainders",
+        "sums"
+      ],
+      "choices": {
+        "A": "2",
+        "B": "3",
+        "C": "5",
+        "D": "7",
+        "E": "12"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "When the positive integer \\(n\\) is divided by 9, the remainder is 5. What is the remainder when\n\\(n + 7\\) is divided by 9?",
+      "explanation": "You don't need to know \\(n\\) itself. Adding 7 to \\(n\\) adds 7 to the remainder, and then you reduce\nmod 9:\n\n\\[\\text{remainder of } (n+7) = \\text{remainder of } (5 + 7) = \\text{remainder of } 12 = 12 - 9 = 3.\\]\n\nYou can confirm with the smallest such \\(n\\): \\(n = 5\\) gives \\(n + 7 = 12\\), and \\(12 \\div 9\\) leaves\nremainder **3**.\n\n**Traps:** answer **E (12)** is \\(5 + 7\\) with the final reduction forgotten. Answer **D (7)** is\njust the number being added, as if the original remainder didn't matter. Answer **C (5)** assumes\nadding 7 leaves the remainder unchanged. Answer **A (2)** comes from *subtracting* the remainders\n(\\(7 - 5\\)) instead of adding.",
+      "hints": [
+        "You never need the actual value of \\(n\\) — work with its remainder.",
+        "Add the remainders, then reduce mod 9 one more time."
+      ]
+    },
+    {
+      "id": "quant-number-properties-remainders-0025",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "remainders",
+        "division-identity"
+      ],
+      "choices": {
+        "A": "31",
+        "B": "42",
+        "C": "46",
+        "D": "52",
+        "E": "66"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "When a positive integer is divided by 6, the quotient is 7 and the remainder is 4. What is the\ninteger?",
+      "explanation": "Every division fits the identity\n\n\\[n = (\\text{divisor})(\\text{quotient}) + (\\text{remainder}).\\]\n\nHere the divisor is 6, the quotient is 7, and the remainder is 4, so\n\n\\[n = 6 \\times 7 + 4 = 42 + 4 = 46.\\]\n\nCheck: \\(46 \\div 6 = 7\\) with \\(46 - 42 = 4\\) left over. ✓\n\n**Traps:** answer **A (31)** swaps the quotient and remainder (\\(6 \\times 4 + 7\\)). Answer\n**B (42)** is \\(6 \\times 7\\) with the remainder dropped. Answer **D (52)** multiplies by one too many\n(\\(6 \\times 8 + 4\\)). Answer **E (66)** treats the problem as \\(6 \\times (7 + 4)\\).",
+      "hints": [
+        "Use \\(n = \\text{divisor} \\times \\text{quotient} + \\text{remainder}\\).",
+        "Keep the quotient and the remainder in their correct roles — one gets multiplied, one gets added."
+      ]
+    },
+    {
+      "id": "quant-number-properties-remainders-0026",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "remainders",
+        "products"
+      ],
+      "choices": {
+        "A": "1",
+        "B": "2",
+        "C": "4",
+        "D": "8",
+        "E": "11"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "When the positive integers \\(a\\) and \\(b\\) are each divided by 12, the remainders are 5 and 8\nrespectively. What is the remainder when the product \\(ab\\) is divided by 12?",
+      "explanation": "To get the remainder of a product, multiply the remainders and then reduce mod 12:\n\n\\[\\text{remainder of } ab = \\text{remainder of } (5 \\times 8) = \\text{remainder of } 40.\\]\n\nSince \\(40 = 12 \\times 3 + 4\\), the remainder is **4**.\n\nConfirm with the smallest values: \\(a = 5\\), \\(b = 8\\), so \\(ab = 40\\), and \\(40 \\div 12\\) leaves\nremainder **4**. ✓\n\n**Traps:** answer **A (1)** *adds* the remainders (\\(5 + 8 = 13\\), then \\(13 - 12 = 1\\)) instead of\nmultiplying them. Answer **D (8)** and answer **E (11)** come from botching the final reduction of\n40 (subtracting 32 or 29 rather than a multiple of 12). The key discipline: combine the remainders\nthe same way the numbers are combined — here, multiply — then reduce one last time.",
+      "hints": [
+        "The remainder of a product is the remainder of (remainder \\(\\times\\) remainder).",
+        "\\(5 \\times 8 = 40\\) is not the final answer — reduce 40 modulo 12."
+      ]
+    },
+    {
+      "id": "quant-number-properties-remainders-0027",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "remainders",
+        "powers",
+        "cyclicity",
+        "units-digit"
+      ],
+      "choices": {
+        "A": "1",
+        "B": "3",
+        "C": "7",
+        "D": "9",
+        "E": "0"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "What is the remainder when \\(7^{20}\\) is divided by 10?",
+      "explanation": "Dividing by 10 asks for the **units digit**. The units digit of powers of 7 repeats in a cycle of\nlength 4:\n\n\\[7^1 \\to 7, \\quad 7^2 \\to 9, \\quad 7^3 \\to 3, \\quad 7^4 \\to 1, \\quad 7^5 \\to 7, \\ \\dots\\]\n\nTo find where \\(7^{20}\\) lands, divide the exponent by the cycle length: \\(20 = 4 \\times 5\\), so 20 is\na multiple of 4 and \\(7^{20}\\) sits at the **end** of a cycle — the same spot as \\(7^4\\), whose units\ndigit is **1**. The remainder when \\(7^{20}\\) is divided by 10 is **1**.\n\n**Traps:** answer **C (7)** treats \"\\(20 \\div 4\\) has remainder 0\" as if it pointed to the *first*\nentry (7) — but a remainder of 0 means the *last* entry of the cycle, not the first. Answers\n**D (9)** and **B (3)** are the units digits of \\(7^2\\) and \\(7^3\\), chosen by miscounting the\nposition. Answer **E (0)** wrongly assumes any large power is divisible by 10.",
+      "hints": [
+        "Remainder mod 10 is just the units digit.",
+        "Units digits of \\(7^n\\) cycle 7, 9, 3, 1; use \"exponent mod 4,\" and treat a result of 0 as the end"
+      ]
+    },
+    {
+      "id": "quant-number-properties-remainders-0028",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "remainders",
+        "simultaneous-conditions"
+      ],
+      "choices": {
+        "A": "12",
+        "B": "16",
+        "C": "17",
+        "D": "22",
+        "E": "31"
+      },
+      "answer": "D",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "The positive integer \\(n\\) leaves a remainder of 2 when divided by 5 and a remainder of 1 when\ndivided by 3. What is the least possible value of \\(n\\) that is greater than 10?",
+      "explanation": "List the numbers that satisfy each condition and find the overlap.\n\n- Remainder 2 when divided by 5: \\(2, 7, 12, 17, 22, 27, \\dots\\)\n- Remainder 1 when divided by 3: \\(1, 4, 7, 10, 13, 16, 19, 22, 25, \\dots\\)\n\nThe values in **both** lists are \\(7, 22, 37, \\dots\\) — they repeat every \\(\\text{lcm}(5, 3) = 15\\).\nThe smallest one greater than 10 is **22**.\n\nCheck: \\(22 = 5 \\times 4 + 2\\) (remainder 2) and \\(22 = 3 \\times 7 + 1\\) (remainder 1). ✓\n\n**Traps:** answer **A (12)** satisfies only the mod-5 condition (\\(12 \\div 3\\) leaves remainder 0).\nAnswer **B (16)** satisfies only the mod-3 condition (\\(16 \\div 5\\) leaves remainder 1). Answer\n**C (17)** leaves remainder 2 mod 5 but remainder 2 mod 3, so it fails the second condition. Answer\n**E (31)** leaves remainder 1 mod 5, satisfying neither cleanly. Only 22 meets both.",
+      "hints": [
+        "Write out the numbers that fit each condition separately, then look for the first common value.",
+        "Above the first overlap, solutions repeat every lcm of the two divisors."
+      ]
+    },
+    {
+      "id": "quant-number-properties-remainders-0029",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "remainders",
+        "squares",
+        "could-be"
+      ],
+      "choices": {
+        "A": "2",
+        "B": "3",
+        "C": "4",
+        "D": "6",
+        "E": "0"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "The positive integer \\(n\\) is such that \\(n^2\\) leaves a remainder of 4 when divided by 7. Which of\nthe following could be the remainder when \\(n\\) is divided by 7?",
+      "explanation": "The remainder of \\(n^2\\) depends only on the remainder of \\(n\\), so test every possible remainder\n\\(r = 0, 1, 2, \\dots, 6\\) and see which give \\(r^2 \\equiv 4 \\pmod 7\\):\n\n\\[0^2 \\to 0,\\quad 1^2 \\to 1,\\quad 2^2 \\to 4,\\quad 3^2 \\to 2,\\quad 4^2 \\to 2,\\quad 5^2 \\to 4,\\quad 6^2 \\to 1.\\]\n\n(For example, \\(5^2 = 25 = 7 \\times 3 + 4\\).) The value 4 appears when \\(r = 2\\) or \\(r = 5\\). Of the\nanswer choices, only **2** is possible.\n\nCheck: if \\(n = 2\\), then \\(n^2 = 4\\), and \\(4 \\div 7\\) leaves remainder 4. ✓\n\n**Traps:** answer **C (4)** assumes \\(n\\) leaves the *same* remainder as \\(n^2\\) — but a number and\nits square rarely share a remainder (here \\(4^2 = 16\\) leaves remainder 2, not 4). Answers **B (3)**,\n**D (6)**, and **E (0)** all fail: their squares leave remainders 2, 1, and 0 respectively.",
+      "hints": [
+        "The remainder of \\(n^2\\) depends only on the remainder of \\(n\\), so just test \\(r = 0\\) through 6.",
+        "Don't assume \\(n\\) and \\(n^2\\) leave the same remainder — square each candidate and check."
+      ]
+    },
+    {
+      "id": "quant-number-properties-remainders-0030",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "remainders",
+        "simultaneous-conditions",
+        "lcm"
+      ],
+      "choices": {
+        "A": "1",
+        "B": "5",
+        "C": "7",
+        "D": "9",
+        "E": "11"
+      },
+      "answer": "E",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "When the positive integer \\(n\\) is divided by 4, the remainder is 3, and when \\(n\\) is divided by 6,\nthe remainder is 5. What is the remainder when \\(n\\) is divided by 12?",
+      "explanation": "Find a number that satisfies both conditions, then reduce mod 12. List each:\n\n- Remainder 3 when divided by 4: \\(3, 7, 11, 15, 19, 23, \\dots\\)\n- Remainder 5 when divided by 6: \\(5, 11, 17, 23, 29, \\dots\\)\n\nThe first common value is **11**. Both divisors go into \\(\\text{lcm}(4, 6) = 12\\), so the solutions\nrepeat every 12: \\(11, 23, 35, \\dots\\), all of which leave remainder **11** when divided by 12.\n\nCheck: \\(11 = 4 \\times 2 + 3\\) (remainder 3) and \\(11 = 6 \\times 1 + 5\\) (remainder 5). ✓\n\n**Traps:** answers **B (5)** and **C (7)** grab one of the given remainders (or their sum minus one)\nwithout combining the conditions. Answer **D (9)** and answer **A (1)** come from guessing a value\nthat fits only one of the two divisions. A shortcut worth noticing: each condition says \\(n\\) is one\nshort of a multiple (\\(n + 1\\) is divisible by 4 **and** by 6, hence by 12), so \\(n\\) is one less than\na multiple of 12 — remainder **11**.",
+      "hints": [
+        "Find the smallest \\(n\\) satisfying both conditions, then take its remainder mod 12.",
+        "Notice that \\(n + 1\\) is divisible by both 4 and 6 — what does that say about \\(n + 1\\) and 12?"
+      ]
+    },
+    {
+      "id": "quant-statistics-descriptive-0009",
+      "section": "quant",
+      "topic": "statistics",
+      "subtopic": "descriptive",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "weighted-average",
+        "mean"
+      ],
+      "choices": {
+        "A": "84",
+        "B": "85",
+        "C": "86",
+        "D": "87",
+        "E": "88"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "In a class, 10 students scored an average (arithmetic mean) of 80 on a test, and the other 15\nstudents scored an average of 90. What is the average score of all 25 students?",
+      "explanation": "You cannot simply average 80 and 90 — the two groups are different sizes, so the overall mean must be\na **weighted average**. Work with total points, not averages.\n\n- The 10 students contribute \\(10 \\times 80 = 800\\) points.\n- The 15 students contribute \\(15 \\times 90 = 1350\\) points.\n- Total points \\(= 800 + 1350 = 2150\\), spread over \\(25\\) students.\n\n\\[\\text{average} = \\frac{2150}{25} = 86\\]\n\nThe answer is **86**.\n\n**The trap:** answer **B (85)** is the simple average \\(\\tfrac{80 + 90}{2}\\), which ignores that the\nlarger group (15 students) scored higher. A weighted average is always pulled toward the bigger group,\nso it lands *above* the midpoint — closer to 90 than to 80.",
+      "hints": [
+        "Don't average the two averages — the groups aren't the same size.",
+        "Convert each group's average into a total number of points, add the totals, then divide by the total number of students."
+      ]
+    },
+    {
+      "id": "quant-statistics-descriptive-0010",
+      "section": "quant",
+      "topic": "statistics",
+      "subtopic": "descriptive",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "mean",
+        "median",
+        "outliers"
+      ],
+      "choices": {
+        "A": "6",
+        "B": "9",
+        "C": "15",
+        "D": "24",
+        "E": "66"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "For the set \\(\\{4, 7, 9, 10, 90\\}\\), by how much does the mean exceed the median?",
+      "explanation": "Compute the two measures separately.\n\n**Mean:**\n\n\\[\\frac{4 + 7 + 9 + 10 + 90}{5} = \\frac{120}{5} = 24\\]\n\n**Median:** the set is already sorted and has 5 values, so the median is the middle (3rd) value:\n\n\\[4,\\ 7,\\ \\underline{9},\\ 10,\\ 90 \\quad\\Rightarrow\\quad \\text{median} = 9\\]\n\nThe mean exceeds the median by \\(24 - 9 = 15\\).\n\nThe answer is **15**.\n\n**Why they differ:** the outlier \\(90\\) sits far above the rest. It drags the **mean** upward (it adds\n90 to the sum) but has no effect on the **median**, which only cares about the *position* of the middle\nvalue. Whenever a set has a large outlier, expect the mean to sit well above the median.\n\n**The traps:** **D (24)** is the mean itself and **B (9)** is the median itself — both are values you\ncompute on the way, not the difference the question asks for.",
+      "hints": [
+        "Find the mean and the median as two separate numbers, then subtract.",
+        "The set is already in order — the median of five values is just the middle one."
+      ]
+    },
+    {
+      "id": "quant-statistics-descriptive-0011",
+      "section": "quant",
+      "topic": "statistics",
+      "subtopic": "descriptive",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "median",
+        "even-count"
+      ],
+      "choices": {
+        "A": "8",
+        "B": "10",
+        "C": "12",
+        "D": "15",
+        "E": "17"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "The six numbers \\(3,\\ 5,\\ 8,\\ 12,\\ 15,\\ x\\) have a median of 10. What is the least possible value\nof \\(x\\)?",
+      "explanation": "With **six** numbers, the median is the average of the **3rd and 4th values** once the list is sorted.\nWe need that average to equal 10, so the 3rd and 4th values must sum to 20.\n\nSort the five fixed numbers: \\(3, 5, 8, 12, 15\\). Now see where \\(x\\) can land.\n\n- If \\(x \\ge 12\\), the sorted order starts \\(3, 5, 8, 12, \\dots\\), so the 3rd and 4th values are\n  \\(8\\) and \\(12\\): median \\(= \\tfrac{8 + 12}{2} = 10\\). ✓\n- If \\(8 < x < 12\\), the 3rd and 4th values are \\(8\\) and \\(x\\), giving median \\(\\tfrac{8 + x}{2}\\),\n  which is **less than** 10.\n- If \\(x \\le 8\\), the 3rd and 4th values are \\(5\\) and \\(8\\) (or smaller), giving a median of at most\n  \\(6.5\\).\n\nSo the median equals 10 exactly when \\(x \\ge 12\\). Checking the boundary \\(x = 12\\): the set becomes\n\\(3, 5, 8, 12, 12, 15\\), whose middle pair is \\(8\\) and \\(12\\) — median \\(10\\). ✓\n\nThe least possible value is **12**.\n\n**The trap:** answer **B (10)** assumes \\(x\\) *is* the median. But with an even count the median is a\nvalue *between* two data points, and here it happens to require \\(x\\) to be at least 12, not 10.",
+      "hints": [
+        "With six values the median is the average of the 3rd and 4th numbers in sorted order — so they must add to 20.",
+        "Try placing \\(x\\) in different positions relative to 8 and 12, and find the smallest \\(x\\) that pushes the 4th value up to 12."
+      ]
+    },
+    {
+      "id": "quant-statistics-descriptive-0012",
+      "section": "quant",
+      "topic": "statistics",
+      "subtopic": "descriptive",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "mean",
+        "sum",
+        "average-change"
+      ],
+      "choices": {
+        "A": "25",
+        "B": "28",
+        "C": "31",
+        "D": "43",
+        "E": "47"
+      },
+      "answer": "D",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "The average (arithmetic mean) of a set of 6 numbers is 22. When a seventh number is added to the set,\nthe average becomes 25. What is the seventh number?",
+      "explanation": "Averages are easiest to handle through **totals**, using sum \\(=\\) mean \\(\\times\\) count.\n\n- Sum of the original 6 numbers: \\(22 \\times 6 = 132\\).\n- Sum of all 7 numbers: \\(25 \\times 7 = 175\\).\n\nThe seventh number is the difference between the two sums:\n\n\\[175 - 132 = 43\\]\n\nThe answer is **43**.\n\n**Sanity check:** the new number must be large enough not only to sit above the old average of 22 but\nto pull the average of *all seven* values up by 3. Raising 7 numbers' average by 3 requires\n\\(7 \\times 3 = 21\\) extra points beyond the new average, so \\(25 + 21 = 46\\)… let's confirm directly:\nadding 43 raises the sum by \\(43\\), and \\(132 + 43 = 175 = 25 \\times 7\\). ✓\n\n**The traps:** **A (25)** is just the new average, and **B (28)** is the tempting \"\\(25 + 3\\)\" guess.\nBoth underestimate how far above the average a single new value must be to move the mean of the whole\nset.",
+      "hints": [
+        "Convert both averages into sums: sum = mean × count, for 6 numbers and then for 7.",
+        "The seventh number is simply (new sum) − (old sum)."
+      ]
+    },
+    {
+      "id": "quant-statistics-standard-deviation-0013",
+      "section": "quant",
+      "topic": "statistics",
+      "subtopic": "standard-deviation",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "standard-deviation",
+        "spread",
+        "conceptual"
+      ],
+      "choices": {
+        "A": "{2, 4, 6, 8, 10}",
+        "B": "{102, 104, 106, 108, 110}",
+        "C": "{6, 6, 6, 6, 6}",
+        "D": "{1, 5, 6, 7, 11}",
+        "E": "{5, 6, 6, 6, 7}"
+      },
+      "answer": "D",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "Each set below contains five numbers. Which set has the **greatest** standard deviation?\n\n- **A.** \\(\\{2, 4, 6, 8, 10\\}\\)\n- **B.** \\(\\{102, 104, 106, 108, 110\\}\\)\n- **C.** \\(\\{6, 6, 6, 6, 6\\}\\)\n- **D.** \\(\\{1, 5, 6, 7, 11\\}\\)\n- **E.** \\(\\{5, 6, 6, 6, 7\\}\\)",
+      "explanation": "Standard deviation measures how far the values sit from their **mean** — not how large the numbers\nare. Every set here has mean 6 except where noted, so compare the spreads directly. (You don't need\nthe actual SD; the sum of squared distances from the mean orders them correctly.)\n\n- **A** — mean 6, distances \\(-4, -2, 0, 2, 4\\); squared: \\(16+4+0+4+16 = 40\\).\n- **B** — mean 106, distances \\(-4, -2, 0, 2, 4\\); squared: \\(40\\). *Identical spread to A — just shifted up by 100.*\n- **C** — all values equal, so every distance is 0. Standard deviation \\(= 0\\), the smallest possible.\n- **D** — mean 6, distances \\(-5, -1, 0, 1, 5\\); squared: \\(25+1+0+1+25 = 52\\).\n- **E** — mean 6, distances \\(-1, 0, 0, 0, 1\\); squared: \\(2\\).\n\nSet **D** has the largest total squared distance (52), so it has the **greatest** standard deviation.\n\n**The trap:** answer **B** contains by far the largest *numbers*, which tempts you to pick it. But\nadding 100 to every value in A just slides the whole set up the number line without changing how spread\nout it is — A and B have the **same** standard deviation. Magnitude is not spread.",
+      "hints": [
+        "Standard deviation depends only on how far values are from their own mean, not on the size of the numbers — so a set of large numbers can have a small SD.",
+        "Each set's mean is easy to spot; compare the sums of squared distances from the mean rather than computing the actual standard deviations."
+      ]
+    },
+    {
+      "id": "quant-statistics-standard-deviation-0014",
+      "section": "quant",
+      "topic": "statistics",
+      "subtopic": "standard-deviation",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "standard-deviation",
+        "transformation",
+        "conceptual"
+      ],
+      "choices": {
+        "A": "6",
+        "B": "10",
+        "C": "16",
+        "D": "60",
+        "E": "Cannot be determined from the information given"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A set of numbers has a standard deviation of 6. If 10 is added to every number in the set, what is the\nstandard deviation of the resulting set?",
+      "explanation": "Standard deviation measures how far the values are spread from their **mean**. Adding the same\nconstant to every value shifts the entire set — and its mean — by that constant, so every value stays\nexactly as far from the (also-shifted) mean as before. The spread is unchanged.\n\nConcretely, if a value was \\(d\\) above the old mean, it is still \\(d\\) above the new mean after adding\n10 to everything. All the distances from the mean are identical, so the standard deviation is\nunchanged:\n\n\\[\\text{new standard deviation} = 6\\]\n\nThe answer is **6**.\n\n**The trap:** answer **C (16)** adds 10 to the standard deviation, and **B (10)** confuses the shift\namount with the spread. Adding a constant moves the data but does not stretch or compress it. (Contrast\nthis with *multiplying* every value by 10, which **would** multiply the standard deviation by 10.)",
+      "hints": [
+        "Standard deviation is about distance from the mean. What happens to those distances when you slide every value up by the same amount?",
+        "Adding a constant relocates the whole set; only stretching or shrinking it (multiplying) changes the spread."
+      ]
+    },
+    {
+      "id": "quant-word-problems-interest-0045",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "interest",
+        "simple-interest"
+      ],
+      "choices": {
+        "A": "$24",
+        "B": "$120",
+        "C": "$240",
+        "D": "$249.60",
+        "E": "$1,740"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "Priya deposits $1,500 in an account paying 8% simple annual interest. How much interest does the deposit earn in 2 years?",
+      "explanation": "Simple interest is a fixed slice of the *original* principal each year:\n\n\\[I = P \\cdot r \\cdot t = 1500 \\times 0.08 \\times 2 = 240\\]\n\nSo the deposit earns **$240**.\n\n**The traps:**\n- **B ($120)** computes the interest for a single year (\\(1500 \\times 0.08\\)) and forgets to multiply by the 2 years.\n- **E ($1,740)** is the final *balance* \\(P + I = 1500 + 240\\), not the interest the question asked for.\n- **D ($249.60)** compounds the interest annually (\\(1500 \\times 1.08^2 - 1500\\)); the account pays simple interest, so nothing compounds.\n- **A ($24)** slips the decimal, using \\(0.008\\) instead of \\(0.08\\) for the rate.",
+      "hints": [
+        "Simple interest uses the original principal every year: \\(I = P \\cdot r \\cdot t\\).",
+        "The question asks for the interest earned, not the total in the account."
+      ]
+    },
+    {
+      "id": "quant-word-problems-interest-0046",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "interest",
+        "simple-interest",
+        "balance"
+      ],
+      "choices": {
+        "A": "$192",
+        "B": "$848",
+        "C": "$992",
+        "D": "$1,010",
+        "E": "$2,720"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "An account holds $800 and earns 6% simple annual interest. What is the balance in the account after 4 years?",
+      "explanation": "First find the simple interest, then add it back to the principal:\n\n\\[I = P \\cdot r \\cdot t = 800 \\times 0.06 \\times 4 = 192\\]\n\\[A = P + I = 800 + 192 = 992\\]\n\nThe balance is **$992**.\n\n**The traps:**\n- **A ($192)** is the interest alone — the question asks for the balance, so the $800 principal must be added back.\n- **B ($848)** adds only one year of interest (\\(800 + 48\\)) instead of four.\n- **D ($1,010)** compounds annually (\\(800 \\times 1.06^4 \\approx 1{,}010\\)); simple interest does not compound.\n- **E ($2,720)** slips the decimal, treating 6% as \\(0.6\\): \\(800 + 800(0.6)(4)\\).",
+      "hints": [
+        "Compute the interest with \\(I = P \\cdot r \\cdot t\\), then add the principal for the balance.",
+        "Simple interest is the same amount each year — no compounding."
+      ]
+    },
+    {
+      "id": "quant-word-problems-interest-0047",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "interest",
+        "compound-interest"
+      ],
+      "choices": {
+        "A": "$500",
+        "B": "$1,000",
+        "C": "$1,050",
+        "D": "$1,500",
+        "E": "$6,050"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "$5,000 is invested at 10% compounded annually. How much interest does the investment earn over 2 years?",
+      "explanation": "Compound interest adds each year's interest to the balance before the next year. The balance after 2 years is\n\n\\[A = P(1 + r)^{t} = 5000(1.10)^2 = 5000 \\times 1.21 = 6050\\]\n\nThe interest *earned* is the balance minus the principal:\n\n\\[A - P = 6050 - 5000 = 1050\\]\n\nSo the investment earns **$1,050**. (The first year earns $500; the second year earns 10% of the new $5,500 balance, or $550.)\n\n**The traps:**\n- **B ($1,000)** uses simple interest, \\(5000 \\times 0.10 \\times 2\\), ignoring that the first year's interest also earns interest.\n- **E ($6,050)** is the final *balance*, not the interest earned — the $5,000 principal must be subtracted.\n- **A ($500)** counts only the first year's interest.\n- **D ($1,500)** applies simple interest over 3 years instead of 2.",
+      "hints": [
+        "Compound balance is \\(P(1+r)^t\\); the interest earned is that balance minus \\(P\\).",
+        "Over the second year, the interest is charged on the *new* balance, not the original principal."
+      ]
+    },
+    {
+      "id": "quant-word-problems-interest-0048",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "interest",
+        "simple-interest",
+        "rate"
+      ],
+      "choices": {
+        "A": "6%",
+        "B": "9%",
+        "C": "18%",
+        "D": "54%",
+        "E": "0.6%"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "An investment of $1,500 earns $270 in simple interest over 3 years. What is the annual interest rate?",
+      "explanation": "Start from \\(I = P \\cdot r \\cdot t\\) and solve for the rate:\n\n\\[r = \\frac{I}{P \\cdot t} = \\frac{270}{1500 \\times 3} = \\frac{270}{4500} = 0.06\\]\n\nSo the annual rate is **6%**.\n\n**The traps:**\n- **C (18%)** divides the interest by the principal but forgets the 3 years: \\(270 / 1500 = 0.18\\).\n- **B (9%)** divides by only 2 years instead of 3: \\(270 / (1500 \\times 2)\\).\n- **D (54%)** multiplies by the time instead of dividing: \\(270 \\times 3 / 1500\\).\n- **E (0.6%)** slips a decimal place in the final rate.",
+      "hints": [
+        "Rearrange \\(I = P \\cdot r \\cdot t\\) to \\(r = \\dfrac{I}{P \\cdot t}\\).",
+        "Divide by *both* the principal and the number of years."
+      ]
+    },
+    {
+      "id": "quant-word-problems-interest-0049",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "interest",
+        "compound-interest",
+        "simple-interest",
+        "comparison"
+      ],
+      "choices": {
+        "A": "$4",
+        "B": "$40",
+        "C": "$80",
+        "D": "$400",
+        "E": "$840"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "$4,000 is invested for 2 years at 10% per year. How much more interest does the money earn if the interest is compounded annually rather than paid as simple interest?",
+      "explanation": "Compute both and subtract.\n\nSimple interest: \\(I = 4000 \\times 0.10 \\times 2 = 800\\).\n\nCompound interest: \\(A = 4000(1.10)^2 = 4000 \\times 1.21 = 4840\\), so the interest is \\(4840 - 4000 = 840\\).\n\nThe difference is \\(840 - 800 = 40\\).\n\nA faster route: over exactly 2 years the compound-minus-simple gap is \\(P r^2 = 4000 \\times (0.10)^2 = 4000 \\times 0.01 = 40\\) — the interest that the first year's $400 earns in the second year.\n\nSo compounding earns **$40** more.\n\n**The traps:**\n- **E ($840)** is the total compound interest, not the *difference* from simple interest.\n- **D ($400)** is one year's interest (\\(4000 \\times 0.10\\)), a common stand-in for the gap.\n- **C ($80)** doubles the correct gap, as if the $40 recurred in both years.\n- **A ($4)** slips a decimal, using \\(r^2 = 0.001\\) in the shortcut.",
+      "hints": [
+        "Find the simple interest and the compound interest separately, then subtract.",
+        "Over 2 years the gap equals \\(P r^2\\) — the interest earned on the first year's interest."
+      ]
+    },
+    {
+      "id": "quant-word-problems-interest-0050",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "interest",
+        "compound-interest",
+        "semiannual-compounding"
+      ],
+      "choices": {
+        "A": "$400",
+        "B": "$800",
+        "C": "$816",
+        "D": "$1,600",
+        "E": "$1,664"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "$10,000 is invested at a nominal annual rate of 8%, compounded semiannually. How much interest does the investment earn in one year?",
+      "explanation": "Semiannual compounding means two periods per year, each at half the annual rate: \\(8\\% / 2 = 4\\%\\) per period, for \\(2\\) periods.\n\n\\[A = P\\left(1 + \\frac{R}{n}\\right)^{ny} = 10000(1.04)^2 = 10000 \\times 1.0816 = 10816\\]\n\nThe interest earned is \\(10816 - 10000 = 816\\).\n\nSo the investment earns **$816**. (First half-year: $400 of interest; second half-year: 4% of the new $10,400 balance, or $416.)\n\n**The traps:**\n- **B ($800)** ignores compounding and applies a flat 8% for the year (\\(10000 \\times 0.08\\)).\n- **A ($400)** counts only one semiannual period (\\(10000 \\times 0.04\\)).\n- **D ($1,600)** applies the full 8% to *each* half-year as simple interest (\\(10000 \\times 0.08 \\times 2\\)).\n- **E ($1,664)** compounds but forgets to halve the rate, using 8% per period: \\(10000(1.08)^2 - 10000\\).",
+      "hints": [
+        "With semiannual compounding, halve the rate and double the number of periods.",
+        "Two periods at 4% is \\(P(1.04)^2\\) — then subtract the principal for the interest."
+      ]
+    },
+    {
+      "id": "quant-word-problems-interest-0051",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "interest",
+        "compound-interest",
+        "reverse",
+        "principal"
+      ],
+      "choices": {
+        "A": "$7,744",
+        "B": "$8,000",
+        "C": "$8,067",
+        "D": "$8,800",
+        "E": "$10,648"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A sum of money invested at 10% compounded annually grows to $9,680 after 2 years. What was the amount originally invested?",
+      "explanation": "Compound growth multiplies the principal by \\((1+r)^t\\), so to recover the principal you *divide* by that factor:\n\n\\[P = \\frac{A}{(1 + r)^t} = \\frac{9680}{(1.10)^2} = \\frac{9680}{1.21} = 8000\\]\n\nCheck: \\(8000 \\times 1.10 = 8800\\), and \\(8800 \\times 1.10 = 9680\\). ✓\n\nSo the original investment was **$8,000**.\n\n**The traps:**\n- **D ($8,800)** reverses only one of the two years, dividing by 1.10 once.\n- **A ($7,744)** takes a flat 20% off the final amount (\\(9680 \\times 0.8\\)) — but subtracting a percentage is not the inverse of compound growth.\n- **C ($8,067)** divides by 1.20, treating the growth as simple interest over 2 years.\n- **E ($10,648)** grows the amount by another year (\\(9680 \\times 1.10\\)) instead of working backward.",
+      "hints": [
+        "To undo compound growth, divide the final amount by \\((1+r)^t\\), not subtract a percentage.",
+        "Over 2 years at 10%, the factor is \\(1.10^2 = 1.21\\)."
+      ]
+    },
+    {
+      "id": "quant-word-problems-mixtures-0038",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "mixtures",
+        "concentration",
+        "percents"
+      ],
+      "choices": {
+        "A": "2.7",
+        "B": "6",
+        "C": "15",
+        "D": "34",
+        "E": "60"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A 40-liter salt solution is 15% salt by volume. How many liters of salt does the solution contain?",
+      "explanation": "The amount of a substance is its concentration times the total: multiply the percent (as a decimal)\nby the total volume.\n\n\\[0.15 \\times 40 = 6 \\text{ liters}\\]\n\nSo the solution contains **6 liters** of salt.\n\n**The traps:**\n- **C (15)** just repeats the percent number — but 15% of 40 is not 15.\n- **D (34)** is the amount of *water* (\\(40 - 6\\)), not the salt.\n- **A (2.7)** divides instead of multiplying (\\(40 \\div 15\\)).\n- **E (60)** reads \"15%\" as \"times 1.5\" (\\(40 \\times 1.5\\)).",
+      "hints": [
+        "Amount of substance = concentration (as a decimal) × total volume.",
+        "15% means \\(0.15\\), so compute \\(0.15 \\times 40\\)."
+      ]
+    },
+    {
+      "id": "quant-word-problems-mixtures-0039",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "type": "problem-solving",
+      "difficulty": "easy",
+      "tags": [
+        "mixtures",
+        "part-whole",
+        "percents"
+      ],
+      "choices": {
+        "A": "20%",
+        "B": "25%",
+        "C": "33 1/3 %",
+        "D": "50%",
+        "E": "75%"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A trail mix is made from 3 kilograms of peanuts and 1 kilogram of raisins. What percent of the\nmixture, by weight, is raisins?",
+      "explanation": "Concentration is part over **whole**. The whole mixture weighs \\(3 + 1 = 4\\) kilograms, and the\nraisins are 1 of those kilograms:\n\n\\[\\frac{1}{4} = 0.25 = 25\\%\\]\n\nSo raisins make up **25%** of the mixture.\n\n**The traps:**\n- **C (33⅓%)** divides raisins by *peanuts* (\\(\\tfrac{1}{3}\\)) instead of by the whole (\\(\\tfrac{1}{4}\\)).\n- **E (75%)** is the *peanut* share, not the raisin share.\n- **D (50%)** assumes an even split, ignoring the 3-to-1 amounts.\n- **A (20%)** uses a total of 5 (\\(\\tfrac{1}{5}\\)) — miscounting the whole.",
+      "hints": [
+        "Percent of the mixture = (that ingredient) ÷ (total of all ingredients).",
+        "The denominator is the *whole* mixture: \\(3 + 1\\), not just the peanuts."
+      ]
+    },
+    {
+      "id": "quant-word-problems-mixtures-0040",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "mixtures",
+        "weighted-average",
+        "concentration"
+      ],
+      "choices": {
+        "A": "48%",
+        "B": "50%",
+        "C": "55%",
+        "D": "60%",
+        "E": "110%"
+      },
+      "answer": "D",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "20 liters of a 30% acid solution is mixed with 30 liters of an 80% acid solution. What is the acid\nconcentration of the resulting mixture?",
+      "explanation": "Add up the actual **amount of acid**, then divide by the **total volume** — never average the\npercents directly.\n\nAcid from each solution:\n\n\\[0.30 \\times 20 = 6 \\text{ L} \\qquad 0.80 \\times 30 = 24 \\text{ L}\\]\n\nTotal acid \\(= 6 + 24 = 30\\) L in a total volume of \\(20 + 30 = 50\\) L:\n\n\\[\\frac{30}{50} = 60\\%\\]\n\nSo the mixture is **60%** acid. (It's above the halfway point because there's more of the strong\nsolution.)\n\n**The traps:**\n- **C (55%)** is the unweighted average \\(\\tfrac{30 + 80}{2}\\) — wrong because the volumes differ.\n- **E (110%)** adds the two percentages, which can't exceed 100%.\n- **B (50%)** swaps the volumes and concentrations (\\(0.30 \\times 30 + 0.80 \\times 20 = 25\\), then \\(\\tfrac{25}{50}\\)).\n- **A (48%)** counts only the strong solution's acid (\\(\\tfrac{24}{50}\\)), forgetting the 6 L from the weak one.",
+      "hints": [
+        "Compute the liters of acid in each solution separately, then add.",
+        "Concentration = total acid ÷ total volume; weight by the amounts, don't average the percents."
+      ]
+    },
+    {
+      "id": "quant-word-problems-mixtures-0041",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "mixtures",
+        "dilution",
+        "concentration"
+      ],
+      "choices": {
+        "A": "4",
+        "B": "6",
+        "C": "10",
+        "D": "15",
+        "E": "16"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "How many liters of water must be added to 10 liters of a 40% salt solution to dilute it to a 25%\nsalt solution?",
+      "explanation": "Adding water changes the total volume but not the amount of salt, so the salt is the quantity that\nstays fixed. Start by finding it:\n\n\\[\\text{salt} = 0.40 \\times 10 = 4 \\text{ liters}\\]\n\nAfter adding \\(w\\) liters of water, the salt is 25% of the new total \\((10 + w)\\):\n\n\\[\\frac{4}{10 + w} = 0.25 \\;\\Rightarrow\\; 10 + w = \\frac{4}{0.25} = 16 \\;\\Rightarrow\\; w = 6\\]\n\nSo you must add **6 liters** of water.\n\n**The traps:**\n- **E (16)** is the new *total* volume — you still have to subtract the original 10 liters.\n- **A (4)** is the amount of salt, not the water added.\n- **D (15)** treats the 15-percentage-point drop as \\(1.5 \\times 10\\) liters.\n- **C (10)** guesses that you simply double the volume.",
+      "hints": [
+        "The salt stays constant while you add water — find that fixed amount first.",
+        "Set (salt) ÷ (new total) = 0.25, solve for the new total, then subtract the original 10 liters."
+      ]
+    },
+    {
+      "id": "quant-word-problems-mixtures-0042",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "mixtures",
+        "alligation",
+        "price-mixture"
+      ],
+      "choices": {
+        "A": "5",
+        "B": "8",
+        "C": "10",
+        "D": "12",
+        "E": "15"
+      },
+      "answer": "A",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A shopkeeper blends coffee costing $8 per kilogram with coffee costing $12 per kilogram to make\n20 kilograms of a blend that costs $9 per kilogram. How many kilograms of the $12 coffee are in the\nblend?",
+      "explanation": "Let \\(x\\) be the kilograms of $12 coffee; then \\(20 - x\\) kilograms are the $8 coffee. The total cost\nof the blend is $9 × 20 = $180, so match the costs:\n\n\\[8(20 - x) + 12x = 180\\]\n\\[160 - 8x + 12x = 180 \\;\\Rightarrow\\; 4x = 20 \\;\\Rightarrow\\; x = 5\\]\n\nSo the blend contains **5 kilograms** of the $12 coffee. (Alligation agrees: the ratio of $12 to $8\ncoffee is \\((9-8):(12-9) = 1:3\\), so the $12 coffee is \\(\\tfrac14 \\times 20 = 5\\) kg.)\n\n**The traps:**\n- **E (15)** is the amount of the *$8* coffee (the other part), or the alligation ratio read upside down.\n- **C (10)** assumes a 50-50 split of the 20 kg.\n- **B (8)** and **D (12)** are the per-kilogram *prices*, not weights — the question asks for kilograms.",
+      "hints": [
+        "Let \\(x\\) be the $12 coffee and \\(20 - x\\) the $8 coffee; total cost is \\(9 \\times 20 = 180\\).",
+        "Alligation: the ratio of the two coffees is the inverse of their distances from $9."
+      ]
+    },
+    {
+      "id": "quant-word-problems-mixtures-0043",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "mixtures",
+        "replacement",
+        "successive-dilution"
+      ],
+      "choices": {
+        "A": "3",
+        "B": "6",
+        "C": "6.75",
+        "D": "7.5",
+        "E": "9"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A 12-liter tank is full of pure acid. 3 liters are drained off and replaced with water; the tank is\nstirred, then 3 liters of the mixture are drained off and replaced with water again. How many liters\nof pure acid remain in the tank?",
+      "explanation": "Each step removes the same **fraction** of whatever acid is present, then tops up with water. Draining\n3 of 12 liters removes \\(\\tfrac{3}{12} = \\tfrac14\\) of the contents, leaving \\(\\tfrac34\\) of the acid.\nBecause the tank is refilled to 12 liters each time, the *second* draining also removes \\(\\tfrac14\\)\nof the acid then present.\n\nSo the acid is multiplied by \\(\\tfrac34\\) twice:\n\n\\[12 \\times \\left(\\frac{3}{4}\\right)^2 = 12 \\times \\frac{9}{16} = \\frac{108}{16} = 6.75 \\text{ liters}\\]\n\nSo **6.75 liters** of pure acid remain.\n\n**The traps:**\n- **E (9)** stops after the first replacement (\\(12 \\times \\tfrac34 = 9\\)).\n- **D (7.5)** subtracts the fractions instead of multiplying: \\(12\\left(1 - \\tfrac14 - \\tfrac14\\right) = 12 \\times \\tfrac12\\).\n- **B (6)** subtracts 3 liters of acid twice (\\(12 - 3 - 3\\)), as if each draining removed pure acid.\n- **A (3)** is the amount of water added, not the acid remaining.",
+      "hints": [
+        "After the first swap the liquid is a mixture, so the second draining removes acid *and* water — think in fractions, not fixed liters.",
+        "Each replacement multiplies the remaining acid by \\(\\tfrac34\\); apply it twice."
+      ]
+    },
+    {
+      "id": "quant-word-problems-mixtures-0044",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "type": "problem-solving",
+      "difficulty": "hard",
+      "tags": [
+        "mixtures",
+        "weighted-average",
+        "solving"
+      ],
+      "choices": {
+        "A": "4",
+        "B": "5",
+        "C": "10",
+        "D": "25",
+        "E": "80"
+      },
+      "answer": "B",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "A chemist has a 60% alcohol solution and a 10% alcohol solution. How many liters of the 60% solution\nmust be mixed with 20 liters of the 10% solution to produce a 20% alcohol solution?",
+      "explanation": "Let \\(x\\) be the liters of 60% solution added. Track the actual alcohol on each side. The alcohol\nbefore mixing must equal the alcohol after:\n\n\\[0.60x + 0.10(20) = 0.20(x + 20)\\]\n\\[0.60x + 2 = 0.20x + 4\\]\n\\[0.40x = 2 \\;\\Rightarrow\\; x = 5\\]\n\nSo you need **5 liters** of the 60% solution.\n\nCheck: alcohol \\(= 0.60(5) + 0.10(20) = 3 + 2 = 5\\) L in \\(5 + 20 = 25\\) L, and \\(\\tfrac{5}{25} = 20\\%\\). ✓\n\n**The traps:**\n- **E (80)** inverts the alligation ratio. The correct ratio of 60% to 10% solution is\n  \\((20-10):(60-20) = 10:40 = 1:4\\); flipping it to 4:1 gives \\(4 \\times 20 = 80\\).\n- **A (4)** uses the ratio's \"1 part\" as if it were already the number of liters.\n- **D (25)** is the *total* volume of the final mixture, not the amount added.\n- **C (10)** guesses half of the 20 liters.",
+      "hints": [
+        "Write \"alcohol in = alcohol out\": \\(0.60x + 0.10(20) = 0.20(x + 20)\\).",
+        "Or use alligation: the amounts are in the ratio \\((20-10):(60-20)\\), with the 10% solution being the larger share."
+      ]
+    },
+    {
+      "id": "quant-word-problems-overlapping-sets-0006",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "overlapping-sets",
+      "type": "problem-solving",
+      "difficulty": "medium",
+      "tags": [
+        "overlapping-sets",
+        "inclusion-exclusion"
+      ],
+      "choices": {
+        "A": "3",
+        "B": "5",
+        "C": "8",
+        "D": "10",
+        "E": "12"
+      },
+      "answer": "C",
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "prompt": "In a class of 30 students, 18 study French and 15 study Spanish. If 5 students study neither\nlanguage, how many students study both French and Spanish?",
+      "explanation": "Use the two-set inclusion-exclusion formula. First, the number studying **at least one**\nlanguage is everyone minus those studying neither:\n\n\\[30 - 5 = 25 \\text{ study at least one language}\\]\n\nNow apply:\n\n\\[\\text{(at least one)} = \\text{French} + \\text{Spanish} - \\text{both}\\]\n\\[25 = 18 + 15 - \\text{both} \\implies \\text{both} = 33 - 25 = 8\\]\n\nSo **8** students study both.\n\n**Why subtract \"both\":** the 18 French and 15 Spanish counts each include the both-language\nstudents, so adding 18 + 15 double-counts them once. Subtracting \"both\" removes the duplicate.",
       "hints": [
-        "Don't list divisors — factor the number into primes first.",
-        "If \\(N = p^a q^b\\), the divisor count is \\((a+1)(b+1)\\). Remember the \"+1\" accounts for using the prime zero times."
+        "First find how many study at least one language (total minus \"neither\").",
+        "Then use: (at least one) = A + B − (both), and solve for \"both\"."
       ]
     },
     {
-      "id": "quant-statistics-descriptive-0009",
+      "id": "quant-word-problems-rates-speed-distance-0004",
       "section": "quant",
-      "topic": "statistics",
-      "subtopic": "descriptive",
+      "topic": "word-problems",
+      "subtopic": "rates-speed-distance",
       "type": "problem-solving",
-      "difficulty": "medium",
+      "difficulty": "hard",
       "tags": [
-        "weighted-average",
-        "mean"
+        "average-speed",
+        "harmonic-mean"
       ],
       "choices": {
-        "A": "84",
-        "B": "85",
-        "C": "86",
-        "D": "87",
-        "E": "88"
+        "A": "45 mph",
+        "B": "48 mph",
+        "C": "50 mph",
+        "D": "52 mph",
+        "E": "55 mph"
       },
-      "answer": "C",
+      "answer": "B",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "In a class, 10 students scored an average (arithmetic mean) of 80 on a test, and the other 15\nstudents scored an average of 90. What is the average score of all 25 students?",
-      "explanation": "You cannot simply average 80 and 90 — the two groups are different sizes, so the overall mean must be\na **weighted average**. Work with total points, not averages.\n\n- The 10 students contribute \\(10 \\times 80 = 800\\) points.\n- The 15 students contribute \\(15 \\times 90 = 1350\\) points.\n- Total points \\(= 800 + 1350 = 2150\\), spread over \\(25\\) students.\n\n\\[\\text{average} = \\frac{2150}{25} = 86\\]\n\nThe answer is **86**.\n\n**The trap:** answer **B (85)** is the simple average \\(\\tfrac{80 + 90}{2}\\), which ignores that the\nlarger group (15 students) scored higher. A weighted average is always pulled toward the bigger group,\nso it lands *above* the midpoint — closer to 90 than to 80.",
+      "prompt": "A car travels from town A to town B at an average speed of 60 mph, and then returns along the\nsame road at an average speed of 40 mph. What is the car's average speed for the entire round\ntrip?",
+      "explanation": "Average speed is **total distance ÷ total time** — never the simple average of the two speeds.\n\nBecause the two legs cover the *same distance*, pick a convenient distance. Let each leg be 120\nmiles (a common multiple of 60 and 40):\n\n- Going: \\(120 \\div 60 = 2\\) hours\n- Returning: \\(120 \\div 40 = 3\\) hours\n\nTotal distance \\(= 240\\) miles; total time \\(= 5\\) hours.\n\n\\[\\text{average speed} = \\frac{240}{5} = 48 \\text{ mph}\\]\n\n**The trap:** answer **C (50 mph)** is the plain average of 60 and 40. That's wrong because the\ncar spends *more time* at the slower speed (3 hours at 40 vs. 2 hours at 60), which pulls the\naverage below 50.",
       "hints": [
-        "Don't average the two averages — the groups aren't the same size.",
-        "Convert each group's average into a total number of points, add the totals, then divide by the total number of students."
+        "Average speed = total distance / total time. Set the two speeds' simple average aside.",
+        "Choose a distance that both speeds divide evenly (e.g. 120), compute the time for each leg, then combine."
       ]
     },
     {
-      "id": "quant-statistics-descriptive-0010",
+      "id": "quant-word-problems-translation-0066",
       "section": "quant",
-      "topic": "statistics",
-      "subtopic": "descriptive",
+      "topic": "word-problems",
+      "subtopic": "translation",
       "type": "problem-solving",
-      "difficulty": "medium",
+      "difficulty": "easy",
       "tags": [
-        "mean",
-        "median",
-        "outliers"
+        "translation",
+        "linear-equations",
+        "setup"
       ],
       "choices": {
-        "A": "6",
-        "B": "9",
-        "C": "15",
-        "D": "24",
-        "E": "66"
+        "A": "7",
+        "B": "10",
+        "C": "13",
+        "D": "14",
+        "E": "26"
       },
-      "answer": "C",
+      "answer": "A",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "For the set \\(\\{4, 7, 9, 10, 90\\}\\), by how much does the mean exceed the median?",
-      "explanation": "Compute the two measures separately.\n\n**Mean:**\n\n\\[\\frac{4 + 7 + 9 + 10 + 90}{5} = \\frac{120}{5} = 24\\]\n\n**Median:** the set is already sorted and has 5 values, so the median is the middle (3rd) value:\n\n\\[4,\\ 7,\\ \\underline{9},\\ 10,\\ 90 \\quad\\Rightarrow\\quad \\text{median} = 9\\]\n\nThe mean exceeds the median by \\(24 - 9 = 15\\).\n\nThe answer is **15**.\n\n**Why they differ:** the outlier \\(90\\) sits far above the rest. It drags the **mean** upward (it adds\n90 to the sum) but has no effect on the **median**, which only cares about the *position* of the middle\nvalue. Whenever a set has a large outlier, expect the mean to sit well above the median.\n\n**The traps:** **D (24)** is the mean itself and **B (9)** is the median itself — both are values you\ncompute on the way, not the difference the question asks for.",
+      "prompt": "Twice a number, increased by 6, is equal to 20. What is the number?",
+      "explanation": "Translate the sentence piece by piece. Let \\(n\\) be the number.\n\n- \"Twice a number\" → \\(2n\\)\n- \"increased by 6\" → \\(+\\,6\\)\n- \"is equal to 20\" → \\(=\\,20\\)\n\nSo the equation is:\n\n\\[2n + 6 = 20\\]\n\nSolve it in the usual order — undo the addition first, then the multiplication:\n\n\\[2n = 20 - 6 = 14, \\qquad n = \\frac{14}{2} = 7\\]\n\nThe number is **7**. (Check: \\(2(7) + 6 = 20\\). ✓)\n\n**The traps:**\n- **D (14)** stops one step early: it's \\(20 - 6\\), forgetting to divide by 2.\n- **B (10)** ignores the \"\\(+6\\)\" entirely and just computes \\(20 \\div 2\\).\n- **C (13)** *adds* 6 instead of subtracting: \\((20 + 6)/2\\). Undo operations in reverse — the last\n  thing done to \\(n\\) was \"add 6,\" so subtract 6 first.\n- **E (26)** is simply \\(20 + 6\\), with no equation solved at all.",
       "hints": [
-        "Find the mean and the median as two separate numbers, then subtract.",
-        "The set is already in order — the median of five values is just the middle one."
+        "Let \\(n\\) be the number and turn each phrase into a symbol: \"twice\" is \\(\\times 2\\), \"increased by\"",
+        "To undo \\(2n + 6 = 20\\), subtract 6 first, then divide by 2."
       ]
     },
     {
-      "id": "quant-statistics-descriptive-0011",
+      "id": "quant-word-problems-translation-0067",
       "section": "quant",
-      "topic": "statistics",
-      "subtopic": "descriptive",
+      "topic": "word-problems",
+      "subtopic": "translation",
       "type": "problem-solving",
-      "difficulty": "hard",
+      "difficulty": "easy",
       "tags": [
-        "median",
-        "even-count"
+        "translation",
+        "consecutive-integers",
+        "setup"
       ],
       "choices": {
-        "A": "8",
-        "B": "10",
-        "C": "12",
-        "D": "15",
-        "E": "17"
+        "A": "23",
+        "B": "24",
+        "C": "25",
+        "D": "26",
+        "E": "72"
       },
       "answer": "C",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "The six numbers \\(3,\\ 5,\\ 8,\\ 12,\\ 15,\\ x\\) have a median of 10. What is the least possible value\nof \\(x\\)?",
-      "explanation": "With **six** numbers, the median is the average of the **3rd and 4th values** once the list is sorted.\nWe need that average to equal 10, so the 3rd and 4th values must sum to 20.\n\nSort the five fixed numbers: \\(3, 5, 8, 12, 15\\). Now see where \\(x\\) can land.\n\n- If \\(x \\ge 12\\), the sorted order starts \\(3, 5, 8, 12, \\dots\\), so the 3rd and 4th values are\n  \\(8\\) and \\(12\\): median \\(= \\tfrac{8 + 12}{2} = 10\\). ✓\n- If \\(8 < x < 12\\), the 3rd and 4th values are \\(8\\) and \\(x\\), giving median \\(\\tfrac{8 + x}{2}\\),\n  which is **less than** 10.\n- If \\(x \\le 8\\), the 3rd and 4th values are \\(5\\) and \\(8\\) (or smaller), giving a median of at most\n  \\(6.5\\).\n\nSo the median equals 10 exactly when \\(x \\ge 12\\). Checking the boundary \\(x = 12\\): the set becomes\n\\(3, 5, 8, 12, 12, 15\\), whose middle pair is \\(8\\) and \\(12\\) — median \\(10\\). ✓\n\nThe least possible value is **12**.\n\n**The trap:** answer **B (10)** assumes \\(x\\) *is* the median. But with an even count the median is a\nvalue *between* two data points, and here it happens to require \\(x\\) to be at least 12, not 10.",
+      "prompt": "The sum of three consecutive integers is 72. What is the largest of the three integers?",
+      "explanation": "Represent consecutive integers with one variable. Let the smallest be \\(n\\); the next two are\n\\(n + 1\\) and \\(n + 2\\). \"The sum … is 72\" translates to:\n\n\\[n + (n + 1) + (n + 2) = 72\\]\n\nCombine like terms — note the constants add up to \\(+3\\), not \\(0\\):\n\n\\[3n + 3 = 72 \\quad\\Rightarrow\\quad 3n = 69 \\quad\\Rightarrow\\quad n = 23\\]\n\nSo the three integers are \\(23,\\ 24,\\ 25\\), and the **largest** is **25**.\n(Check: \\(23 + 24 + 25 = 72\\). ✓)\n\n**The traps:**\n- **D (26)** comes from forgetting the \"\\(+3\\)\" correction: solving \\(3n = 72\\) gives \\(n = 24\\),\n  and calling that the smallest makes the largest \\(26\\). But the three constants sum to \\(3\\), so\n  \\(3n + 3 = 72\\).\n- **B (24)** is \\(72 \\div 3\\), the *middle* integer (the average), not the largest.\n- **A (23)** is the smallest integer — the value of \\(n\\) — not what the question asked for.\n- **E (72)** is the given total, untouched.",
       "hints": [
-        "With six values the median is the average of the 3rd and 4th numbers in sorted order — so they must add to 20.",
-        "Try placing \\(x\\) in different positions relative to 8 and 12, and find the smallest \\(x\\) that pushes the 4th value up to 12."
+        "Call the smallest integer \\(n\\); the others are \\(n+1\\) and \\(n+2\\). Their sum is \\(3n + 3\\), not",
+        "After you find \\(n\\), the question wants the *largest* value, \\(n + 2\\) — re-read which one it asks"
       ]
     },
     {
-      "id": "quant-statistics-descriptive-0012",
+      "id": "quant-word-problems-translation-0068",
       "section": "quant",
-      "topic": "statistics",
-      "subtopic": "descriptive",
+      "topic": "word-problems",
+      "subtopic": "translation",
       "type": "problem-solving",
       "difficulty": "medium",
       "tags": [
-        "mean",
-        "sum",
-        "average-change"
+        "translation",
+        "ages",
+        "setup"
       ],
       "choices": {
-        "A": "25",
-        "B": "28",
-        "C": "31",
-        "D": "43",
-        "E": "47"
+        "A": "10",
+        "B": "20",
+        "C": "22",
+        "D": "24",
+        "E": "28"
       },
       "answer": "D",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "The average (arithmetic mean) of a set of 6 numbers is 22. When a seventh number is added to the set,\nthe average becomes 25. What is the seventh number?",
-      "explanation": "Averages are easiest to handle through **totals**, using sum \\(=\\) mean \\(\\times\\) count.\n\n- Sum of the original 6 numbers: \\(22 \\times 6 = 132\\).\n- Sum of all 7 numbers: \\(25 \\times 7 = 175\\).\n\nThe seventh number is the difference between the two sums:\n\n\\[175 - 132 = 43\\]\n\nThe answer is **43**.\n\n**Sanity check:** the new number must be large enough not only to sit above the old average of 22 but\nto pull the average of *all seven* values up by 3. Raising 7 numbers' average by 3 requires\n\\(7 \\times 3 = 21\\) extra points beyond the new average, so \\(25 + 21 = 46\\)… let's confirm directly:\nadding 43 raises the sum by \\(43\\), and \\(132 + 43 = 175 = 25 \\times 7\\). ✓\n\n**The traps:** **A (25)** is just the new average, and **B (28)** is the tempting \"\\(25 + 3\\)\" guess.\nBoth underestimate how far above the average a single new value must be to move the mean of the whole\nset.",
+      "prompt": "Maria is 4 years older than twice her brother's age. If the sum of their ages is 34, how old is\nMaria?",
+      "explanation": "Express both ages with a single variable. Let \\(b\\) be the brother's age. Then:\n\n- \"twice her brother's age\" → \\(2b\\)\n- \"4 years older than …\" → \\(2b + 4\\), so Maria \\(= 2b + 4\\)\n\n\"The sum of their ages is 34\" translates to:\n\n\\[b + (2b + 4) = 34\\]\n\nSolve for \\(b\\):\n\n\\[3b + 4 = 34 \\quad\\Rightarrow\\quad 3b = 30 \\quad\\Rightarrow\\quad b = 10\\]\n\nThe brother is 10, so Maria is \\(2(10) + 4 = 24\\).\n(Check: \\(10 + 24 = 34\\). ✓)\n\nThe answer is **24**.\n\n**The traps:**\n- **A (10)** is the brother's age — the variable you solved for, but not the person the question\n  asks about. Always report the requested quantity.\n- **B (20)** is \\(2b\\), forgetting the \"\\(+4\\).\"\n- **E (28)** solves the sum incorrectly as \\(3b = 36 \\Rightarrow b = 12\\) and then computes\n  \\(2b + 4\\) — a slip in isolating \\(b\\).\n- **C (22)** splits 34 loosely (near half) without setting up the relationship at all.",
       "hints": [
-        "Convert both averages into sums: sum = mean × count, for 6 numbers and then for 7.",
-        "The seventh number is simply (new sum) − (old sum)."
+        "Let \\(b\\) be the brother's age and write Maria's age in terms of \\(b\\): \"4 more than twice \\(b\\)\"",
+        "Add the two expressions, set the sum to 34, solve for \\(b\\) — then compute *Maria's* age, not the"
       ]
     },
     {
-      "id": "quant-statistics-standard-deviation-0013",
+      "id": "quant-word-problems-translation-0069",
       "section": "quant",
-      "topic": "statistics",
-      "subtopic": "standard-deviation",
+      "topic": "word-problems",
+      "subtopic": "translation",
       "type": "problem-solving",
-      "difficulty": "hard",
+      "difficulty": "medium",
       "tags": [
-        "standard-deviation",
-        "spread",
-        "conceptual"
+        "translation",
+        "coins",
+        "systems",
+        "setup"
       ],
       "choices": {
-        "A": "{2, 4, 6, 8, 10}",
-        "B": "{102, 104, 106, 108, 110}",
-        "C": "{6, 6, 6, 6, 6}",
-        "D": "{1, 5, 6, 7, 11}",
-        "E": "{5, 6, 6, 6, 7}"
+        "A": "8",
+        "B": "14",
+        "C": "15",
+        "D": "16",
+        "E": "20"
       },
       "answer": "D",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "Each set below contains five numbers. Which set has the **greatest** standard deviation?\n\n- **A.** \\(\\{2, 4, 6, 8, 10\\}\\)\n- **B.** \\(\\{102, 104, 106, 108, 110\\}\\)\n- **C.** \\(\\{6, 6, 6, 6, 6\\}\\)\n- **D.** \\(\\{1, 5, 6, 7, 11\\}\\)\n- **E.** \\(\\{5, 6, 6, 6, 7\\}\\)",
-      "explanation": "Standard deviation measures how far the values sit from their **mean** — not how large the numbers\nare. Every set here has mean 6 except where noted, so compare the spreads directly. (You don't need\nthe actual SD; the sum of squared distances from the mean orders them correctly.)\n\n- **A** — mean 6, distances \\(-4, -2, 0, 2, 4\\); squared: \\(16+4+0+4+16 = 40\\).\n- **B** — mean 106, distances \\(-4, -2, 0, 2, 4\\); squared: \\(40\\). *Identical spread to A — just shifted up by 100.*\n- **C** — all values equal, so every distance is 0. Standard deviation \\(= 0\\), the smallest possible.\n- **D** — mean 6, distances \\(-5, -1, 0, 1, 5\\); squared: \\(25+1+0+1+25 = 52\\).\n- **E** — mean 6, distances \\(-1, 0, 0, 0, 1\\); squared: \\(2\\).\n\nSet **D** has the largest total squared distance (52), so it has the **greatest** standard deviation.\n\n**The trap:** answer **B** contains by far the largest *numbers*, which tempts you to pick it. But\nadding 100 to every value in A just slides the whole set up the number line without changing how spread\nout it is — A and B have the **same** standard deviation. Magnitude is not spread.",
+      "prompt": "A jar contains only nickels (5 cents each) and dimes (10 cents each). There are 30 coins in the jar\nwith a total value of $2.30. How many dimes are in the jar?",
+      "explanation": "Two facts give two equations. Let \\(n\\) be the number of nickels and \\(d\\) the number of dimes, and\nwork in cents ($2.30 = 230 cents).\n\n- \"30 coins\" → \\(n + d = 30\\)\n- \"total value is 230 cents\" → \\(5n + 10d = 230\\)\n\nFrom the first equation, \\(n = 30 - d\\). Substitute into the value equation:\n\n\\[5(30 - d) + 10d = 230\\]\n\\[150 - 5d + 10d = 230\\]\n\\[150 + 5d = 230 \\quad\\Rightarrow\\quad 5d = 80 \\quad\\Rightarrow\\quad d = 16\\]\n\nSo there are **16 dimes** (and \\(30 - 16 = 14\\) nickels).\n(Check: \\(14 \\times 5 + 16 \\times 10 = 70 + 160 = 230\\) cents. ✓)\n\n**The traps:**\n- **B (14)** is the number of *nickels* — the other unknown. The question asks for dimes.\n- **C (15)** assumes the coins split evenly, \\(30 \\div 2\\), ignoring the value equation.\n- **E (20)** and **A (8)** come from mixing up the coin values (e.g. treating nickels as 10 cents\n  and dimes as 5) or a sign slip in the substitution. Keep 5-cent and 10-cent coins straight.",
       "hints": [
-        "Standard deviation depends only on how far values are from their own mean, not on the size of the numbers — so a set of large numbers can have a small SD.",
-        "Each set's mean is easy to spot; compare the sums of squared distances from the mean rather than computing the actual standard deviations."
+        "Set up two equations: one counting coins (\\(n + d = 30\\)) and one totaling value in cents",
+        "Substitute \\(n = 30 - d\\) into the value equation, then make sure you report *dimes*, not nickels."
       ]
     },
     {
-      "id": "quant-statistics-standard-deviation-0014",
+      "id": "quant-word-problems-translation-0070",
       "section": "quant",
-      "topic": "statistics",
-      "subtopic": "standard-deviation",
+      "topic": "word-problems",
+      "subtopic": "translation",
       "type": "problem-solving",
-      "difficulty": "easy",
+      "difficulty": "medium",
       "tags": [
-        "standard-deviation",
-        "transformation",
-        "conceptual"
+        "translation",
+        "ratio",
+        "revenue",
+        "setup"
       ],
       "choices": {
-        "A": "6",
-        "B": "10",
-        "C": "16",
+        "A": "20",
+        "B": "28",
+        "C": "40",
         "D": "60",
-        "E": "Cannot be determined from the information given"
+        "E": "84"
       },
-      "answer": "A",
+      "answer": "D",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "A set of numbers has a standard deviation of 6. If 10 is added to every number in the set, what is the\nstandard deviation of the resulting set?",
-      "explanation": "Standard deviation measures how far the values are spread from their **mean**. Adding the same\nconstant to every value shifts the entire set — and its mean — by that constant, so every value stays\nexactly as far from the (also-shifted) mean as before. The spread is unchanged.\n\nConcretely, if a value was \\(d\\) above the old mean, it is still \\(d\\) above the new mean after adding\n10 to everything. All the distances from the mean are identical, so the standard deviation is\nunchanged:\n\n\\[\\text{new standard deviation} = 6\\]\n\nThe answer is **6**.\n\n**The trap:** answer **C (16)** adds 10 to the standard deviation, and **B (10)** confuses the shift\namount with the spread. Adding a constant moves the data but does not stretch or compress it. (Contrast\nthis with *multiplying* every value by 10, which **would** multiply the standard deviation by 10.)",
+      "prompt": "A theater sold three times as many student tickets as adult tickets. Student tickets cost $6 each\nand adult tickets cost $10 each. If total ticket revenue was $560, how many student tickets were\nsold?",
+      "explanation": "Tie the two quantities to one variable using the ratio. Let \\(a\\) be the number of adult tickets.\n\"Three times as many student tickets as adult tickets\" means student tickets \\(= 3a\\).\n\nNow translate the revenue:\n\n\\[\\underbrace{6(3a)}_{\\text{student \\$}} + \\underbrace{10a}_{\\text{adult \\$}} = 560\\]\n\n\\[18a + 10a = 560 \\quad\\Rightarrow\\quad 28a = 560 \\quad\\Rightarrow\\quad a = 20\\]\n\nSo there were 20 adult tickets and \\(3a = 3(20) = 60\\) student tickets.\n(Check: \\(60 \\times \\$6 + 20 \\times \\$10 = \\$360 + \\$200 = \\$560\\). ✓)\n\nThe answer is **60**.\n\n**The traps:**\n- **A (20)** is the number of *adult* tickets, \\(a\\). The question asks for student tickets, which\n  is \\(3a\\) — don't stop at the variable you solved for.\n- **C (40)** doubles \\(a\\) instead of tripling it (\\(2a\\) rather than \\(3a\\)).\n- **B (28)** is the coefficient \\(28\\) from \\(28a = 560\\), reported by mistake.\n- **E (84)** ignores the prices and just splits the ratio across a wrong total.",
       "hints": [
-        "Standard deviation is about distance from the mean. What happens to those distances when you slide every value up by the same amount?",
-        "Adding a constant relocates the whole set; only stretching or shrinking it (multiplying) changes the spread."
+        "Let \\(a\\) = adult tickets, so student tickets \\(= 3a\\). \"Three times as many students\" multiplies",
+        "Write revenue as \\(6(3a) + 10a = 560\\); after solving for \\(a\\), the question wants \\(3a\\)."
       ]
     },
     {
-      "id": "quant-word-problems-overlapping-sets-0006",
+      "id": "quant-word-problems-translation-0071",
       "section": "quant",
       "topic": "word-problems",
-      "subtopic": "overlapping-sets",
+      "subtopic": "translation",
       "type": "problem-solving",
-      "difficulty": "medium",
+      "difficulty": "hard",
       "tags": [
-        "overlapping-sets",
-        "inclusion-exclusion"
+        "translation",
+        "fractions",
+        "before-after",
+        "setup"
       ],
       "choices": {
-        "A": "3",
-        "B": "5",
-        "C": "8",
-        "D": "10",
-        "E": "12"
+        "A": "24",
+        "B": "36",
+        "C": "40",
+        "D": "60",
+        "E": "72"
       },
-      "answer": "C",
+      "answer": "D",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "In a class of 30 students, 18 study French and 15 study Spanish. If 5 students study neither\nlanguage, how many students study both French and Spanish?",
-      "explanation": "Use the two-set inclusion-exclusion formula. First, the number studying **at least one**\nlanguage is everyone minus those studying neither:\n\n\\[30 - 5 = 25 \\text{ study at least one language}\\]\n\nNow apply:\n\n\\[\\text{(at least one)} = \\text{French} + \\text{Spanish} - \\text{both}\\]\n\\[25 = 18 + 15 - \\text{both} \\implies \\text{both} = 33 - 25 = 8\\]\n\nSo **8** students study both.\n\n**Why subtract \"both\":** the 18 French and 15 Spanish counts each include the both-language\nstudents, so adding 18 + 15 double-counts them once. Subtracting \"both\" removes the duplicate.",
+      "prompt": "Two-fifths of the marbles in a bag are blue. If 12 more blue marbles were added to the bag, then\nblue marbles would make up one-half of all the marbles in the bag. How many marbles are in the bag\noriginally?",
+      "explanation": "The key is that adding blue marbles raises **both** the blue count and the total. Let \\(t\\) be the\noriginal total number of marbles. Then the original number of blue marbles is \\(\\tfrac{2}{5}t\\).\n\nAfter adding 12 blue marbles:\n\n- new blue count \\(= \\tfrac{2}{5}t + 12\\)\n- new total \\(= t + 12\\)\n\n\"Blue would make up one-half of all the marbles\" translates to:\n\n\\[\\frac{2}{5}t + 12 = \\frac{1}{2}\\,(t + 12)\\]\n\nClear the fractions by multiplying every term by 10:\n\n\\[4t + 120 = 5(t + 12) = 5t + 60\\]\n\\[120 - 60 = 5t - 4t \\quad\\Rightarrow\\quad t = 60\\]\n\nSo the bag originally holds **60** marbles.\n(Check: originally \\(\\tfrac{2}{5}(60) = 24\\) blue out of 60. Add 12 blue: \\(36\\) blue out of\n\\(72\\), and \\(36/72 = \\tfrac{1}{2}\\). ✓)\n\n**The traps:**\n- **A (24)** is the original number of *blue* marbles, \\(\\tfrac{2}{5}t\\), not the total.\n- **E (72)** is the *new* total after adding 12 — a \"which total?\" mix-up.\n- **B (36)** is the new blue count after adding 12.\n- **C (40)** results from adding 12 only to the blue side while forgetting to add it to the total,\n  i.e. solving \\(\\tfrac{2}{5}t + 12 = \\tfrac{1}{2}t\\).",
       "hints": [
-        "First find how many study at least one language (total minus \"neither\").",
-        "Then use: (at least one) = A + B − (both), and solve for \"both\"."
+        "Let \\(t\\) be the original total; blue \\(= \\tfrac{2}{5}t\\). Adding 12 blue marbles changes *both*",
+        "Set \\(\\tfrac{2}{5}t + 12 = \\tfrac{1}{2}(t + 12)\\), then clear fractions by multiplying through by"
       ]
     },
     {
-      "id": "quant-word-problems-rates-speed-distance-0004",
+      "id": "quant-word-problems-translation-0072",
       "section": "quant",
       "topic": "word-problems",
-      "subtopic": "rates-speed-distance",
+      "subtopic": "translation",
       "type": "problem-solving",
       "difficulty": "hard",
       "tags": [
-        "average-speed",
-        "harmonic-mean"
+        "translation",
+        "systems",
+        "less-than",
+        "setup"
       ],
       "choices": {
-        "A": "45 mph",
-        "B": "48 mph",
-        "C": "50 mph",
-        "D": "52 mph",
-        "E": "55 mph"
+        "A": "23",
+        "B": "57",
+        "C": "61",
+        "D": "65",
+        "E": "69"
       },
-      "answer": "B",
+      "answer": "C",
       "author": "openmat",
       "reviewers": [],
       "status": "in-review",
-      "prompt": "A car travels from town A to town B at an average speed of 60 mph, and then returns along the\nsame road at an average speed of 40 mph. What is the car's average speed for the entire round\ntrip?",
-      "explanation": "Average speed is **total distance ÷ total time** — never the simple average of the two speeds.\n\nBecause the two legs cover the *same distance*, pick a convenient distance. Let each leg be 120\nmiles (a common multiple of 60 and 40):\n\n- Going: \\(120 \\div 60 = 2\\) hours\n- Returning: \\(120 \\div 40 = 3\\) hours\n\nTotal distance \\(= 240\\) miles; total time \\(= 5\\) hours.\n\n\\[\\text{average speed} = \\frac{240}{5} = 48 \\text{ mph}\\]\n\n**The trap:** answer **C (50 mph)** is the plain average of 60 and 40. That's wrong because the\ncar spends *more time* at the slower speed (3 hours at 40 vs. 2 hours at 60), which pulls the\naverage below 50.",
+      "prompt": "The sum of two numbers is 84. The larger number is 8 less than 3 times the smaller number. What is\nthe larger number?",
+      "explanation": "Two relationships, so express both numbers with one variable. Let \\(s\\) be the smaller number and\n\\(L\\) the larger.\n\nThe phrase \"8 less than 3 times the smaller number\" must be read right-to-left: take \\(3s\\), then\nsubtract 8. So:\n\n\\[L = 3s - 8\\]\n\n\"The sum of the two numbers is 84\" gives \\(s + L = 84\\). Substitute:\n\n\\[s + (3s - 8) = 84\\]\n\\[4s - 8 = 84 \\quad\\Rightarrow\\quad 4s = 92 \\quad\\Rightarrow\\quad s = 23\\]\n\nThen the larger number is:\n\n\\[L = 3(23) - 8 = 69 - 8 = 61\\]\n\n(Check: \\(23 + 61 = 84\\), and \\(61\\) is \\(8\\) less than \\(3 \\times 23 = 69\\). ✓)\n\nThe answer is **61**.\n\n**The traps:**\n- **A (23)** is the *smaller* number, \\(s\\). The question asks for the larger.\n- **E (69)** is \\(3s\\) — forgetting to subtract the 8 at the end.\n- **D (65)** reverses the sign, reading \"8 *more* than 3 times the smaller\" as \\(L = 3s + 8\\), which\n  gives \\(s = 19\\) and \\(L = 65\\). \"Less than\" subtracts.\n- **B (57)** mis-groups the phrase as \\(3(s - 8)\\) instead of \\(3s - 8\\), giving \\(s = 27\\) and\n  \\(L = 57\\). The \"3 times\" multiplies only the smaller number, not the whole \"\\(s - 8\\).\"",
       "hints": [
-        "Average speed = total distance / total time. Set the two speeds' simple average aside.",
-        "Choose a distance that both speeds divide evenly (e.g. 120), compute the time for each leg, then combine."
+        "Let \\(s\\) be the smaller number. \"8 less than 3 times \\(s\\)\" is \\(3s - 8\\) (subtract *after*",
+        "Substitute into \\(s + L = 84\\), solve for \\(s\\), then compute the *larger* value \\(L\\)."
       ]
     },
     {
@@ -968,6 +2649,42 @@
       "status": "in-review",
       "body": "## Overview\n\nA quadratic equation has the form \\(ax^2 + bx + c = 0\\). On GMAT Focus Quant, most quadratics are\nbuilt to **factor cleanly**, so factoring — not the quadratic formula — is usually the fast route.\n\n## Core concepts\n\n**Factoring \\(x^2 + bx + c\\).** Find two numbers that **multiply to \\(c\\)** and **add to \\(b\\)**.\nThen \\(x^2 + bx + c = (x + p)(x + q)\\), and the solutions make each factor zero.\n\nExample: \\(x^2 - 5x + 6\\). Two numbers multiplying to \\(+6\\) and adding to \\(-5\\) are \\(-2\\) and\n\\(-3\\), so \\((x-2)(x-3) = 0\\) and \\(x = 2\\) or \\(x = 3\\).\n\n**Sum and product shortcut.** For \\(x^2 + bx + c = 0\\):\n\n\\[\\text{sum of roots} = -b, \\qquad \\text{product of roots} = c\\]\n\nThis lets you answer \"sum/product of solutions\" questions without fully solving.\n\n**Special products** worth recognizing instantly:\n\n\\[a^2 - b^2 = (a-b)(a+b), \\quad (a+b)^2 = a^2 + 2ab + b^2, \\quad (a-b)^2 = a^2 - 2ab + b^2\\]\n\n**The quadratic formula** (fallback when factoring is ugly):\n\n\\[x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}\\]\n\n## Worked examples\n\n**Factor and solve.** \\(x^2 + 2x - 15 = 0 \\Rightarrow (x+5)(x-3) = 0 \\Rightarrow x = -5 \\text{ or } 3\\).\n\n**Difference of squares.** \\(x^2 - 49 = 0 \\Rightarrow (x-7)(x+7) = 0 \\Rightarrow x = \\pm 7\\).\n\n## Common traps\n\n- **Confusing sum and product of roots.** Sum is \\(-b\\), product is \\(c\\). Read which the question wants.\n- **Dropping a solution.** A quadratic usually has *two* roots — check whether the question wants both, their sum, or only positive values.\n- **Sign errors in factoring.** Verify by expanding your factors back out.\n\n## Key takeaways\n\n- Factor into \\((x + p)(x + q)\\) where \\(pq = c\\) and \\(p + q = b\\); each factor set to zero gives a root.\n- Sum of roots \\(= -b\\), product \\(= c\\) — a fast shortcut.\n- Recognize the difference of squares and perfect-square forms on sight."
     },
+    {
+      "id": "quant-algebra-sequences-functions",
+      "section": "quant",
+      "topic": "algebra",
+      "subtopic": "sequences-functions",
+      "title": "Sequences & Functions",
+      "tags": [
+        "sequences",
+        "arithmetic-sequence",
+        "geometric-sequence",
+        "recursion",
+        "functions",
+        "function-notation"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nTwo closely related ideas show up together on GMAT Focus Quant: **sequences** (ordered lists of\nnumbers built by a rule) and **functions** (a rule that turns an input into an output). Both reward\nthe same discipline — read the rule exactly, mind the indexing, and don't do more arithmetic than the\nquestion needs. The traps are almost always **off-by-one** errors and **misreading the rule**, not\nhard computation.\n\n## Core concepts\n\n**Arithmetic sequences** add a fixed **common difference** \\(d\\) each step. If the first term is\n\\(a_1\\), the \\(n\\)th term is\n\n\\[a_n = a_1 + (n-1)\\,d.\\]\n\nThe multiplier is \\((n-1)\\), **not** \\(n\\): you take \\(d\\) one fewer time than the term number,\nbecause the first term takes zero steps. To **sum** the first \\(n\\) terms, average the first and last\nterm and multiply by how many there are:\n\n\\[S_n = \\frac{n(a_1 + a_n)}{2}.\\]\n\n**Geometric sequences** multiply by a fixed **common ratio** \\(r\\) each step:\n\n\\[a_n = a_1 \\, r^{\\,n-1}.\\]\n\nAgain the exponent is \\(n-1\\), not \\(n\\). Terms grow (or shrink) by a constant *factor*, so they\nchange far faster than an arithmetic sequence's constant *steps*.\n\n**Recursive sequences** define each term from the previous one, e.g. \\(a_{n+1} = 2a_n - 1\\) with a\ngiven starting value. There is no shortcut unless you spot a pattern — just apply the rule one step at\na time, and **stop at the right term** (writing out \\(a_1, a_2, a_3, \\dots\\) with labels prevents the\nusual off-by-one slip).\n\n**Functions** are rules written as \\(f(x)\\). To evaluate \\(f(\\text{something})\\), substitute that\n\"something\" for **every** \\(x\\) in the definition. For a **composition** \\(f(f(x))\\), work\n**inside-out**: compute the inner \\(f\\) first, then feed its result back into \\(f\\).\n\n## Worked examples\n\n**Arithmetic term.** First term \\(a_1 = 7\\), common difference \\(d = 4\\). The 20th term is\n\\(a_{20} = 7 + (20-1)\\cdot 4 = 7 + 76 = 83\\) — not \\(7 + 20\\cdot 4 = 87\\). The \\((n-1)\\) is the whole\ngame.\n\n**Geometric term.** First term \\(3\\), ratio \\(2\\). The 6th term is \\(3\\cdot 2^{6-1} = 3\\cdot 32 = 96\\),\nnot \\(3\\cdot 2^6 = 192\\).\n\n**Recursion.** With \\(a_1 = 2\\) and \\(a_{n+1} = 2a_n - 1\\): \\(a_2 = 3,\\; a_3 = 5,\\; a_4 = 9,\\;\na_5 = 17\\). Label each step so you hand back \\(a_5\\), not \\(a_4\\) or \\(a_6\\).\n\n**Composition.** If \\(f(x) = 2x - 3\\), then \\(f(f(4))\\): first \\(f(4) = 2\\cdot 4 - 3 = 5\\), then\n\\(f(5) = 2\\cdot 5 - 3 = 7\\). Inner first, then outer — and don't drop the \\(-3\\) either time.\n\n## Common traps\n\n- **Off-by-one in the term formula.** Using \\(a_1 + n\\,d\\) or \\(a_1 r^{\\,n}\\) instead of the\n  \\((n-1)\\) version overshoots by exactly one step. This is the single most common error here.\n- **Forgetting the first term.** Computing \\((n-1)d\\) or \\(r^{\\,n-1}\\) but not adding/multiplying the\n  first term.\n- **Dropping ÷2 in the arithmetic sum,** or using the wrong count of terms (the number of terms from\n  \\(a\\) to \\(b\\) with step \\(d\\) is \\(\\frac{b-a}{d} + 1\\), not \\(\\frac{b-a}{d}\\)).\n- **Answering the wrong quantity.** A problem that gives two terms and asks for the *first term* is\n  easy to finish by reporting \\(d\\) instead. Reread what's being asked.\n- **Composition outside-in or stopping early.** \\(f(f(x))\\) is not \\(f(x)\\), and it is not\n  \\([f(x)]^2\\); evaluate the inner function first, then the outer.\n\n## Key takeaways\n\n- Arithmetic: \\(a_n = a_1 + (n-1)d\\); sum \\(= \\dfrac{n(a_1+a_n)}{2}\\).\n- Geometric: \\(a_n = a_1 r^{\\,n-1}\\); terms change by a constant factor, so they move fast.\n- Recursive sequences: apply the rule step by step and **stop at the right index** — label your terms.\n- Functions: substitute for every \\(x\\); compose **inside-out** and keep the constant term.\n- Nearly every miss here is an **off-by-one** or a **misread rule**, not hard arithmetic — slow down on\n  the indexing."
+    },
+    {
+      "id": "quant-arithmetic-exponents-roots",
+      "section": "quant",
+      "topic": "arithmetic",
+      "subtopic": "exponents-roots",
+      "title": "Exponents & Roots",
+      "tags": [
+        "exponents",
+        "roots",
+        "radicals",
+        "powers"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nMost GMAT Focus exponent questions reward you for **rewriting everything in the same base** and\nthen just tracking the exponents. Almost nothing here needs a calculator — the whole game is a\nhandful of rules applied cleanly and, above all, **not confusing multiplication of bases with\naddition of exponents**.\n\n## Core concepts\n\n**The power rules.** For any nonzero base \\(a\\):\n\n\\[a^m \\cdot a^n = a^{m+n}, \\qquad \\frac{a^m}{a^n} = a^{m-n}, \\qquad (a^m)^n = a^{mn}\\]\n\nRead them carefully: you **add** exponents when you *multiply* like bases, and you **multiply**\nexponents when you *raise a power to a power*. Mixing these two up is the single most common\nexponent error on the test.\n\n**Zero and negative exponents.**\n\n\\[a^0 = 1 \\ (a \\neq 0), \\qquad a^{-n} = \\frac{1}{a^n}\\]\n\nA negative exponent means \"reciprocal,\" not \"negative number\": \\(2^{-3} = \\tfrac{1}{8}\\), not \\(-8\\).\n\n**Fractional exponents are roots.**\n\n\\[a^{1/n} = \\sqrt[n]{a}, \\qquad a^{m/n} = \\left(\\sqrt[n]{a}\\right)^{m} = \\sqrt[n]{a^{m}}\\]\n\nSo \\(27^{2/3} = \\left(\\sqrt[3]{27}\\right)^2 = 3^2 = 9\\). Take the root first when it keeps the numbers\nsmall.\n\n**Roots simplify by pulling out perfect squares.** \\(\\sqrt{ab} = \\sqrt{a}\\,\\sqrt{b}\\), so\n\n\\[\\sqrt{50} = \\sqrt{25 \\cdot 2} = 5\\sqrt{2}\\]\n\nYou can only **add or subtract** radicals when the part under the root matches:\n\\(5\\sqrt{2} + 3\\sqrt{2} = 8\\sqrt{2}\\). You may **never** merge \\(\\sqrt{a} + \\sqrt{b}\\) into\n\\(\\sqrt{a+b}\\).\n\n**Factoring out a common power** cracks the hardest questions. Terms like \\(2^n + 2^n\\) are not a\nnew base — they collapse:\n\n\\[2^n + 2^n = 2 \\cdot 2^n = 2^{n+1}, \\qquad 3^{k+1} - 3^{k} = 3^{k}(3 - 1) = 2\\cdot 3^{k}\\]\n\n## Worked examples\n\n**Same-base equation.** Solve \\(2^{x+1} = 8^{x-1}\\). Rewrite \\(8 = 2^3\\), so the right side is\n\\(2^{3(x-1)} = 2^{3x-3}\\). Same base ⟹ equal exponents: \\(x + 1 = 3x - 3\\), giving \\(2x = 4\\) and\n\\(x = 2\\).\n\n**Collapse a sum of equal powers.** If \\(2^n + 2^n + 2^n + 2^n = 2^{12}\\), the left side is\n\\(4 \\cdot 2^n = 2^2 \\cdot 2^n = 2^{n+2}\\). So \\(n + 2 = 12\\) and \\(n = 10\\).\n\n**Ratio of powers.** With \\(x = 3^{20}\\) and \\(y = 3^{18}\\), \\(\\dfrac{x}{y} = 3^{20-18} = 3^2 = 9\\).\n\n## Common traps\n\n- **Multiplying exponents when you should add them** (and vice versa). \\(2^5 \\cdot 2^3 = 2^8\\), not\n  \\(2^{15}\\).\n- **Reading a negative exponent as a negative number.** \\(3^{-2} = \\tfrac{1}{9}\\), and \\(3^0 = 1\\)\n  (not \\(0\\)).\n- **Merging unlike radicals.** \\(\\sqrt{50} + \\sqrt{18} \\neq \\sqrt{68}\\); simplify each first, then add.\n- **Treating a sum of powers as a new base.** \\(2^n + 2^n\\) is \\(2^{n+1}\\), not \\(4^n\\) or \\(2^{2n}\\).\n\n## Key takeaways\n\n- Get everything into a **common base**, then just manipulate exponents.\n- Multiply like bases → **add** exponents; raise a power to a power → **multiply** exponents.\n- \\(a^0 = 1\\), \\(a^{-n} = 1/a^n\\), and \\(a^{m/n} = \\sqrt[n]{a^m}\\).\n- Simplify radicals by pulling out perfect squares; only combine radicals with the same radicand.\n- Sums of equal powers **factor**: \\(k\\) copies of \\(a^n\\) is \\(k \\cdot a^n\\)."
+    },
     {
       "id": "quant-arithmetic-percents-ratios",
       "section": "quant",
@@ -984,6 +2701,58 @@
       "status": "in-review",
       "body": "## Overview\n\nPercents and ratios are the most frequently tested arithmetic ideas on GMAT Focus Quant, and\nthey hide inside word problems everywhere. Master three moves: converting between forms, handling\npercent *change*, and scaling ratios.\n\n## Core concepts\n\n**Percent as a factor.** A percent is just a number over 100. The fastest way to apply a percent\nchange is to turn it into a **multiplier**:\n\n- Increase by \\(r\\%\\): multiply by \\(1 + \\tfrac{r}{100}\\). (+25% → \\(\\times 1.25\\))\n- Decrease by \\(r\\%\\): multiply by \\(1 - \\tfrac{r}{100}\\). (−20% → \\(\\times 0.80\\))\n\n**Successive changes multiply.** Two changes in a row are multiplied, never added:\n\n\\[(+25\\%)\\text{ then }(-20\\%): \\quad 1.25 \\times 0.80 = 1.00 \\quad (\\text{no net change})\\]\n\n**Percent change formula.**\n\n\\[\\text{percent change} = \\frac{\\text{new} - \\text{old}}{\\text{old}} \\times 100\\%\\]\n\nAlways divide by the **original** value, not the new one.\n\n**Ratios scale together.** A ratio \\(a : b\\) means the quantities are \\(ak\\) and \\(bk\\) for some\nmultiplier \\(k\\). If boys : girls \\(= 3 : 5\\) and there are 24 boys, then \\(k = 8\\), so there are\n\\(5 \\times 8 = 40\\) girls.\n\n## Worked examples\n\n**Percent change.** A stock rises from $80 to $100. Percent increase \\(= \\tfrac{100 - 80}{80} = \\tfrac{20}{80} = 25\\%\\).\nNote it later falls from $100 back to $80: that's \\(\\tfrac{-20}{100} = -20\\%\\) — a *smaller* percent, because the base is now larger.\n\n**Ratio scaling.** A recipe uses flour : sugar \\(= 7 : 2\\). To use 21 cups of flour, \\(k = 3\\), so\nyou need \\(2 \\times 3 = 6\\) cups of sugar.\n\n## Common traps\n\n- **Adding successive percents.** +25% then −20% is *not* +5%; it's \\(1.25 \\times 0.80 = 1.00\\).\n- **Wrong base.** Percent change always divides by the original amount. A rise then an equal-percent fall does not return to the start.\n- **Ratio ≠ actual count.** \\(3 : 5\\) does not mean 3 and 5 — it means \\(3k\\) and \\(5k\\). Find \\(k\\) first.\n\n## Key takeaways\n\n- Convert percent changes to multipliers and multiply them for successive changes.\n- Percent change = (new − old) / old.\n- A ratio \\(a : b\\) represents \\(ak\\) and \\(bk\\); solve for the multiplier \\(k\\), then scale."
     },
+    {
+      "id": "quant-counting-probability-combinatorics",
+      "section": "quant",
+      "topic": "counting-probability",
+      "subtopic": "combinatorics",
+      "title": "Combinatorics: Counting Arrangements and Selections",
+      "tags": [
+        "combinatorics",
+        "permutations",
+        "combinations",
+        "factorial",
+        "counting"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nCombinatorics is the art of counting outcomes **without listing them all**. On GMAT Focus\nQuant the whole game is deciding a single question: **does order matter?** Get that right and\npick the matching tool, and most counting problems collapse into one line of arithmetic.\n\n## Core concepts\n\n**The factorial.** The number of ways to arrange \\(n\\) distinct objects in a row is\n\\(n!\\) (read \"\\(n\\) factorial\"):\n\n\\[n! = n \\times (n-1) \\times (n-2) \\times \\cdots \\times 2 \\times 1\\]\n\nSo \\(5! = 5\\cdot4\\cdot3\\cdot2\\cdot1 = 120\\). By convention \\(0! = 1\\).\n\n**Permutations — order matters.** The number of ways to arrange \\(k\\) objects chosen from\n\\(n\\) distinct objects, where the order of the chosen objects matters, is\n\n\\[P(n,k) = \\frac{n!}{(n-k)!} = n(n-1)\\cdots(n-k+1)\\]\n\nUse permutations for rankings, seatings, passwords, or codes — anything where \"A then B\"\ndiffers from \"B then A.\"\n\n**Combinations — order does NOT matter.** The number of ways to *choose* \\(k\\) objects from\n\\(n\\), ignoring order, is\n\n\\[C(n,k) = \\binom{n}{k} = \\frac{n!}{k!\\,(n-k)!}\\]\n\nUse combinations for committees, teams, handshakes, or any unordered selection. Notice\n\\(C(n,k) = \\dfrac{P(n,k)}{k!}\\): a combination is a permutation with the \\(k!\\) reorderings of\nthe same group divided out.\n\n**Two governing principles.**\n\n- **Multiplication (AND).** If a task is a sequence of independent stages, multiply the counts.\n  Choosing 3 seniors *and* 2 juniors: multiply \\(C(5,3)\\times C(4,2)\\).\n- **Addition (OR).** If outcomes fall into mutually exclusive cases, add the counts.\n\n**Identical items.** When some objects repeat, divide by the factorial of each repeat count.\nDistinct arrangements of a word with letter counts \\(n_1, n_2, \\dots\\) totalling \\(n\\) letters:\n\n\\[\\frac{n!}{n_1!\\,n_2!\\cdots}\\]\n\n**Circular arrangements.** Seating \\(n\\) people around a round table gives \\((n-1)!\\)\narrangements, because rotations of the same circle are not distinct — fix one person, then\narrange the rest.\n\n## Worked examples\n\n**Combination.** A book club picks 3 of its 8 members to lead a discussion. Order does not\nmatter, so \\(C(8,3) = \\dfrac{8\\cdot7\\cdot6}{3!} = \\dfrac{336}{6} = 56\\) ways.\n\n**Permutation.** Those same 8 members award a gold, silver, and bronze badge — now order\nmatters, so \\(P(8,3) = 8\\cdot7\\cdot6 = 336\\) ways. Same people, \\(3! = 6\\) times as many\noutcomes because the badges are distinguishable.\n\n**Complement (\"at least\").** From 4 men and 3 women, choose a 3-person team with at least one\nwoman. Count the total and subtract the unwanted case: \\(C(7,3) - C(4,3) = 35 - 4 = 31\\).\n\n## Common traps\n\n- **Order confusion.** The single biggest error. A committee is unordered (combination); a\n  lineup or ranking is ordered (permutation). Using \\(P\\) where \\(C\\) belongs overcounts by\n  exactly \\(k!\\).\n- **Adding instead of multiplying.** \"3 seniors AND 2 juniors\" multiplies: \\(C(5,3)\\times C(4,2)\\),\n  not \\(C(5,3)+C(4,2)\\).\n- **Forgetting repeated items.** Arranging the letters of a word with repeats requires dividing\n  \\(n!\\) by the factorial of *each* repeated letter's count — miss one and you overcount.\n- **\"At least one\" head-on.** Counting cases directly often double-counts. Use the complement:\n  total minus \"none.\"\n- **Blocks that must stay together.** Glue the block into one unit, arrange the units, then\n  multiply by the internal arrangements of the block.\n\n## Key takeaways\n\n- Ask first: **does order matter?** Yes → permutation \\(P(n,k)\\); no → combination \\(C(n,k)\\).\n- \\(C(n,k) = \\dfrac{P(n,k)}{k!}\\) — dividing out the reorderings is the whole difference.\n- Independent stages **multiply**; mutually exclusive cases **add**.\n- Repeated identical items → divide by each repeat's factorial; round tables → \\((n-1)!\\).\n- For \"at least one,\" count the complement and subtract."
+    },
+    {
+      "id": "quant-number-properties-odds-evens-signs",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "odds-evens-signs",
+      "title": "Odds, Evens & Signs",
+      "tags": [
+        "parity",
+        "odd-even",
+        "signs",
+        "number-properties"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nParity (whether a number is odd or even) and sign rules are the quiet workhorses of GMAT Focus\nnumber-properties questions. On their own they look trivial, but the exam hides them inside\n\"must be true,\" \"could be true,\" and Data-Insights logic problems where you never learn the actual\nnumbers. Knowing the rules cold lets you answer those without any arithmetic at all.\n\n## Core concepts\n\n**Parity under addition and subtraction.** The result is even exactly when the two numbers share\nthe same parity:\n\n- even ± even = **even**\n- odd ± odd = **even**\n- even ± odd = **odd**\n\nA useful shortcut: adding or subtracting an **even** number never changes parity; adding or\nsubtracting an **odd** number always flips it.\n\n**Parity under multiplication.** A product is even as soon as **one** factor is even:\n\n- even × (anything) = **even**\n- odd × odd = **odd**\n\nSo a product of integers is odd only if *every* factor is odd.\n\n**Sums of several odd numbers.** Adding \\(k\\) odd numbers gives an **odd** total when \\(k\\) is odd,\nand an **even** total when \\(k\\) is even. (Each odd number flips the parity once.) Even numbers in\nthe sum never affect the parity.\n\n**Consecutive integers.** Among any two consecutive integers \\(n\\) and \\(n+1\\), exactly one is\neven, so the product \\(n(n+1)\\) is **always even**. This is why expressions like \\(n^2 + n\\) are\nguaranteed even.\n\n**Sign rules for products.** The sign of a product depends only on **how many negative factors**\nthere are:\n\n- an **even** number of negative factors → the product is **positive**\n- an **odd** number of negative factors → the product is **negative**\n- any factor of \\(0\\) → the product is \\(0\\)\n\n**Sign logic with two variables.** Two facts appear constantly:\n\n- \\(xy > 0\\) means \\(x\\) and \\(y\\) have the **same sign** (both positive or both negative).\n- \\(xy < 0\\) means they have **opposite signs**.\n\nCombine that with a fact about \\(x + y\\): if \\(xy > 0\\) and \\(x + y < 0\\), the shared sign must be\nnegative, so **both are negative**.\n\n## Worked examples\n\n**\"Must be odd.\"** If \\(n\\) is even, is \\(n + 1\\) odd? Adding the odd number \\(1\\) flips even to\nodd, so yes — always. By contrast \\(2n\\), \\(3n\\), \\(n^2\\), and \\(n + 2\\) are all still even.\n\n**Counting signs.** Evaluate \\((-3)(-4)(-2)\\). The magnitude is \\(3 \\cdot 4 \\cdot 2 = 24\\); there\nare three negative factors, an odd count, so the product is negative: \\(-24\\).\n\n**Parity from a product.** If \\(mn\\) is odd, what do you know? A product is odd only when every\nfactor is odd, so both \\(m\\) and \\(n\\) are odd — and therefore \\(m + n\\) is even.\n\n## Common traps\n\n- **Confusing \"adds up\" with \"multiplies.\"** Four copies of an odd number add to an even total\n  (even count of odds), even though each is odd.\n- **Losing track of signs.** With three negative factors the product is negative, not positive —\n  count the negatives, don't just take the magnitude.\n- **Assuming \\(n^2\\) is even.** \\(n^2\\) has the same parity as \\(n\\); it is odd whenever \\(n\\) is\n  odd.\n- **Forgetting zero.** \"\\(xy > 0\\)\" excludes \\(x = 0\\) or \\(y = 0\\); \"same sign\" quietly rules zero\n  out.\n\n## Key takeaways\n\n- Add/subtract: same parity → even, different parity → odd. Adding an even number never changes\n  parity.\n- A product is even if any factor is even; odd only if every factor is odd.\n- Product sign: even number of negatives → positive, odd number → negative, any zero → zero.\n- \\(n(n+1)\\) is always even — consecutive integers guarantee an even factor.\n- \\(xy > 0\\) means same sign; pair it with the sign of \\(x + y\\) to pin down both."
+    },
+    {
+      "id": "quant-number-properties-remainders",
+      "section": "quant",
+      "topic": "number-properties",
+      "subtopic": "remainders",
+      "title": "Remainders",
+      "tags": [
+        "remainders",
+        "divisibility",
+        "modular-arithmetic",
+        "number-properties"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nRemainder questions look like they need long division, but almost none of them do. The whole\ntopic rests on one identity and a few rules for how remainders behave when you **add**,\n**multiply**, and take **powers**. Master those and you can answer most remainder questions in\nunder a minute — often just by picking the smallest number that fits and testing it.\n\n## Core concepts\n\n**The division identity.** When a positive integer \\(n\\) is divided by a positive divisor \\(d\\),\nthere is a unique quotient \\(q\\) and remainder \\(r\\) with\n\n\\[n = dq + r, \\qquad 0 \\le r < d.\\]\n\nThe remainder is always **less than the divisor** (a remainder of 7 is impossible when dividing\nby 5). Reading a word problem, \"quotient 7, remainder 4, divisor 6\" becomes\n\\(n = 6\\cdot 7 + 4 = 46\\) — nothing more.\n\n**Remainders add and multiply, then reduce.** If you only care about the remainder mod \\(d\\), you\nmay replace any number by its remainder at every step, as long as you **reduce again at the end**:\n\n\\[\\text{rem}(a+b) = \\text{rem}\\big(\\text{rem}(a) + \\text{rem}(b)\\big), \\qquad\n\\text{rem}(ab) = \\text{rem}\\big(\\text{rem}(a)\\cdot \\text{rem}(b)\\big).\\]\n\nSo if \\(a\\) leaves remainder 5 and \\(b\\) leaves remainder 8 when divided by 12, then \\(ab\\) leaves\nthe remainder of \\(5\\times 8 = 40\\), which is \\(40 - 36 = 4\\). The single most common mistake is\nstopping at 40 (or at \\(5+8=13\\)) and forgetting the final reduction.\n\n**Powers cycle.** Remainders of \\(a^1, a^2, a^3, \\dots\\) repeat in a short cycle. Dividing by 10\nis the same as asking for the **units digit**, and units digits cycle with period at most 4:\n\n\\[7^1 \\to 7,\\quad 7^2 \\to 9,\\quad 7^3 \\to 3,\\quad 7^4 \\to 1,\\quad 7^5 \\to 7,\\ \\dots\\]\n\nTo find \\(7^{20} \\bmod 10\\), note the cycle length is 4 and \\(20\\) is a multiple of 4, so \\(7^{20}\\)\nlands on the **last** entry of the cycle: remainder **1**. Match the exponent to its position with\n\"exponent mod cycle-length,\" and remember that a remainder of 0 means the *end* of the cycle, not\nthe start.\n\n**Pick the smallest number that fits.** \"\\(n\\) leaves remainder 2 when divided by 5\" just means\n\\(n\\) is one of \\(2, 7, 12, 17, \\dots\\). For two conditions at once, list each and find the first\noverlap; the pattern then repeats every \\(\\text{lcm}\\) of the divisors.\n\n## Worked examples\n\n**Translate and compute.** A positive integer divided by 6 gives quotient 7 and remainder 4.\nThe integer is \\(6\\cdot 7 + 4 = 46\\). (Swapping quotient and remainder — \\(6\\cdot 4 + 7 = 31\\) — is\nthe classic trap.)\n\n**Combine two conditions.** Find the smallest integer above 10 that leaves remainder 2 mod 5 and\nremainder 1 mod 3. Numbers \\(\\equiv 2 \\pmod 5\\): \\(2, 7, 12, 17, 22, \\dots\\); of these, the ones\n\\(\\equiv 1 \\pmod 3\\) are \\(7, 22, 37, \\dots\\) (they repeat every \\(\\text{lcm}(5,3)=15\\)). The first\none above 10 is **22**.\n\n**Reason with squares.** If \\(n^2\\) leaves remainder 4 when divided by 7, what can \\(n\\) leave?\nTest each possible remainder \\(r = 0,1,\\dots,6\\): \\(r^2 \\bmod 7\\) gives \\(0,1,4,2,2,4,1\\). The value\n4 appears at \\(r = 2\\) and \\(r = 5\\), so \\(n\\) leaves remainder **2 or 5** — *not* 4. The remainder of\n\\(n\\) is generally **not** the same as the remainder of \\(n^2\\).\n\n## Common traps\n\n- **Forgetting to reduce at the end.** \\(5\\times 8 = 40\\) is not the answer when dividing by 12; the\n  remainder is \\(40 - 36 = 4\\).\n- **Adding remainders when the numbers are multiplied** (or vice versa). Combine remainders the\n  same way you combine the numbers, then reduce.\n- **Swapping quotient and remainder** in \\(n = dq + r\\).\n- **A remainder as large as the divisor.** Dividing by \\(d\\), the remainder must be \\(0\\) to \\(d-1\\).\n- **Assuming \\(n\\) and \\(n^2\\) (or \\(n\\) and \\(2n\\)) share a remainder.** They usually don't — test it.\n\n## Key takeaways\n\n- Everything flows from \\(n = dq + r\\) with \\(0 \\le r < d\\).\n- To combine numbers, combine their remainders **and reduce again**: works for sums and products.\n- Powers of a number cycle; use \"exponent mod cycle-length,\" and treat a result of 0 as the end of\n  the cycle.\n- For \"which could be\" and two-condition problems, list the smallest values that fit and look for\n  the overlap; it repeats every lcm of the divisors."
+    },
     {
       "id": "quant-statistics-descriptive",
       "section": "quant",
@@ -1004,6 +2773,41 @@
       "status": "in-review",
       "body": "## Overview\n\nDescriptive statistics on GMAT Focus Quant is almost entirely about two ideas: where a data set is\n**centered** (mean, median, mode) and how **spread out** it is (range, standard deviation). The\narithmetic is light — the exam tests whether you understand what each measure does, especially how\noutliers and transformations affect them. Master a handful of moves and this topic becomes free points.\n\n## Core concepts\n\n**Mean (arithmetic average).**\n\n\\[\\text{mean} = \\frac{\\text{sum of values}}{\\text{count}}\\]\n\nThe single most useful rearrangement is **sum = mean × count**. Most \"average\" problems are really\nabout the *sum*: find it, adjust it, and divide again.\n\n**Weighted average.** When groups have different sizes, you cannot just average the group averages —\nweight each by its count:\n\n\\[\\text{overall mean} = \\frac{n_1\\bar{x}_1 + n_2\\bar{x}_2}{n_1 + n_2}\\]\n\nThe result always lands *between* the two group averages, pulled toward the larger group.\n\n**Median.** The middle value once the data is sorted.\n\n- **Odd count:** the single middle value.\n- **Even count:** the average of the two middle values.\n\nAlways sort first. The median ignores *how far away* extreme values are — only their position matters.\n\n**Mode.** The value that appears most often. A set can have no mode, one mode, or several.\n\n**Range.** \\(\\text{range} = \\text{max} - \\text{min}\\). A crude measure of spread that depends on only\ntwo numbers.\n\n**Standard deviation (SD).** Measures the *typical distance of values from the mean*. You will almost\nnever compute it by hand on the GMAT; you reason about it:\n\n- SD is **0** only when every value is identical.\n- SD depends on **spread from the mean, not on the size of the numbers**. \\(\\{1,2,3\\}\\) and\n  \\(\\{101,102,103\\}\\) have the *same* SD.\n- **Adding a constant** to every value shifts the set but leaves the spread — and therefore the SD —\n  **unchanged**.\n- **Multiplying** every value by \\(k\\) multiplies the SD by \\(|k|\\).\n\n## Worked examples\n\n**Sum trick.** The average of 6 numbers is 22. A seventh number is added and the average becomes 25.\nThe seventh number? Old sum \\(= 22 \\times 6 = 132\\); new sum \\(= 25 \\times 7 = 175\\); the added number\nis \\(175 - 132 = 43\\). (Notice it is far above 25 — it had to *pull the average up* across all seven\nvalues.)\n\n**Outlier: mean vs. median.** For \\(\\{4, 7, 9, 10, 90\\}\\), the mean is \\(120/5 = 24\\) but the median is\n\\(9\\). The lone large value \\(90\\) drags the mean far above the median while leaving the median\nuntouched. When a distribution has a long tail, mean \\(\\ne\\) median.\n\n**Spread, not magnitude.** Which has the larger SD, \\(\\{2,4,6,8,10\\}\\) or \\(\\{102,104,106,108,110\\}\\)?\nNeither — they are identical shapes shifted by 100, so the SD is the same. Bigger numbers do not mean\nbigger spread.\n\n## Common traps\n\n- **Averaging averages.** With unequal group sizes, the simple average of two means is wrong — you must\n  weight by count.\n- **Forgetting to sort before taking the median.** The median is positional; unsorted data gives a\n  meaningless \"middle.\"\n- **Assuming mean = median.** True only for symmetric data. Any outlier or skew separates them, and the\n  mean moves toward the tail.\n- **Thinking \"big numbers ⇒ big standard deviation.\"** SD measures distance from the mean, not the\n  numbers' magnitude. Adding a constant changes the mean but never the SD.\n\n## Key takeaways\n\n- Turn average problems into **sum** problems: sum = mean × count.\n- Weighted averages weight by group size and always fall between the group means.\n- Sort first, then take the median; the median ignores how extreme the outliers are, but the mean does not.\n- Standard deviation is about spread from the mean: it is 0 only for identical values, unchanged by\n  adding a constant, and independent of how large the numbers themselves are."
     },
+    {
+      "id": "quant-word-problems-interest",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "interest",
+      "title": "Simple and Compound Interest",
+      "tags": [
+        "interest",
+        "simple-interest",
+        "compound-interest",
+        "growth",
+        "percents"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nInterest problems ask what a sum of money becomes after it earns a percentage return over time.\nThere are only two engines you need: **simple interest**, where the interest is a fixed slice of\nthe original principal every period, and **compound interest**, where each period's interest is\nitself added to the balance and earns interest afterward. Almost every GMAT interest question is a\nmatter of picking the right engine and being careful about what the question asks for — the\n**interest earned** or the **final balance**. Confusing those two is the single most common mistake.\n\n## Core concepts\n\n**Notation.** Let \\(P\\) be the principal (the starting amount), \\(r\\) the interest rate per period\nwritten as a decimal, and \\(t\\) the number of periods.\n\n**Simple interest** pays \\(r\\) of the *original* principal each period. The interest is\n\n\\[I = P \\cdot r \\cdot t\\]\n\nand the final balance (or \"amount\") is \\(A = P + I = P(1 + rt)\\). Because only the principal ever\nearns, simple interest grows in a straight line.\n\n**Compound interest** adds each period's interest to the balance before the next period, so the\nbalance grows geometrically:\n\n\\[A = P(1 + r)^{t}\\]\n\nThe **interest earned** is \\(A - P = P\\big[(1 + r)^{t} - 1\\big]\\). Notice \\(A\\) is the *balance*,\nnot the interest — subtract \\(P\\) whenever the question asks how much was *earned*.\n\n**Compounding more than once a year.** If a stated annual rate \\(R\\) is compounded \\(n\\) times per\nyear for \\(y\\) years, split the rate and multiply the periods:\n\n\\[A = P\\left(1 + \\frac{R}{n}\\right)^{n y}\\]\n\nSo 8% compounded semiannually for one year is two periods at 4%: \\(P(1.04)^2\\), **not** \\(P(1.08)\\).\n\n**Simple vs. compound over exactly 2 years — a shortcut.** For the same \\(P\\) and \\(r\\), compound\ninterest beats simple interest after 2 years by exactly\n\n\\[P r^{2}\\]\n\n(the interest that the first year's interest earns in the second year). It's a fast way to compute\nthe gap without expanding both formulas.\n\n## Worked examples\n\n**Simple interest.** $2,000 at 5% simple annual interest for 3 years earns\n\\(I = 2000 \\times 0.05 \\times 3 = 300\\), for a balance of \\(2000 + 300 = 2300\\).\n\n**Compound interest.** $5,000 at 10% compounded annually for 2 years grows to\n\\(5000(1.1)^2 = 5000 \\times 1.21 = 6050\\). The interest *earned* is \\(6050 - 5000 = 1050\\) — more\nthan the \\(1000\\) simple interest would give, because the first year's $500 itself earns 10%.\n\n**The 2-year gap.** With \\(P = 4000\\) and \\(r = 0.10\\), the difference between compound and simple\ninterest over 2 years is \\(P r^2 = 4000 \\times 0.01 = 40\\). Check: compound earns\n\\(4000(1.21) - 4000 = 840\\); simple earns \\(4000 \\times 0.1 \\times 2 = 800\\); the gap is \\(40\\). ✓\n\n## Common traps\n\n- **Balance vs. interest earned.** \\(P(1+r)^t\\) and \\(P(1+rt)\\) are *balances*. If the question\n  asks for interest, subtract the principal. Reading \"how much interest\" and answering with the\n  total is the classic trap.\n- **Compounding periods.** A rate compounded semiannually or quarterly must be divided by the\n  number of periods per year, and the exponent multiplied by it. Applying the full annual rate each\n  sub-period massively overstates the result.\n- **Simple where compound is meant (and vice versa).** Over one period they agree; after that they\n  diverge, and the difference is exactly what many questions test.\n- **Decimal slips.** 6% is \\(0.06\\), not \\(0.6\\) or \\(0.006\\). A misplaced decimal changes the\n  answer by a factor of ten.\n- **Reversing growth by subtraction.** To find the principal behind a compounded balance you must\n  *divide* by \\((1+r)^t\\); taking a flat percentage off the final amount is not the inverse.\n\n## Key takeaways\n\n- Simple interest: \\(I = Prt\\), balance \\(= P(1 + rt)\\) — linear, only the principal earns.\n- Compound interest: balance \\(= P(1+r)^t\\), interest earned \\(= P[(1+r)^t - 1]\\) — geometric.\n- Sub-annual compounding: rate \\(\\to R/n\\), periods \\(\\to ny\\).\n- The 2-year compound-minus-simple gap is \\(P r^2\\).\n- Always confirm whether the question wants the **interest** or the **final balance**."
+    },
+    {
+      "id": "quant-word-problems-mixtures",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "mixtures",
+      "title": "Mixtures",
+      "tags": [
+        "mixtures",
+        "concentration",
+        "alligation",
+        "weighted-average"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nMixture problems ask you to combine things of different \"strengths\" — salt in water, alcohol in a\nsolution, cheap coffee with expensive coffee — and reason about the result. Almost every one of\nthem yields to a single idea: **track the pure quantity of the thing you care about, not the\npercentages.** Percentages don't add; amounts do.\n\n## Core concepts\n\n**Concentration is a part-over-whole.** For a solution,\n\n\\[\\text{concentration} = \\frac{\\text{amount of substance}}{\\text{total amount}}\\]\n\nSo 40 liters of a 15% salt solution contains \\(0.15 \\times 40 = 6\\) liters of salt. When you mix or\ndilute, the **amount of substance** is what you carry from line to line.\n\n**Mixing is a weighted average.** Combine \\(a\\) units at concentration \\(p\\) with \\(b\\) units at\nconcentration \\(q\\). Add the *amounts*, then divide by the *total*:\n\n\\[\\text{result} = \\frac{p\\,a + q\\,b}{a + b}\\]\n\nThe answer always lands **between** \\(p\\) and \\(q\\), pulled toward whichever quantity is larger.\nA simple average \\(\\tfrac{p+q}{2}\\) is only right when \\(a = b\\).\n\n**Diluting or concentrating.** Adding pure water adds to the total but **not** to the substance.\nAdding the pure substance adds to both. Set up an equation on the quantity that stays fixed:\n\n> Add \\(w\\) liters of water to 10 liters of 40% salt to reach 25%. The salt is fixed at\n> \\(0.40 \\times 10 = 4\\) liters, so \\(\\dfrac{4}{10 + w} = 0.25 \\Rightarrow 10 + w = 16 \\Rightarrow w = 6\\).\n\n**Alligation (a shortcut).** The ratio of the two quantities in a mix is the *inverse* of their\ndistances from the blend value:\n\n\\[\\frac{a}{b} = \\frac{q - m}{m - p}\\]\n\nwhere \\(m\\) is the mixture's concentration. It's the weighted average solved for the ratio — handy\nwhen a question asks \"in what ratio?\"\n\n## Worked examples\n\n**Weighted average.** Mix 20 L of 30% acid with 30 L of 80% acid. Acid \\(= 0.30(20) + 0.80(30) =\n6 + 24 = 30\\) L in \\(50\\) L total, so \\(\\tfrac{30}{50} = 60\\%\\). (Note it's above the simple average\nof 55%, because more of the stronger solution is present.)\n\n**Alligation.** In what ratio should $8/kg and $12/kg coffee be mixed to sell at $9/kg?\n\\(\\dfrac{\\text{amt of \\$8}}{\\text{amt of \\$12}} = \\dfrac{12 - 9}{9 - 8} = \\dfrac{3}{1}\\) — three parts\ncheap to one part expensive.\n\n## Common traps\n\n- **Averaging the percentages.** \\(\\tfrac{30\\% + 80\\%}{2} = 55\\%\\) is wrong unless the volumes are\n  equal. Weight by the amounts.\n- **Adding water to the substance too.** Diluting with water raises the total but leaves the amount\n  of substance unchanged — that fixed amount is your equation.\n- **Answering the wrong part.** If you solve for one component, re-read to check whether the\n  question wanted that one or the other (or the total).\n- **Inverting the alligation ratio.** The ratio of amounts is inverse to the distances — the\n  quantity *farther* from the blend value is the *smaller* share.\n\n## Key takeaways\n\n- Track the **amount** of the substance (part = concentration × total), not the percentages.\n- A mixture is a weighted average; the result sits between the inputs, nearer the larger quantity.\n- Dilution/concentration: write an equation on the quantity that stays fixed.\n- Alligation gives the ratio directly: distances from the blend value, inverted."
+    },
     {
       "id": "quant-word-problems-rates-speed-distance",
       "section": "quant",
@@ -1021,6 +2825,23 @@
       "status": "in-review",
       "body": "## Overview\n\nNearly every rate problem rests on one relationship. Learn it, learn to organize the numbers,\nand learn the one trap that catches most test-takers: average speed.\n\n## Core concepts\n\n**The master equation.**\n\n\\[\\text{distance} = \\text{rate} \\times \\text{time}\\]\n\nRearranged as needed: \\(\\text{time} = \\dfrac{\\text{distance}}{\\text{rate}}\\) and\n\\(\\text{rate} = \\dfrac{\\text{distance}}{\\text{time}}\\).\n\n**Organize with a table.** For multi-part trips, make a row for each leg with columns\nDistance / Rate / Time. Fill in what you know, use \\(d = rt\\) to get the rest.\n\n**Average speed is total-over-total.**\n\n\\[\\text{average speed} = \\frac{\\text{total distance}}{\\text{total time}}\\]\n\nThis is almost never the simple average of the speeds, because time spent at each speed differs.\n\n## Worked examples\n\n**Basic.** A train travels 300 miles at 50 mph. Time \\(= \\tfrac{300}{50} = 6\\) hours.\n\n**Round trip (the classic trap).** You drive 120 miles at 60 mph, then return the same 120 miles\nat 40 mph. Times are \\(2\\) hours and \\(3\\) hours; total distance \\(240\\) miles, total time \\(5\\)\nhours:\n\n\\[\\text{average speed} = \\frac{240}{5} = 48 \\text{ mph}\\]\n\nNot 50. You spend more time at the slower speed, so the average is pulled below the midpoint.\n\n## Common traps\n\n- **Averaging the speeds.** For equal distances at two speeds, the average speed is the *harmonic*\n  mean, always **less** than the arithmetic mean. Compute total distance ÷ total time instead.\n- **Mismatched units.** Convert minutes to hours (or vice versa) before dividing; mixing them is a silent error.\n- **Combined/closing speed.** Two objects moving toward each other close the gap at the *sum* of\n  their speeds; moving apart, the same; one chasing the other, the *difference*.\n\n## Key takeaways\n\n- Everything comes from \\(d = rt\\); rearrange it for what you need.\n- Use a Distance/Rate/Time table for multi-leg trips.\n- Average speed = total distance / total time — never just the average of the speeds."
     },
+    {
+      "id": "quant-word-problems-translation",
+      "section": "quant",
+      "topic": "word-problems",
+      "subtopic": "translation",
+      "title": "Translating Words into Equations",
+      "tags": [
+        "word-problem",
+        "translation",
+        "algebra",
+        "setup"
+      ],
+      "author": "openmat",
+      "reviewers": [],
+      "status": "in-review",
+      "body": "## Overview\n\nMost GMAT Focus word problems are not hard *algebra* — they are hard *reading*. The whole game is\nturning an English sentence into a correct equation. Once the equation is right, the solving is\nusually routine. This lesson is a phrasebook: it maps the words you actually see on the exam onto\nthe symbols they stand for, and it flags the phrasings that are engineered to be misread.\n\n## Core concepts\n\n**Name the unknown first.** Before writing any equation, decide what the variable is and write it\ndown: \"let \\(n\\) = the number,\" \"let \\(a\\) = adult tickets.\" Nearly every wrong answer on a word\nproblem is the value of *some* quantity in the problem — just not the one that was asked. Naming the\nunknown keeps you honest about which number to report at the end.\n\n**A translation dictionary.** These map almost every arithmetic phrase you will meet:\n\n| English | Symbol |\n|---------|--------|\n| is, was, will be, gives, results in | \\(=\\) |\n| sum, more than, increased by, added to, exceeds by | \\(+\\) |\n| difference, less than, decreased by, fewer than | \\(-\\) |\n| of, times, product, twice (\\(\\times 2\\)), triple (\\(\\times 3\\)) | \\(\\times\\) |\n| per, out of, ratio, quotient | \\(\\div\\) |\n| a number, an unknown | a variable, e.g. \\(n\\) |\n\n**Order matters for subtraction and division.** \"5 **less than** \\(x\\)\" is \\(x - 5\\), **not**\n\\(5 - x\\) — the thing after \"less than\" is what you subtract *from*. Likewise \"8 less than 3 times a\nnumber\" is \\(3n - 8\\). Read these right-to-left.\n\n**Grouping: watch what the multiplier attaches to.** \"Twice a number increased by 6\" is\n\\(2n + 6\\). \"Twice the sum of a number and 6\" is \\(2(n + 6)\\). The words *the sum of* signal a\ngroup — a set of parentheses.\n\n**Consecutive integers.** Consecutive integers are \\(n,\\ n+1,\\ n+2,\\dots\\); consecutive **even** or\n**odd** integers are \\(n,\\ n+2,\\ n+4,\\dots\\). Their sum is not \\(3n\\) — the \\(+1,+2\\) corrections\nmatter.\n\n**Two unknowns, one relationship.** When a problem describes one quantity *in terms of* another\n(\"the larger is 8 less than 3 times the smaller\"), write both in terms of a single variable, then\nuse the remaining fact (usually a sum or total) to get one equation in one unknown.\n\n## Worked examples\n\n**Basic.** \"Twice a number, increased by 6, is 20.\" Let \\(n\\) be the number:\n\\(2n + 6 = 20 \\Rightarrow 2n = 14 \\Rightarrow n = 7\\).\n\n**Two people, one variable.** \"Maria is 4 years older than twice her brother's age; their ages sum\nto 34.\" Let \\(b\\) = brother's age, so Maria \\(= 2b + 4\\). Then\n\\(b + (2b + 4) = 34 \\Rightarrow 3b = 30 \\Rightarrow b = 10\\), and Maria \\(= 2(10) + 4 = 24\\). The\nquestion asks for **Maria**, so the answer is 24 — not the 10 you solved for.\n\n**\"Less than\" order.** \"The larger of two numbers is 8 less than 3 times the smaller; the two sum to\n84.\" Smaller \\(= s\\), larger \\(= 3s - 8\\). Then \\(s + (3s - 8) = 84 \\Rightarrow 4s = 92\n\\Rightarrow s = 23\\), and the larger is \\(3(23) - 8 = 61\\).\n\n## Common traps\n\n- **Answering the wrong quantity.** You solved for the smaller number, the brother's age, or the\n  number of adults — but the question asked for the larger, Maria, or the students. Re-read the last\n  line before you bubble.\n- **Reversing subtraction/division.** \"\\(x\\) less than \\(y\\)\" is \\(y - x\\). \"The quotient of \\(x\\)\n  and \\(y\\)\" is \\(x / y\\). Getting the order backward flips the sign or inverts the fraction.\n- **Misplacing the multiplier.** \\(2n + 6\\) vs. \\(2(n + 6)\\) — \"the sum of\" is your cue to group.\n- **Forgetting the consecutive-integer correction.** Three consecutive integers summing to 72 give\n  \\(3n + 3 = 72\\), so \\(n = 23\\) — not \\(72 / 3 = 24\\).\n\n## Key takeaways\n\n- Name the unknown in words first; report the quantity the question actually asks for.\n- Use the phrase-to-symbol dictionary, and read \"less than\" / \"fewer than\" right-to-left.\n- \"The sum of …\" means parentheses — group before you multiply.\n- Express everything in one variable, then spend the last fact (a total or sum) to close the system."
+    },
     {
       "id": "verbal-critical-reasoning-strengthen-weaken",
       "section": "verbal",