From 1eaf91da96b4370a2c7ba3156e7ca614c1104ba4 Mon Sep 17 00:00:00 2001 From: Claude Date: Tue, 28 Jul 2026 09:09:58 +0000 Subject: [PATCH] Add exponents & roots bundle (quant/arithmetic) Covers the previously-empty arithmetic/exponents-roots subtopic: - Lesson: laws of exponents, zero/negative/fractional exponents, simplifying radicals, common-base equations, factoring powers. - 7 original questions (0017-0023), easy -> hard: same-base product, power of a power, negative exponent, fractional exponent, radical simplification, common-base equation, and factoring a difference of powers. Each has a worked explanation, two hints, and distractors keyed to specific common errors. All status: in-review; build regenerated with zero warnings. Co-Authored-By: Claude Opus 4.8 Claude-Session: https://claude.ai/code/session_018u7wcXBDwbejPTPmcHj5b1 --- .../quant/quant-arithmetic-exponents-roots.md | 87 +++++++ .../quant-arithmetic-exponents-roots-0017.md | 47 ++++ .../quant-arithmetic-exponents-roots-0018.md | 46 ++++ .../quant-arithmetic-exponents-roots-0019.md | 45 ++++ .../quant-arithmetic-exponents-roots-0020.md | 48 ++++ .../quant-arithmetic-exponents-roots-0021.md | 48 ++++ .../quant-arithmetic-exponents-roots-0022.md | 51 ++++ .../quant-arithmetic-exponents-roots-0023.md | 53 ++++ docs/data/content.json | 235 +++++++++++++++++- 9 files changed, 656 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..d1d947b --- /dev/null +++ b/content/lessons/quant/quant-arithmetic-exponents-roots.md @@ -0,0 +1,87 @@ +--- +id: quant-arithmetic-exponents-roots +section: quant +topic: arithmetic +subtopic: exponents-roots +title: "Exponents & Roots" +tags: [exponents, roots, radicals, powers, laws-of-exponents] +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Overview + +Exponents and roots show up all over GMAT Focus Quant — in number properties, algebra, and word +problems — and almost every mistake comes from misapplying one of a small set of rules. Learn the +laws cold, know that a root is just a fractional exponent, and practice the two power moves the +exam loves: **factoring out a common power** and **rewriting both sides with the same base**. + +## Core concepts + +**The laws of exponents (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: \((a^m)^n = a^{mn}\). +- Power of a product → distribute: \((ab)^n = a^n b^n\). + +The single most common error is mixing these up — adding when you should multiply, or vice versa. +Multiplying like bases adds exponents; raising a power to a power multiplies them. + +**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. \(2^{-3} = \tfrac18\), which is +positive. For a fraction, \(\left(\tfrac{a}{b}\right)^{-n} = \left(\tfrac{b}{a}\right)^{n}\) — flip, +then apply the positive power. + +**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 \(27^{2/3}\) means "cube root, then square": \((\sqrt[3]{27})^2 = 3^2 = 9\). Take the root of the +**base**; the numerator of the fraction is the power. + +**Simplifying radicals.** Pull out perfect-square factors: \(\sqrt{ab} = \sqrt{a}\,\sqrt{b}\). Find +the largest perfect square that divides the number and take its square root out front: + +\[\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}\] + +The number you pull out is the **square root** of the perfect square (5, not 25), and the factor +you leave inside must have no perfect-square factors left. + +## Worked examples + +**Same base, solve for the exponent.** Solve \(4^{x+1} = 8^{x-1}\). Rewrite both sides with base 2: +\(4 = 2^2\) and \(8 = 2^3\), so \(2^{2(x+1)} = 2^{3(x-1)}\). With equal bases the exponents must be +equal: \(2(x+1) = 3(x-1)\Rightarrow 2x + 2 = 3x - 3 \Rightarrow x = 5\). Check: \(4^6 = 8^4 = 4096\). + +**Factor out the common power.** Simplify \(\dfrac{3^{12} - 3^{10}}{3^{10}}\). You cannot subtract +\(3^{12} - 3^{10}\) directly, but you can factor the smaller power out of the numerator: +\(3^{12} - 3^{10} = 3^{10}(3^{2} - 1) = 3^{10}\cdot 8\). Then +\(\dfrac{3^{10}\cdot 8}{3^{10}} = 8\). + +## Common traps + +- **Adding vs. multiplying exponents.** \(2^3 \cdot 2^4 = 2^7\) (add), but \((2^3)^4 = 2^{12}\) + (multiply). Know which rule you're using. +- **Negative exponent ≠ negative number.** \(3^{-2} = \tfrac19\), not \(-9\). It flips to a + reciprocal. +- **Fractional exponent applied to the wrong part.** In \(27^{2/3}\), the 3 is a cube *root* of 27 + and the 2 is a *power* — it is not \(27 \times \tfrac23 = 18\). +- **Leaving the perfect square inside — or pulling the whole thing out.** \(\sqrt{50} = 5\sqrt2\): + you pull out \(\sqrt{25} = 5\), not 25, and 2 has no square factor left to remove. +- **Combining a sum or difference of powers by "canceling."** In \(\tfrac{3^{12}-3^{10}}{3^{10}}\) + you must factor first; you cannot cancel one term of a difference. + +## Key takeaways + +- Same base: multiply → add exponents, divide → subtract, power of a power → multiply. +- \(a^0 = 1\); a negative exponent is a reciprocal; a fractional exponent is a root. +- Simplify radicals by pulling out the square root of the largest perfect-square factor. +- To solve \(a^{f(x)} = b^{g(x)}\), rewrite both sides as the same base and set the exponents equal. +- Sums and differences of powers get simplified by **factoring out the common power**, not canceling. 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..e75b558 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md @@ -0,0 +1,47 @@ +--- +id: quant-arithmetic-exponents-roots-0017 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, laws-of-exponents, same-base] +choices: + A: "729" + B: "243" + C: "6561" + D: "9" + E: "90" +answer: A +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 powers of the **same base**, you **add** the exponents: + +\[3^4 \cdot 3^2 = 3^{4+2} = 3^6\] + +Now evaluate \(3^6\): \(3^4 = 81\) and \(3^2 = 9\), so \(3^6 = 81 \times 9 = 729\). + +**The traps:** + +- **C (6561)** = \(3^8\), from *multiplying* the exponents (\(4 \times 2\)). You multiply exponents + only when raising a power to a power, \((3^4)^2\) — not when multiplying two powers. +- **D (9)** = \(3^2\), from *subtracting* the exponents; that's the rule for division, not + multiplication. +- **E (90)** = \(81 + 9\), from adding the two values instead of combining the powers. +- **B (243)** = \(3^5\), a miscount of \(4 + 2\). + +## Hints + +- Same base, and the powers are multiplied — which exponent rule applies? +- \(a^m \cdot a^n = a^{m+n}\). Add the exponents, then evaluate. 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..130d51f --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md @@ -0,0 +1,46 @@ +--- +id: quant-arithmetic-exponents-roots-0018 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, power-of-a-power] +choices: + A: "128" + B: "24" + C: "4096" + D: "81" + E: "512" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(\left(2^3\right)^4\)? + +## Explanation + +To raise a power to a power, **multiply** the exponents: + +\[\left(2^3\right)^4 = 2^{3 \times 4} = 2^{12}\] + +And \(2^{12} = 4096\) (since \(2^{10} = 1024\), and \(2^{12} = 1024 \times 4\)). + +**The traps:** + +- **A (128)** = \(2^{3+4} = 2^7\), from *adding* the exponents. Adding is for multiplying like + bases (\(2^3 \cdot 2^4\)); a power of a power multiplies them. +- **B (24)** = \(2 \times 3 \times 4\), from multiplying all three numbers together. +- **D (81)** = \(3^4\), from raising the *exponent* to the 4th power and dropping the base. +- **E (512)** = \(2^9\), a miscount of \(3 \times 4\). + +## Hints + +- A power raised to another power — do you add or multiply the exponents? +- \((a^m)^n = a^{mn}\). Multiply, then evaluate the single power. 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..a766511 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0019.md @@ -0,0 +1,45 @@ +--- +id: quant-arithmetic-exponents-roots-0019 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [exponents, negative-exponents, fractions] +choices: + A: "4/9" + B: "-9/4" + C: "3/2" + D: "-4/9" + E: "9/4" +answer: E +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(\left(\dfrac{2}{3}\right)^{-2}\)? + +## Explanation + +A **negative exponent** means take the reciprocal, then apply the positive power. For a fraction, +flip it: + +\[\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^{2} = \frac{3^2}{2^2} = \frac{9}{4}\] + +**The traps:** + +- **A (4/9)** = \(\left(\tfrac23\right)^2\), from squaring without first taking the reciprocal — + ignoring the negative sign on the exponent. +- **C (3/2)** = the reciprocal without squaring — you flipped but forgot the exponent is 2. +- **B (−9/4) and D (−4/9)** treat the negative exponent as making the *value* negative. A negative + exponent flips the base to a reciprocal; it never changes the sign of the result. + +## Hints + +- A negative exponent does not make the answer negative — it means "reciprocal." +- Flip the fraction first, then apply the positive power of 2. 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..5ec8f5a --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0020.md @@ -0,0 +1,48 @@ +--- +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: "3" + B: "9" + C: "18" + D: "729" + E: "6" +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 is a root and a power: the **denominator** is the root, the **numerator** is +the power. So \(27^{2/3}\) means "take the cube root of 27, then square it": + +\[27^{2/3} = \left(\sqrt[3]{27}\right)^{2} = 3^2 = 9\] + +(Doing it in the other order gives the same result: \(\sqrt[3]{27^2} = \sqrt[3]{729} = 9\).) + +**The traps:** + +- **A (3)** = \(\sqrt[3]{27}\) — you took the cube root but forgot to square it. +- **C (18)** = \(27 \times \tfrac23\), from *multiplying* the base by the fraction instead of + treating \(\tfrac23\) as a root-and-power. +- **D (729)** = \(27^2\), from squaring but never taking the cube root. +- **E (6)** = \(3 \times 2\), from taking the cube root (3) and multiplying by the numerator (2) + instead of raising to that power. + +## Hints + +- In \(a^{m/n}\), the denominator \(n\) is a root and the numerator \(m\) is a power. +- Take the cube root of 27 first (it's a perfect cube), then apply the power of 2. 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..68cca37 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0021.md @@ -0,0 +1,48 @@ +--- +id: quant-arithmetic-exponents-roots-0021 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [roots, radicals, simplifying-radicals] +choices: + A: "25√2" + B: "2√5" + C: "5√10" + D: "5√2" + E: "10√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 \(\sqrt{50}\)? + +## Explanation + +To simplify a radical, pull out the largest **perfect-square** factor. The largest perfect square +dividing 50 is 25: + +\[\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}\] + +You take the *square root* of the perfect square (\(\sqrt{25} = 5\)) out front, and 2 has no +perfect-square factor left, so \(5\sqrt2\) is fully simplified. (Check: \((5\sqrt2)^2 = 25 \cdot 2 = 50\).) + +**The traps:** + +- **A (25√2)** pulls the 25 out front without taking its square root — you must move out + \(\sqrt{25} = 5\), not 25. +- **C (5√10)** factors 50 as \(5 \times 10\) and pulls out the 5, but 5 isn't a perfect square, so + nothing legitimately comes out and \(\sqrt{10}\) is not simplified. +- **B (2√5)** = \(\sqrt{20}\), and **E (10√2)** = \(\sqrt{200}\); neither equals \(\sqrt{50}\). + +## Hints + +- Find the largest perfect square that divides 50. +- Pull out its *square root*, not the perfect square itself. 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..e55d31d --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0022.md @@ -0,0 +1,51 @@ +--- +id: quant-arithmetic-exponents-roots-0022 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: hard +tags: [exponents, common-base, equations] +choices: + A: "1" + B: "2" + C: "5" + D: "-5" + E: "3" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(4^{\,x+1} = 8^{\,x-1}\), what is the value of \(x\)? + +## Explanation + +The bases 4 and 8 aren't equal, but both are powers of 2. Rewrite everything with base 2, using +\(4 = 2^2\) and \(8 = 2^3\): + +\[4^{\,x+1} = \left(2^2\right)^{x+1} = 2^{\,2(x+1)}, \qquad 8^{\,x-1} = \left(2^3\right)^{x-1} = 2^{\,3(x-1)}\] + +With the same base, the exponents must be equal: + +\[2(x+1) = 3(x-1) \;\Rightarrow\; 2x + 2 = 3x - 3 \;\Rightarrow\; x = 5\] + +Check: \(4^{6} = 4096\) and \(8^{4} = 4096\). ✓ + +**The traps:** + +- **E (3)** comes from writing \(8 = 2^4\) instead of \(2^3\): then \(2(x+1) = 4(x-1)\) gives + \(x = 3\). Remember \(8 = 2^3\). +- **B (2)** comes from not distributing the multiplier over the constant — writing + \(2x + 1 = 3x - 1\) instead of \(2(x+1) = 3(x-1)\). +- **A (1)** and **D (−5)** are sign slips when moving terms across the equation. + +## Hints + +- Rewrite both sides as powers of the same base (2), then set the exponents equal. +- Be careful to distribute: \((2^2)^{x+1} = 2^{2(x+1)}\), and \(2(x+1)\) means \(2x+2\). 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..a8d5cfa --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0023.md @@ -0,0 +1,53 @@ +--- +id: quant-arithmetic-exponents-roots-0023 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: hard +tags: [exponents, factoring, common-base] +choices: + A: "9" + B: "2" + C: "8·3^10" + D: "80" + E: "8" +answer: E +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(\dfrac{3^{12} - 3^{10}}{3^{10}}\)? + +## Explanation + +You can't subtract \(3^{12} - 3^{10}\) directly (they're not like terms), but you can **factor out +the smaller power** — the common factor \(3^{10}\) — from the numerator: + +\[3^{12} - 3^{10} = 3^{10}\left(3^{2} - 1\right) = 3^{10}(9 - 1) = 3^{10} \cdot 8\] + +Now divide: + +\[\frac{3^{10} \cdot 8}{3^{10}} = 8\] + +(Equivalently, split the fraction: \(\dfrac{3^{12}}{3^{10}} - \dfrac{3^{10}}{3^{10}} = 3^{2} - 1 = 8\).) + +**The traps:** + +- **A (9)** = \(3^2\), from computing \(\tfrac{3^{12}}{3^{10}} = 9\) but forgetting to subtract the + second term (the \(-1\)). +- **B (2)** just subtracts the exponents (\(12 - 10\)) and stops. +- **C (8·3^10)** correctly factors the numerator but forgets to divide by the \(3^{10}\) in the + denominator. +- **D (80)** = \(3^4 - 1\), from miscounting the exponent difference as 4 instead of 2. + +## Hints + +- You cannot combine \(3^{12} - 3^{10}\) into a single power — factor out the common \(3^{10}\) + first. +- After factoring, the \(3^{10}\) in the numerator cancels the denominator. What's left? diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..afb3b47 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 +301,215 @@ "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", + "laws-of-exponents", + "same-base" + ], + "choices": { + "A": "729", + "B": "243", + "C": "6561", + "D": "9", + "E": "90" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(3^4 \\cdot 3^2\\)?", + "explanation": "When you multiply 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\\): \\(3^4 = 81\\) and \\(3^2 = 9\\), so \\(3^6 = 81 \\times 9 = 729\\).\n\n**The traps:**\n\n- **C (6561)** = \\(3^8\\), from *multiplying* the exponents (\\(4 \\times 2\\)). You multiply exponents\n only when raising a power to a power, \\((3^4)^2\\) — not when multiplying two powers.\n- **D (9)** = \\(3^2\\), from *subtracting* the exponents; that's the rule for division, not\n multiplication.\n- **E (90)** = \\(81 + 9\\), from adding the two values instead of combining the powers.\n- **B (243)** = \\(3^5\\), a miscount of \\(4 + 2\\).", + "hints": [ + "Same base, and the powers are multiplied — which exponent rule applies?", + "\\(a^m \\cdot a^n = a^{m+n}\\). Add the exponents, then evaluate." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0018", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "exponents", + "power-of-a-power" + ], + "choices": { + "A": "128", + "B": "24", + "C": "4096", + "D": "81", + "E": "512" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(\\left(2^3\\right)^4\\)?", + "explanation": "To raise a power to a power, **multiply** the exponents:\n\n\\[\\left(2^3\\right)^4 = 2^{3 \\times 4} = 2^{12}\\]\n\nAnd \\(2^{12} = 4096\\) (since \\(2^{10} = 1024\\), and \\(2^{12} = 1024 \\times 4\\)).\n\n**The traps:**\n\n- **A (128)** = \\(2^{3+4} = 2^7\\), from *adding* the exponents. Adding is for multiplying like\n bases (\\(2^3 \\cdot 2^4\\)); a power of a power multiplies them.\n- **B (24)** = \\(2 \\times 3 \\times 4\\), from multiplying all three numbers together.\n- **D (81)** = \\(3^4\\), from raising the *exponent* to the 4th power and dropping the base.\n- **E (512)** = \\(2^9\\), a miscount of \\(3 \\times 4\\).", + "hints": [ + "A power raised to another power — do you add or multiply the exponents?", + "\\((a^m)^n = a^{mn}\\). Multiply, then evaluate the single power." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0019", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "exponents", + "negative-exponents", + "fractions" + ], + "choices": { + "A": "4/9", + "B": "-9/4", + "C": "3/2", + "D": "-4/9", + "E": "9/4" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(\\left(\\dfrac{2}{3}\\right)^{-2}\\)?", + "explanation": "A **negative exponent** means take the reciprocal, then apply the positive power. For a fraction,\nflip it:\n\n\\[\\left(\\frac{2}{3}\\right)^{-2} = \\left(\\frac{3}{2}\\right)^{2} = \\frac{3^2}{2^2} = \\frac{9}{4}\\]\n\n**The traps:**\n\n- **A (4/9)** = \\(\\left(\\tfrac23\\right)^2\\), from squaring without first taking the reciprocal —\n ignoring the negative sign on the exponent.\n- **C (3/2)** = the reciprocal without squaring — you flipped but forgot the exponent is 2.\n- **B (−9/4) and D (−4/9)** treat the negative exponent as making the *value* negative. A negative\n exponent flips the base to a reciprocal; it never changes the sign of the result.", + "hints": [ + "A negative exponent does not make the answer negative — it means \"reciprocal.\"", + "Flip the fraction first, then apply the positive power of 2." + ] + }, + { + "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": "3", + "B": "9", + "C": "18", + "D": "729", + "E": "6" + }, + "answer": "B", + "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: the **denominator** is the root, the **numerator** is\nthe power. So \\(27^{2/3}\\) means \"take the cube root of 27, then square it\":\n\n\\[27^{2/3} = \\left(\\sqrt[3]{27}\\right)^{2} = 3^2 = 9\\]\n\n(Doing it in the other order gives the same result: \\(\\sqrt[3]{27^2} = \\sqrt[3]{729} = 9\\).)\n\n**The traps:**\n\n- **A (3)** = \\(\\sqrt[3]{27}\\) — you took the cube root but forgot to square it.\n- **C (18)** = \\(27 \\times \\tfrac23\\), from *multiplying* the base by the fraction instead of\n treating \\(\\tfrac23\\) as a root-and-power.\n- **D (729)** = \\(27^2\\), from squaring but never taking the cube root.\n- **E (6)** = \\(3 \\times 2\\), from taking the cube root (3) and multiplying by the numerator (2)\n instead of raising to that power.", + "hints": [ + "In \\(a^{m/n}\\), the denominator \\(n\\) is a root and the numerator \\(m\\) is a power.", + "Take the cube root of 27 first (it's a perfect cube), then apply the power of 2." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0021", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "roots", + "radicals", + "simplifying-radicals" + ], + "choices": { + "A": "25√2", + "B": "2√5", + "C": "5√10", + "D": "5√2", + "E": "10√2" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "Which of the following is equal to \\(\\sqrt{50}\\)?", + "explanation": "To simplify a radical, pull out the largest **perfect-square** factor. The largest perfect square\ndividing 50 is 25:\n\n\\[\\sqrt{50} = \\sqrt{25 \\cdot 2} = \\sqrt{25}\\,\\sqrt{2} = 5\\sqrt{2}\\]\n\nYou take the *square root* of the perfect square (\\(\\sqrt{25} = 5\\)) out front, and 2 has no\nperfect-square factor left, so \\(5\\sqrt2\\) is fully simplified. (Check: \\((5\\sqrt2)^2 = 25 \\cdot 2 = 50\\).)\n\n**The traps:**\n\n- **A (25√2)** pulls the 25 out front without taking its square root — you must move out\n \\(\\sqrt{25} = 5\\), not 25.\n- **C (5√10)** factors 50 as \\(5 \\times 10\\) and pulls out the 5, but 5 isn't a perfect square, so\n nothing legitimately comes out and \\(\\sqrt{10}\\) is not simplified.\n- **B (2√5)** = \\(\\sqrt{20}\\), and **E (10√2)** = \\(\\sqrt{200}\\); neither equals \\(\\sqrt{50}\\).", + "hints": [ + "Find the largest perfect square that divides 50.", + "Pull out its *square root*, not the perfect square itself." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0022", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "exponents", + "common-base", + "equations" + ], + "choices": { + "A": "1", + "B": "2", + "C": "5", + "D": "-5", + "E": "3" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(4^{\\,x+1} = 8^{\\,x-1}\\), what is the value of \\(x\\)?", + "explanation": "The bases 4 and 8 aren't equal, but both are powers of 2. Rewrite everything with base 2, using\n\\(4 = 2^2\\) and \\(8 = 2^3\\):\n\n\\[4^{\\,x+1} = \\left(2^2\\right)^{x+1} = 2^{\\,2(x+1)}, \\qquad 8^{\\,x-1} = \\left(2^3\\right)^{x-1} = 2^{\\,3(x-1)}\\]\n\nWith the same base, the exponents must be equal:\n\n\\[2(x+1) = 3(x-1) \\;\\Rightarrow\\; 2x + 2 = 3x - 3 \\;\\Rightarrow\\; x = 5\\]\n\nCheck: \\(4^{6} = 4096\\) and \\(8^{4} = 4096\\). ✓\n\n**The traps:**\n\n- **E (3)** comes from writing \\(8 = 2^4\\) instead of \\(2^3\\): then \\(2(x+1) = 4(x-1)\\) gives\n \\(x = 3\\). Remember \\(8 = 2^3\\).\n- **B (2)** comes from not distributing the multiplier over the constant — writing\n \\(2x + 1 = 3x - 1\\) instead of \\(2(x+1) = 3(x-1)\\).\n- **A (1)** and **D (−5)** are sign slips when moving terms across the equation.", + "hints": [ + "Rewrite both sides as powers of the same base (2), then set the exponents equal.", + "Be careful to distribute: \\((2^2)^{x+1} = 2^{2(x+1)}\\), and \\(2(x+1)\\) means \\(2x+2\\)." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0023", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "exponents", + "factoring", + "common-base" + ], + "choices": { + "A": "9", + "B": "2", + "C": "8·3^10", + "D": "80", + "E": "8" + }, + "answer": "E", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(\\dfrac{3^{12} - 3^{10}}{3^{10}}\\)?", + "explanation": "You can't subtract \\(3^{12} - 3^{10}\\) directly (they're not like terms), but you can **factor out\nthe smaller power** — the common factor \\(3^{10}\\) — from the numerator:\n\n\\[3^{12} - 3^{10} = 3^{10}\\left(3^{2} - 1\\right) = 3^{10}(9 - 1) = 3^{10} \\cdot 8\\]\n\nNow divide:\n\n\\[\\frac{3^{10} \\cdot 8}{3^{10}} = 8\\]\n\n(Equivalently, split the fraction: \\(\\dfrac{3^{12}}{3^{10}} - \\dfrac{3^{10}}{3^{10}} = 3^{2} - 1 = 8\\).)\n\n**The traps:**\n\n- **A (9)** = \\(3^2\\), from computing \\(\\tfrac{3^{12}}{3^{10}} = 9\\) but forgetting to subtract the\n second term (the \\(-1\\)).\n- **B (2)** just subtracts the exponents (\\(12 - 10\\)) and stops.\n- **C (8·3^10)** correctly factors the numerator but forgets to divide by the \\(3^{10}\\) in the\n denominator.\n- **D (80)** = \\(3^4 - 1\\), from miscounting the exponent difference as 4 instead of 2.", + "hints": [ + "You cannot combine \\(3^{12} - 3^{10}\\) into a single power — factor out the common \\(3^{10}\\)", + "After factoring, the \\(3^{10}\\) in the numerator cancels the denominator. What's left?" + ] + }, { "id": "quant-arithmetic-fractions-decimals-0015", "section": "quant", @@ -968,6 +1177,24 @@ "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", + "laws-of-exponents" + ], + "author": "openmat", + "reviewers": [], + "status": "in-review", + "body": "## Overview\n\nExponents and roots show up all over GMAT Focus Quant — in number properties, algebra, and word\nproblems — and almost every mistake comes from misapplying one of a small set of rules. Learn the\nlaws cold, know that a root is just a fractional exponent, and practice the two power moves the\nexam loves: **factoring out a common power** and **rewriting both sides with the same base**.\n\n## Core concepts\n\n**The laws of exponents (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- Power of a product → distribute: \\((ab)^n = a^n b^n\\).\n\nThe single most common error is mixing these up — adding when you should multiply, or vice versa.\nMultiplying like bases adds exponents; raising a power to a power multiplies them.\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^{-3} = \\tfrac18\\), which is\npositive. For a fraction, \\(\\left(\\tfrac{a}{b}\\right)^{-n} = \\left(\\tfrac{b}{a}\\right)^{n}\\) — flip,\nthen apply the positive power.\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}\\) means \"cube root, then square\": \\((\\sqrt[3]{27})^2 = 3^2 = 9\\). Take the root of the\n**base**; the numerator of the fraction is the power.\n\n**Simplifying radicals.** Pull out perfect-square factors: \\(\\sqrt{ab} = \\sqrt{a}\\,\\sqrt{b}\\). Find\nthe largest perfect square that divides the number and take its square root out front:\n\n\\[\\sqrt{50} = \\sqrt{25 \\cdot 2} = \\sqrt{25}\\,\\sqrt{2} = 5\\sqrt{2}\\]\n\nThe number you pull out is the **square root** of the perfect square (5, not 25), and the factor\nyou leave inside must have no perfect-square factors left.\n\n## Worked examples\n\n**Same base, solve for the exponent.** Solve \\(4^{x+1} = 8^{x-1}\\). Rewrite both sides with base 2:\n\\(4 = 2^2\\) and \\(8 = 2^3\\), so \\(2^{2(x+1)} = 2^{3(x-1)}\\). With equal bases the exponents must be\nequal: \\(2(x+1) = 3(x-1)\\Rightarrow 2x + 2 = 3x - 3 \\Rightarrow x = 5\\). Check: \\(4^6 = 8^4 = 4096\\).\n\n**Factor out the common power.** Simplify \\(\\dfrac{3^{12} - 3^{10}}{3^{10}}\\). You cannot subtract\n\\(3^{12} - 3^{10}\\) directly, but you can factor the smaller power out of the numerator:\n\\(3^{12} - 3^{10} = 3^{10}(3^{2} - 1) = 3^{10}\\cdot 8\\). Then\n\\(\\dfrac{3^{10}\\cdot 8}{3^{10}} = 8\\).\n\n## Common traps\n\n- **Adding vs. multiplying exponents.** \\(2^3 \\cdot 2^4 = 2^7\\) (add), but \\((2^3)^4 = 2^{12}\\)\n (multiply). Know which rule you're using.\n- **Negative exponent ≠ negative number.** \\(3^{-2} = \\tfrac19\\), not \\(-9\\). It flips to a\n reciprocal.\n- **Fractional exponent applied to the wrong part.** In \\(27^{2/3}\\), the 3 is a cube *root* of 27\n and the 2 is a *power* — it is not \\(27 \\times \\tfrac23 = 18\\).\n- **Leaving the perfect square inside — or pulling the whole thing out.** \\(\\sqrt{50} = 5\\sqrt2\\):\n you pull out \\(\\sqrt{25} = 5\\), not 25, and 2 has no square factor left to remove.\n- **Combining a sum or difference of powers by \"canceling.\"** In \\(\\tfrac{3^{12}-3^{10}}{3^{10}}\\)\n you must factor first; you cannot cancel one term of a difference.\n\n## Key takeaways\n\n- Same base: multiply → add exponents, divide → subtract, power of a power → multiply.\n- \\(a^0 = 1\\); a negative exponent is a reciprocal; a fractional exponent is a root.\n- Simplify radicals by pulling out the square root of the largest perfect-square factor.\n- To solve \\(a^{f(x)} = b^{g(x)}\\), rewrite both sides as the same base and set the exponents equal.\n- Sums and differences of powers get simplified by **factoring out the common power**, not canceling." + }, { "id": "quant-arithmetic-percents-ratios", "section": "quant",