From 0a78b54dc5a45140b5b91fdcefab65788fec7127 Mon Sep 17 00:00:00 2001 From: Claude Date: Sat, 1 Aug 2026 09:13:05 +0000 Subject: [PATCH] =?UTF-8?q?Add=20Exponents=20&=20Roots=20bundle:=20lesson?= =?UTF-8?q?=20+=207=20questions=20(0017=E2=80=930023)?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Fills the arithmetic/exponents-roots gap (previously zero coverage): - New lesson quant-arithmetic-exponents-roots (exponent laws, zero/negative and fractional exponents, simplifying/adding radicals, common-base equations). - 7 original problem-solving questions, easy→hard, each with a worked explanation, two hints, and distractors keyed to specific common errors (add-vs-multiply exponents, negative-exponent sign, root-then-power order, adding under the radical, distributing a converted exponent). All status: in-review. node scripts/build.mjs rebuilt docs/data/content.json with zero warnings (39 questions, 9 lessons). Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_01KuhJX8ivBw5ZqBPU4nSzXa --- .../quant/quant-arithmetic-exponents-roots.md | 79 ++++++ .../quant-arithmetic-exponents-roots-0017.md | 44 ++++ .../quant-arithmetic-exponents-roots-0018.md | 41 +++ .../quant-arithmetic-exponents-roots-0019.md | 44 ++++ .../quant-arithmetic-exponents-roots-0020.md | 46 ++++ .../quant-arithmetic-exponents-roots-0021.md | 43 ++++ .../quant-arithmetic-exponents-roots-0022.md | 47 ++++ .../quant-arithmetic-exponents-roots-0023.md | 47 ++++ docs/data/content.json | 235 +++++++++++++++++- 9 files changed, 622 insertions(+), 4 deletions(-) create mode 100644 content/lessons/quant/quant-arithmetic-exponents-roots.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0017.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0018.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0019.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0020.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0021.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0022.md create mode 100644 content/questions/quant/quant-arithmetic-exponents-roots-0023.md diff --git a/content/lessons/quant/quant-arithmetic-exponents-roots.md b/content/lessons/quant/quant-arithmetic-exponents-roots.md new file mode 100644 index 0000000..0617e68 --- /dev/null +++ b/content/lessons/quant/quant-arithmetic-exponents-roots.md @@ -0,0 +1,79 @@ +--- +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 + +Exponents and roots show up throughout GMAT Focus Quant — in pure arithmetic, in algebra, and +hidden inside word problems. Almost every mistake comes from misapplying one of a small set of +rules, so the payoff is learning the rules cold and knowing exactly which ones do **not** exist. + +## Core concepts + +**The exponent rules (same base).** + +- Multiply → **add** exponents: \(a^m \cdot a^n = a^{m+n}\). +- Divide → **subtract** exponents: \(\dfrac{a^m}{a^n} = a^{m-n}\). +- Power of a power → **multiply** exponents: \((a^m)^n = a^{mn}\). + +Notice multiplying powers *adds* the exponents while a power-of-a-power *multiplies* them. Swapping +those two is the single most common exponent error. + +**Zero and negative exponents.** + +\[a^0 = 1 \quad (a \neq 0), \qquad a^{-n} = \frac{1}{a^n}\] + +A negative exponent means **reciprocal**, not a negative number: \(2^{-2} = \tfrac{1}{4}\), not \(-4\). + +**Products and quotients of different bases.** You can only combine bases that are equal. But a +product raised to a power distributes: \((ab)^n = a^n b^n\). That lets you rewrite a mixed base like +\(6^4 = (2\cdot 3)^4 = 2^4 \cdot 3^4\), which is often the key to a cancellation. + +**Roots are fractional exponents.** + +\[\sqrt[n]{a} = a^{1/n}, \qquad a^{m/n} = \left(a^{1/n}\right)^m = \left(\sqrt[n]{a}\right)^m\] + +So \(27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9\). Take the root first (smaller numbers), then the power. + +**Simplifying square roots.** Pull out perfect-square factors: \(\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}\). +Radicals **add only when the radical part matches**: \(5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\), but +\(\sqrt{50} + \sqrt{18}\) must first be simplified to \(5\sqrt{2} + 3\sqrt{2}\) before you can add. + +**Solving equations by matching bases.** If \(a^x = a^y\) then \(x = y\). To solve \(2^{x+3} = 8^{x-1}\), +rewrite everything on one base: \(8 = 2^3\), so \(8^{x-1} = 2^{3(x-1)}\), then set the exponents equal. + +## Worked examples + +**Combine powers.** \(3^4 \cdot 3^2 = 3^{4+2} = 3^6 = 729\). (Not \(3^8\) — you add, not multiply.) + +**Fractional exponent.** \(27^{2/3}\): cube root of 27 is 3, then square it → \(9\). + +**Matching bases.** \(2^{x+3} = 8^{x-1} \Rightarrow 2^{x+3} = 2^{3(x-1)} = 2^{3x-3}\). Set +\(x + 3 = 3x - 3 \Rightarrow 2x = 6 \Rightarrow x = 3\). Check: \(2^6 = 64 = 8^2\). ✓ + +## Common traps + +- **Multiplying the exponents when you multiply powers.** \(3^4 \cdot 3^2 = 3^6\), not \(3^8\). +- **Reading a negative exponent as a negative number.** \(2^{-2} = \tfrac14\), not \(-4\). +- **Thinking \(a^0 = 0\).** Any nonzero base to the zero power is \(1\). +- **Adding under the radical.** \(\sqrt{50} + \sqrt{18} \neq \sqrt{68}\). Simplify each radical first. +- **Multiplying a base by an exponent.** \(27^{2/3}\) is a root-and-power, not \(27 \cdot \tfrac23 = 18\). +- **Forgetting to distribute a converted exponent.** \(8^{x-1} = 2^{3(x-1)} = 2^{3x-3}\), not \(2^{3x-1}\). + +## Key takeaways + +- Multiply powers → add exponents; power of a power → multiply exponents. Don't swap them. +- \(a^0 = 1\) and \(a^{-n} = 1/a^n\) (reciprocal, never negative). +- \(a^{m/n} = (\sqrt[n]{a})^m\): take the root first, then the power. +- Simplify radicals by pulling out perfect squares; add radicals only when the radical part matches. +- To solve exponential equations, rewrite both sides on a common base and set the exponents equal. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0017.md b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md new file mode 100644 index 0000000..d2db641 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md @@ -0,0 +1,44 @@ +--- +id: quant-arithmetic-exponents-roots-0017 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, powers, product-rule] +choices: + A: "6561" + B: "729" + C: "9" + D: "81" + E: "1296" +answer: B +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(3^4 \cdot 3^2\)? + +## Explanation + +When you multiply two powers of the **same base**, you **add** the exponents: + +\[3^4 \cdot 3^2 = 3^{4+2} = 3^6\] + +Now evaluate \(3^6 = 729\). + +**The trap:** choice **A (6561)** is \(3^8\) — the result of *multiplying* the exponents +(\(4 \times 2 = 8\)) instead of adding them. Multiplying exponents is the rule for a power of a +power, \((3^4)^2\), not for a product of powers. Choice **C (9)** is \(3^2\), from subtracting the +exponents; **D (81)** is \(3^4\), from using only one factor; **E (1296)** is \(6^4\), from adding +the bases \((3+3)\). + +## Hints + +- Same base, multiplied together — what do you do with the exponents? +- Add the exponents first, then compute the single power. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0018.md b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md new file mode 100644 index 0000000..5da31dd --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md @@ -0,0 +1,41 @@ +--- +id: quant-arithmetic-exponents-roots-0018 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, powers, power-of-a-power] +choices: + A: "32" + B: "512" + C: "64" + D: "8" + E: "12" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \((2^3)^2\)? + +## Explanation + +A **power raised to a power** means you **multiply** the exponents: + +\[(2^3)^2 = 2^{3 \cdot 2} = 2^6 = 64\] + +**The trap:** choice **A (32)** is \(2^5\), from *adding* the exponents (\(3 + 2\)) — that's the +rule for multiplying powers like \(2^3 \cdot 2^2\), not for a power of a power. Choice **B (512)** +is \(2^9\), from reading the expression as \(2^{(3^2)}\). Choice **D (8)** ignores the outer +exponent entirely (\(2^3\)), and **E (12)** multiplies the digits \(2 \cdot 3 \cdot 2\). + +## Hints + +- The whole quantity \(2^3\) is being squared. +- For \((a^m)^n\), do you add or multiply \(m\) and \(n\)? diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0019.md b/content/questions/quant/quant-arithmetic-exponents-roots-0019.md new file mode 100644 index 0000000..a8c1504 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0019.md @@ -0,0 +1,44 @@ +--- +id: quant-arithmetic-exponents-roots-0019 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [roots, radicals, simplifying-radicals] +choices: + A: '\(8\sqrt{3}\)' + B: '\(6\sqrt{3}\)' + C: '\(2\sqrt{6}\)' + D: '\(6\sqrt{2}\)' + E: '\(36\sqrt{2}\)' +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +Which of the following is equal to \(\sqrt{72}\)? + +## Explanation + +To simplify a square root, factor out the **largest perfect square**. The largest perfect square +dividing 72 is 36: + +\[\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36}\,\sqrt{2} = 6\sqrt{2}\] + +You can sanity-check: \(6\sqrt{2} \approx 6 \times 1.414 = 8.49\), and \(\sqrt{72} \approx 8.49\). ✓ + +**The trap:** choice **E (\(36\sqrt{2}\))** pulls the whole factor 36 *outside* the radical instead +of its square root, 6. Choices **A** and **B** keep the wrong number under the radical (a mis-factor +of 72), and **C (\(2\sqrt{6}\))** equals \(\sqrt{24}\), not \(\sqrt{72}\). Only the perfect-square +factor comes out, and it comes out as its root. + +## Hints + +- Find the biggest perfect square that divides 72. +- \(\sqrt{ab} = \sqrt{a}\,\sqrt{b}\); the perfect square leaves as its square root, not itself. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0020.md b/content/questions/quant/quant-arithmetic-exponents-roots-0020.md new file mode 100644 index 0000000..fac36fc --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0020.md @@ -0,0 +1,46 @@ +--- +id: quant-arithmetic-exponents-roots-0020 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [exponents, zero-exponent, negative-exponent] +choices: + A: '\(\tfrac{5}{4}\)' + B: '\(\tfrac{1}{4}\)' + C: '\(-3\)' + D: '\(\tfrac{17}{4}\)' + E: '\(\tfrac{3}{4}\)' +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(4^0 + 2^{-2}\)? + +## Explanation + +Handle each term with the exponent rules for zero and negative powers: + +- **Zero exponent:** any nonzero base to the zero power is 1, so \(4^0 = 1\). +- **Negative exponent:** a negative exponent means reciprocal, so \(2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}\). + +Add them: + +\[4^0 + 2^{-2} = 1 + \frac{1}{4} = \frac{5}{4}\] + +**The trap:** choice **C (\(-3\))** treats \(2^{-2}\) as the negative number \(-4\) — but a negative +exponent gives a *reciprocal*, never a negative value. Choice **B (\(\tfrac14\))** comes from +\(4^0 = 0\) (wrong — it's 1). Choice **D (\(\tfrac{17}{4}\))** uses \(4^0 = 4\), and **E +(\(\tfrac34\))** reads \(2^{-2}\) as \(-\tfrac14\). + +## Hints + +- What is any nonzero number raised to the power 0? +- A negative exponent flips the base into a fraction; it does not make the value negative. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0021.md b/content/questions/quant/quant-arithmetic-exponents-roots-0021.md new file mode 100644 index 0000000..258b1b5 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0021.md @@ -0,0 +1,43 @@ +--- +id: quant-arithmetic-exponents-roots-0021 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [exponents, fractional-exponents, roots] +choices: + A: "18" + B: "3" + C: "729" + D: "6" + E: "9" +answer: E +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 is a root and a power combined: \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\). The +denominator is the root, the numerator is the power. Take the **root first** to keep the numbers +small: + +\[27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9\] + +**The trap:** choice **A (18)** multiplies the base by the exponent, \(27 \cdot \tfrac{2}{3} = 18\) +— but an exponent is not multiplication. Choice **B (3)** takes the cube root and forgets to square +it. Choice **C (729)** squares 27 and ignores the cube root (\(27^2\)), and **D (6)** takes the cube +root (3) and then doubles it instead of squaring. + +## Hints + +- The denominator of the exponent tells you which root; the numerator tells you the power. +- Cube-root 27 first, then apply the remaining power. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0022.md b/content/questions/quant/quant-arithmetic-exponents-roots-0022.md new file mode 100644 index 0000000..69229cc --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0022.md @@ -0,0 +1,47 @@ +--- +id: quant-arithmetic-exponents-roots-0022 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: hard +tags: [exponents, equations, common-base] +choices: + A: "3" + B: "2" + C: "6" + D: "0" + E: "-1" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(2^{\,x+3} = 8^{\,x-1}\), what is the value of \(x\)? + +## Explanation + +The two sides have different bases, so rewrite them on a **common 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: + +\[x + 3 = 3x - 3 \;\Rightarrow\; 6 = 2x \;\Rightarrow\; x = 3\] + +Check: \(2^{3+3} = 2^6 = 64\) and \(8^{3-1} = 8^2 = 64\). ✓ + +**The trap:** choice **B (2)** comes from distributing the 3 only to the \(x\) and forgetting the +\(-1\): writing \(8^{x-1} = 2^{3x-1}\) gives \(x + 3 = 3x - 1\), so \(x = 2\). You must apply the +exponent 3 to the *entire* \((x-1)\). Choices **C**, **D**, and **E** come from arithmetic slips in +solving the linear equation. + +## Hints + +- Write 8 as a power of 2 so both sides share a base. +- \((2^3)^{x-1} = 2^{3(x-1)}\) — distribute the 3 across the whole exponent before setting them equal. diff --git a/content/questions/quant/quant-arithmetic-exponents-roots-0023.md b/content/questions/quant/quant-arithmetic-exponents-roots-0023.md new file mode 100644 index 0000000..b7945b5 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0023.md @@ -0,0 +1,47 @@ +--- +id: quant-arithmetic-exponents-roots-0023 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [roots, radicals, adding-radicals] +choices: + A: '\(\sqrt{68}\)' + B: '\(15\sqrt{2}\)' + C: '\(8\sqrt{2}\)' + D: "8" + E: '\(2\sqrt{2}\)' +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 cannot add radicals until each one is simplified to the **same radical part**. Pull out the +perfect-square factors: + +\[\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \qquad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\] + +Both are multiples of \(\sqrt{2}\), so add the coefficients: + +\[5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\] + +**The trap:** choice **A (\(\sqrt{68}\))** adds the numbers *under* the radical (\(50 + 18\)) — but +\(\sqrt{a} + \sqrt{b} \neq \sqrt{a+b}\). Choice **B (\(15\sqrt{2}\))** multiplies the coefficients +(\(5 \times 3\)) instead of adding them; **D (8)** drops the \(\sqrt{2}\) altogether; and **E +(\(2\sqrt{2}\))** subtracts the coefficients. Add coefficients only when the radical parts already +match. + +## Hints + +- Simplify each square root first — what perfect squares divide 50 and 18? +- Once both terms are multiples of \(\sqrt{2}\), add their coefficients (don't touch the \(\sqrt{2}\)). diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..1b12e10 100644 --- a/docs/data/content.json +++ b/docs/data/content.json @@ -7,8 +7,8 @@ "stats": { "quant": { "label": "Quantitative", - "questions": 16, - "lessons": 5 + "questions": 23, + "lessons": 6 }, "verbal": { "label": "Verbal", @@ -22,8 +22,8 @@ } }, "counts": { - "questions": 32, - "lessons": 8 + "questions": 39, + "lessons": 9 }, "questions": [ { @@ -301,6 +301,216 @@ "Or use the coefficient shortcut: sum of roots \\(= -b\\), product \\(= c\\)." ] }, + { + "id": "quant-arithmetic-exponents-roots-0017", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "exponents", + "powers", + "product-rule" + ], + "choices": { + "A": "6561", + "B": "729", + "C": "9", + "D": "81", + "E": "1296" + }, + "answer": "B", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(3^4 \\cdot 3^2\\)?", + "explanation": "When you multiply two powers of the **same base**, you **add** the exponents:\n\n\\[3^4 \\cdot 3^2 = 3^{4+2} = 3^6\\]\n\nNow evaluate \\(3^6 = 729\\).\n\n**The trap:** choice **A (6561)** is \\(3^8\\) — the result of *multiplying* the exponents\n(\\(4 \\times 2 = 8\\)) instead of adding them. Multiplying exponents is the rule for a power of a\npower, \\((3^4)^2\\), not for a product of powers. Choice **C (9)** is \\(3^2\\), from subtracting the\nexponents; **D (81)** is \\(3^4\\), from using only one factor; **E (1296)** is \\(6^4\\), from adding\nthe bases \\((3+3)\\).", + "hints": [ + "Same base, multiplied together — what do you do with the exponents?", + "Add the exponents first, then compute the single power." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0018", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "exponents", + "powers", + "power-of-a-power" + ], + "choices": { + "A": "32", + "B": "512", + "C": "64", + "D": "8", + "E": "12" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\((2^3)^2\\)?", + "explanation": "A **power raised to a power** means you **multiply** the exponents:\n\n\\[(2^3)^2 = 2^{3 \\cdot 2} = 2^6 = 64\\]\n\n**The trap:** choice **A (32)** is \\(2^5\\), from *adding* the exponents (\\(3 + 2\\)) — that's the\nrule for multiplying powers like \\(2^3 \\cdot 2^2\\), not for a power of a power. Choice **B (512)**\nis \\(2^9\\), from reading the expression as \\(2^{(3^2)}\\). Choice **D (8)** ignores the outer\nexponent entirely (\\(2^3\\)), and **E (12)** multiplies the digits \\(2 \\cdot 3 \\cdot 2\\).", + "hints": [ + "The whole quantity \\(2^3\\) is being squared.", + "For \\((a^m)^n\\), do you add or multiply \\(m\\) and \\(n\\)?" + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0019", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "roots", + "radicals", + "simplifying-radicals" + ], + "choices": { + "A": "\\(8\\sqrt{3}\\)", + "B": "\\(6\\sqrt{3}\\)", + "C": "\\(2\\sqrt{6}\\)", + "D": "\\(6\\sqrt{2}\\)", + "E": "\\(36\\sqrt{2}\\)" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "Which of the following is equal to \\(\\sqrt{72}\\)?", + "explanation": "To simplify a square root, factor out the **largest perfect square**. The largest perfect square\ndividing 72 is 36:\n\n\\[\\sqrt{72} = \\sqrt{36 \\cdot 2} = \\sqrt{36}\\,\\sqrt{2} = 6\\sqrt{2}\\]\n\nYou can sanity-check: \\(6\\sqrt{2} \\approx 6 \\times 1.414 = 8.49\\), and \\(\\sqrt{72} \\approx 8.49\\). ✓\n\n**The trap:** choice **E (\\(36\\sqrt{2}\\))** pulls the whole factor 36 *outside* the radical instead\nof its square root, 6. Choices **A** and **B** keep the wrong number under the radical (a mis-factor\nof 72), and **C (\\(2\\sqrt{6}\\))** equals \\(\\sqrt{24}\\), not \\(\\sqrt{72}\\). Only the perfect-square\nfactor comes out, and it comes out as its root.", + "hints": [ + "Find the biggest perfect square that divides 72.", + "\\(\\sqrt{ab} = \\sqrt{a}\\,\\sqrt{b}\\); the perfect square leaves as its square root, not itself." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0020", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "exponents", + "zero-exponent", + "negative-exponent" + ], + "choices": { + "A": "\\(\\tfrac{5}{4}\\)", + "B": "\\(\\tfrac{1}{4}\\)", + "C": "\\(-3\\)", + "D": "\\(\\tfrac{17}{4}\\)", + "E": "\\(\\tfrac{3}{4}\\)" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(4^0 + 2^{-2}\\)?", + "explanation": "Handle each term with the exponent rules for zero and negative powers:\n\n- **Zero exponent:** any nonzero base to the zero power is 1, so \\(4^0 = 1\\).\n- **Negative exponent:** a negative exponent means reciprocal, so \\(2^{-2} = \\dfrac{1}{2^2} = \\dfrac{1}{4}\\).\n\nAdd them:\n\n\\[4^0 + 2^{-2} = 1 + \\frac{1}{4} = \\frac{5}{4}\\]\n\n**The trap:** choice **C (\\(-3\\))** treats \\(2^{-2}\\) as the negative number \\(-4\\) — but a negative\nexponent gives a *reciprocal*, never a negative value. Choice **B (\\(\\tfrac14\\))** comes from\n\\(4^0 = 0\\) (wrong — it's 1). Choice **D (\\(\\tfrac{17}{4}\\))** uses \\(4^0 = 4\\), and **E\n(\\(\\tfrac34\\))** reads \\(2^{-2}\\) as \\(-\\tfrac14\\).", + "hints": [ + "What is any nonzero number raised to the power 0?", + "A negative exponent flips the base into a fraction; it does not make the value negative." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0021", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "exponents", + "fractional-exponents", + "roots" + ], + "choices": { + "A": "18", + "B": "3", + "C": "729", + "D": "6", + "E": "9" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(27^{2/3}\\)?", + "explanation": "A fractional exponent is a root and a power combined: \\(a^{m/n} = \\left(\\sqrt[n]{a}\\right)^m\\). The\ndenominator is the root, the numerator is the power. Take the **root first** to keep the numbers\nsmall:\n\n\\[27^{2/3} = \\left(\\sqrt[3]{27}\\right)^2 = 3^2 = 9\\]\n\n**The trap:** choice **A (18)** multiplies the base by the exponent, \\(27 \\cdot \\tfrac{2}{3} = 18\\)\n— but an exponent is not multiplication. Choice **B (3)** takes the cube root and forgets to square\nit. Choice **C (729)** squares 27 and ignores the cube root (\\(27^2\\)), and **D (6)** takes the cube\nroot (3) and then doubles it instead of squaring.", + "hints": [ + "The denominator of the exponent tells you which root; the numerator tells you the power.", + "Cube-root 27 first, then apply the remaining power." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0022", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "exponents", + "equations", + "common-base" + ], + "choices": { + "A": "3", + "B": "2", + "C": "6", + "D": "0", + "E": "-1" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(2^{\\,x+3} = 8^{\\,x-1}\\), what is the value of \\(x\\)?", + "explanation": "The two sides have different bases, so rewrite them on a **common base**. Since \\(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 + 3 = 3x - 3 \\;\\Rightarrow\\; 6 = 2x \\;\\Rightarrow\\; x = 3\\]\n\nCheck: \\(2^{3+3} = 2^6 = 64\\) and \\(8^{3-1} = 8^2 = 64\\). ✓\n\n**The trap:** choice **B (2)** comes from distributing the 3 only to the \\(x\\) and forgetting the\n\\(-1\\): writing \\(8^{x-1} = 2^{3x-1}\\) gives \\(x + 3 = 3x - 1\\), so \\(x = 2\\). You must apply the\nexponent 3 to the *entire* \\((x-1)\\). Choices **C**, **D**, and **E** come from arithmetic slips in\nsolving the linear equation.", + "hints": [ + "Write 8 as a power of 2 so both sides share a base.", + "\\((2^3)^{x-1} = 2^{3(x-1)}\\) — distribute the 3 across the whole exponent before setting them equal." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0023", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "roots", + "radicals", + "adding-radicals" + ], + "choices": { + "A": "\\(\\sqrt{68}\\)", + "B": "\\(15\\sqrt{2}\\)", + "C": "\\(8\\sqrt{2}\\)", + "D": "8", + "E": "\\(2\\sqrt{2}\\)" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is \\(\\sqrt{50} + \\sqrt{18}\\) in simplest form?", + "explanation": "You cannot add radicals until each one is simplified to the **same radical part**. Pull out the\nperfect-square factors:\n\n\\[\\sqrt{50} = \\sqrt{25 \\cdot 2} = 5\\sqrt{2}, \\qquad \\sqrt{18} = \\sqrt{9 \\cdot 2} = 3\\sqrt{2}\\]\n\nBoth are multiples of \\(\\sqrt{2}\\), so add the coefficients:\n\n\\[5\\sqrt{2} + 3\\sqrt{2} = 8\\sqrt{2}\\]\n\n**The trap:** choice **A (\\(\\sqrt{68}\\))** adds the numbers *under* the radical (\\(50 + 18\\)) — but\n\\(\\sqrt{a} + \\sqrt{b} \\neq \\sqrt{a+b}\\). Choice **B (\\(15\\sqrt{2}\\))** multiplies the coefficients\n(\\(5 \\times 3\\)) instead of adding them; **D (8)** drops the \\(\\sqrt{2}\\) altogether; and **E\n(\\(2\\sqrt{2}\\))** subtracts the coefficients. Add coefficients only when the radical parts already\nmatch.", + "hints": [ + "Simplify each square root first — what perfect squares divide 50 and 18?", + "Once both terms are multiples of \\(\\sqrt{2}\\), add their coefficients (don't touch the \\(\\sqrt{2}\\))." + ] + }, { "id": "quant-arithmetic-fractions-decimals-0015", "section": "quant", @@ -968,6 +1178,23 @@ "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-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\nExponents and roots show up throughout GMAT Focus Quant — in pure arithmetic, in algebra, and\nhidden inside word problems. Almost every mistake comes from misapplying one of a small set of\nrules, so the payoff is learning the rules cold and knowing exactly which ones do **not** exist.\n\n## Core concepts\n\n**The exponent rules (same base).**\n\n- Multiply → **add** exponents: \\(a^m \\cdot a^n = a^{m+n}\\).\n- Divide → **subtract** exponents: \\(\\dfrac{a^m}{a^n} = a^{m-n}\\).\n- Power of a power → **multiply** exponents: \\((a^m)^n = a^{mn}\\).\n\nNotice multiplying powers *adds* the exponents while a power-of-a-power *multiplies* them. Swapping\nthose two is the single most common exponent error.\n\n**Zero and negative exponents.**\n\n\\[a^0 = 1 \\quad (a \\neq 0), \\qquad a^{-n} = \\frac{1}{a^n}\\]\n\nA negative exponent means **reciprocal**, not a negative number: \\(2^{-2} = \\tfrac{1}{4}\\), not \\(-4\\).\n\n**Products and quotients of different bases.** You can only combine bases that are equal. But a\nproduct raised to a power distributes: \\((ab)^n = a^n b^n\\). That lets you rewrite a mixed base like\n\\(6^4 = (2\\cdot 3)^4 = 2^4 \\cdot 3^4\\), which is often the key to a cancellation.\n\n**Roots are fractional exponents.**\n\n\\[\\sqrt[n]{a} = a^{1/n}, \\qquad a^{m/n} = \\left(a^{1/n}\\right)^m = \\left(\\sqrt[n]{a}\\right)^m\\]\n\nSo \\(27^{2/3} = (\\sqrt[3]{27})^2 = 3^2 = 9\\). Take the root first (smaller numbers), then the power.\n\n**Simplifying square roots.** Pull out perfect-square factors: \\(\\sqrt{72} = \\sqrt{36 \\cdot 2} = 6\\sqrt{2}\\).\nRadicals **add only when the radical part matches**: \\(5\\sqrt{2} + 3\\sqrt{2} = 8\\sqrt{2}\\), but\n\\(\\sqrt{50} + \\sqrt{18}\\) must first be simplified to \\(5\\sqrt{2} + 3\\sqrt{2}\\) before you can add.\n\n**Solving equations by matching bases.** If \\(a^x = a^y\\) then \\(x = y\\). To solve \\(2^{x+3} = 8^{x-1}\\),\nrewrite everything on one base: \\(8 = 2^3\\), so \\(8^{x-1} = 2^{3(x-1)}\\), then set the exponents equal.\n\n## Worked examples\n\n**Combine powers.** \\(3^4 \\cdot 3^2 = 3^{4+2} = 3^6 = 729\\). (Not \\(3^8\\) — you add, not multiply.)\n\n**Fractional exponent.** \\(27^{2/3}\\): cube root of 27 is 3, then square it → \\(9\\).\n\n**Matching bases.** \\(2^{x+3} = 8^{x-1} \\Rightarrow 2^{x+3} = 2^{3(x-1)} = 2^{3x-3}\\). Set\n\\(x + 3 = 3x - 3 \\Rightarrow 2x = 6 \\Rightarrow x = 3\\). Check: \\(2^6 = 64 = 8^2\\). ✓\n\n## Common traps\n\n- **Multiplying the exponents when you multiply powers.** \\(3^4 \\cdot 3^2 = 3^6\\), not \\(3^8\\).\n- **Reading a negative exponent as a negative number.** \\(2^{-2} = \\tfrac14\\), not \\(-4\\).\n- **Thinking \\(a^0 = 0\\).** Any nonzero base to the zero power is \\(1\\).\n- **Adding under the radical.** \\(\\sqrt{50} + \\sqrt{18} \\neq \\sqrt{68}\\). Simplify each radical first.\n- **Multiplying a base by an exponent.** \\(27^{2/3}\\) is a root-and-power, not \\(27 \\cdot \\tfrac23 = 18\\).\n- **Forgetting to distribute a converted exponent.** \\(8^{x-1} = 2^{3(x-1)} = 2^{3x-3}\\), not \\(2^{3x-1}\\).\n\n## Key takeaways\n\n- Multiply powers → add exponents; power of a power → multiply exponents. Don't swap them.\n- \\(a^0 = 1\\) and \\(a^{-n} = 1/a^n\\) (reciprocal, never negative).\n- \\(a^{m/n} = (\\sqrt[n]{a})^m\\): take the root first, then the power.\n- Simplify radicals by pulling out perfect squares; add radicals only when the radical part matches.\n- To solve exponential equations, rewrite both sides on a common base and set the exponents equal." + }, { "id": "quant-arithmetic-percents-ratios", "section": "quant",