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..6e76ffa --- /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] +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 — inside number-property questions, in +compound-growth word problems, and as fast-simplification steps. Almost every trap here comes +from mixing up **adding**, **multiplying**, and **not touching** an exponent. Learn the handful of +rules below cold and most "scary-looking" powers collapse to a small integer. + +## Core concepts + +**The exponent laws (same base).** These are the whole game. For any nonzero base \(a\): + +\[a^{m}\cdot a^{n} = a^{m+n} \qquad \frac{a^{m}}{a^{n}} = a^{m-n} \qquad \left(a^{m}\right)^{n} = a^{mn}\] + +Read them out loud: **multiply → add exponents**, **divide → subtract exponents**, **power of a +power → multiply exponents**. The last one is the single most common slip: \(\left(a^{m}\right)^{n}\) +is \(a^{mn}\), *not* \(a^{m+n}\). + +**Zero and negative exponents.** These are definitions, not something to "compute": + +\[a^{0} = 1 \ (a\neq 0) \qquad a^{-n} = \frac{1}{a^{n}}\] + +A negative exponent means **reciprocal**, never a negative number. So \(2^{-2} = \tfrac{1}{4}\), +not \(-4\). + +**Roots are fractional exponents.** A root is just an exponent with a fraction: + +\[\sqrt[n]{a} = a^{1/n} \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m} = \sqrt[n]{a^{m}}\] + +So \(27^{2/3} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9\). Take the root **first** (numbers stay +small), then apply the power. + +**Simplifying square roots.** Pull out perfect-square factors: + +\[\sqrt{50} = \sqrt{25\cdot 2} = 5\sqrt{2}\] + +You can only **add or subtract** roots that already have the same radical: \(5\sqrt{2}+3\sqrt{2} = +8\sqrt{2}\). But \(\sqrt{a}+\sqrt{b} \neq \sqrt{a+b}\) — you cannot add under the root. +Multiplication and division *do* combine: \(\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}\) and +\(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\tfrac{a}{b}}\). + +**Solving exponential equations.** Rewrite both sides with the **same base**, then set the +exponents equal. If \(9^{x} = 3^{x+4}\), write \(9 = 3^{2}\) so the left side is \(3^{2x}\); then +\(2x = x+4\), giving \(x = 4\). + +## Worked examples + +**Combining laws.** \(\dfrac{5^{6}}{5^{2}\cdot 5} = \dfrac{5^{6}}{5^{3}} = 5^{6-3} = 5^{3} = 125.\) +Note the lone \(5\) is \(5^{1}\) — don't forget to count it. + +**Fractional exponent.** \(27^{2/3}\): cube-root first, \(\sqrt[3]{27} = 3\), then square, +\(3^{2} = 9\). Doing it the other way (\(27^{2} = 729\), then \(\sqrt[3]{729} = 9\)) gives the same +answer but with uglier numbers. + +**Factoring out a power.** \(2^{x+2} + 2^{x} = 2^{x}\left(2^{2}+1\right) = 5\cdot 2^{x}\). If that +equals \(40\), then \(2^{x} = 8\), so \(x = 3\). When you see a **sum** of powers, factor out the +smallest one — you cannot add the exponents of a sum. + +## Common traps + +- **Power of a power vs. product.** \(\left(a^{2}\right)^{3} = a^{6}\), but \(a^{2}\cdot a^{3} = a^{5}\). Multiply vs. add. +- **Negative exponent ≠ negative number.** \(3^{-1} = \tfrac{1}{3}\), a positive fraction. +- **\(a^{0} = 1\), not \(0\).** Any nonzero base to the zero power is 1. +- **Adding under a root.** \(\sqrt{9+16} = \sqrt{25} = 5\), which is *not* \(\sqrt{9}+\sqrt{16} = 3+4 = 7\). +- **Base times exponent.** \(2^{5}\) is \(32\), not \(2\times 5 = 10\). +- **Power towers.** \(27^{2}\) means \(27\) squared \(\big(=3^{6}\big)\); it is not \(3^{(3^{2})} = 3^{9}\). + +## Key takeaways + +- Same base: multiply → add exponents, divide → subtract, power of a power → multiply. +- \(a^{0}=1\); \(a^{-n}=\tfrac{1}{a^{n}}\) (reciprocal, never negative). +- Roots are fractional exponents; take the root first to keep numbers small. +- You can add/subtract roots only when the radical matches; you can never add under the root. +- Solve exponential equations by rewriting to a common base and equating exponents. 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..1e93e3a --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0017.md @@ -0,0 +1,49 @@ +--- +id: quant-arithmetic-exponents-roots-0017 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, exponent-laws] +choices: + A: "15" + B: "25" + C: "125" + D: "625" + E: "3125" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(\dfrac{5^{6}}{5^{2}\cdot 5}\)? + +## Explanation + +Combine the powers in the denominator first, then subtract exponents. + +The denominator is \(5^{2}\cdot 5 = 5^{2}\cdot 5^{1} = 5^{2+1} = 5^{3}\). (The lone \(5\) is +\(5^{1}\) — it's easy to skip.) + +Now divide, subtracting exponents: + +\[\frac{5^{6}}{5^{3}} = 5^{6-3} = 5^{3} = 125\] + +So the answer is **125**. + +**The traps:** +- **E (3125 = \(5^{5}\))** comes from *adding* the lone \(5\) instead of subtracting it: \(5^{6-2+1}\). +- **D (625 = \(5^{4}\))** comes from ignoring the lone \(5\) entirely: \(5^{6-2}\). +- **B (25)** comes from *dividing* the exponents (\(6\div 3\)) instead of subtracting them. +- **A (15)** comes from reading \(5^{3}\) as \(5\times 3\). + +## Hints + +- Rewrite the lone \(5\) as \(5^{1}\) and combine the whole denominator into a single power. +- Dividing powers of the same base means **subtracting** exponents, not dividing them. 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..18eadad --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0018.md @@ -0,0 +1,54 @@ +--- +id: quant-arithmetic-exponents-roots-0018 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: easy +tags: [exponents, zero-exponent, negative-exponent] +choices: + A: "-3" + B: "1/4" + C: "5/4" + D: "3/2" + E: "1" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(3^{0} + 2^{-2}\)? + +## Explanation + +Handle each term with its definition — these are not things to "multiply out." + +Any nonzero base raised to the zero power is \(1\): + +\[3^{0} = 1\] + +A negative exponent means the **reciprocal**, not a negative number: + +\[2^{-2} = \frac{1}{2^{2}} = \frac{1}{4}\] + +Add them: + +\[1 + \frac{1}{4} = \frac{5}{4}\] + +So the answer is **5/4**. + +**The traps:** +- **B (1/4)** treats \(3^{0}\) as \(0\) instead of \(1\). +- **A (−3)** treats the negative exponent as a negative value: \(2^{-2} = -4\), then \(1 + (-4)\). +- **D (3/2)** takes \(2^{-2} = \tfrac{1}{2}\), forgetting to square the base. +- **E (1)** correctly gets \(3^{0} = 1\) but drops the \(2^{-2}\) term as if it were \(0\). + +## Hints + +- \(a^{0} = 1\) for any nonzero \(a\); a zero exponent never gives \(0\). +- A negative exponent flips the base into a fraction — it never makes the value negative. 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..a0954f7 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0019.md @@ -0,0 +1,49 @@ +--- +id: quant-arithmetic-exponents-roots-0019 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: medium +tags: [roots, radicals, simplifying] +choices: + A: "8√2" + B: "15√2" + C: "14√2" + D: "2√17" + E: "34√2" +answer: A +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 square roots once they share the same radical, so simplify each one first by +pulling out perfect-square factors. + +\[\sqrt{50} = \sqrt{25\cdot 2} = 5\sqrt{2} \qquad \sqrt{18} = \sqrt{9\cdot 2} = 3\sqrt{2}\] + +Now both terms are multiples of \(\sqrt{2}\), so add the coefficients: + +\[5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}\] + +So the answer is **8√2**. + +**The traps:** +- **D (2√17)** comes from adding *under* the root: \(\sqrt{50+18} = \sqrt{68} = 2\sqrt{17}\). But \(\sqrt{a}+\sqrt{b}\neq\sqrt{a+b}\). +- **B (15√2)** comes from *multiplying* the coefficients (\(5\times 3\)) instead of adding them. +- **C (14√2)** comes from simplifying \(\sqrt{18}\) wrongly as \(9\sqrt{2}\) (dividing the radicand by 2), giving \(5\sqrt{2}+9\sqrt{2}\). +- **E (34√2)** comes from mis-extracting both roots by dividing the radicand by 2: \(25\sqrt{2}+9\sqrt{2}\). + +## Hints + +- Factor each radicand into a perfect square times what's left, e.g. \(50 = 25\times 2\). +- Roots add like "like terms": only combine them when the part under the radical matches. 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..0dc35b5 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0020.md @@ -0,0 +1,49 @@ +--- +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: "6" + C: "9" + D: "18" + E: "27" +answer: C +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 packages a root and a power together: the denominator is the root, the +numerator is the power. Take the **root first** to keep the numbers small: + +\[27^{2/3} = \left(\sqrt[3]{27}\right)^{2}\] + +The cube root of \(27\) is \(3\), and then: + +\[3^{2} = 9\] + +So the answer is **9**. (You can also do it the other way — \(27^{2} = 729\), then \(\sqrt[3]{729} = 9\) — but the numbers are uglier.) + +**The traps:** +- **A (3)** stops at the cube root and forgets to apply the power of \(2\). +- **B (6)** takes the cube root (\(3\)) but then *multiplies* by \(2\) instead of squaring. +- **D (18)** multiplies the base by the exponent: \(27 \times \tfrac{2}{3}\). +- **E (27)** ignores the exponent altogether and leaves the base. + +## Hints + +- Read \(a^{m/n}\) as "take the \(n\)th root, then raise to the \(m\)th power." +- The denominator of the fraction is the **root**; taking it first keeps the arithmetic easy. 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..ad4f965 --- /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: [exponents, exponent-laws] +choices: + A: "35" + B: "96" + C: "128" + D: "256" + E: "512" +answer: D +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(2^{x} = 32\), what is the value of \(2^{x+3}\)? + +## Explanation + +You don't even need to find \(x\) — split the power using the product rule. + +\[2^{x+3} = 2^{x}\cdot 2^{3}\] + +You're told \(2^{x} = 32\), and \(2^{3} = 8\), so: + +\[2^{x+3} = 32 \times 8 = 256\] + +So the answer is **256**. (As a check: \(32 = 2^{5}\), so \(x = 5\) and \(2^{x+3} = 2^{8} = 256\).) + +**The traps:** +- **A (35)** *adds* the exponent as if it were a plain number: \(32 + 3\). +- **B (96)** multiplies by \(3\) instead of by \(2^{3} = 8\): \(32 \times 3\). +- **C (128)** multiplies by only \(2^{2} = 4\) (an off-by-one on the exponent): \(32 \times 4\). +- **E (512)** multiplies by \(2^{4} = 16\), doubling one time too many. + +## Hints + +- \(2^{x+3}\) is not \(2^{x} + 3\); a sum in the exponent means a **product** of powers. +- Rewrite \(2^{x+3}\) as \(2^{x}\cdot 2^{3}\) and substitute the value you're given. 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..af06fee --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0022.md @@ -0,0 +1,50 @@ +--- +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: "2" + B: "4" + C: "5" + D: "6" + E: "8" +answer: C +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +If \(3^{\,x+1} = 27^{2}\), what is the value of \(x\)? + +## Explanation + +The trick is to write both sides with the **same base** — here, base \(3\). + +Since \(27 = 3^{3}\), the right side is a power of a power, so **multiply** the exponents: + +\[27^{2} = \left(3^{3}\right)^{2} = 3^{3\times 2} = 3^{6}\] + +Now both sides are powers of \(3\), so the exponents must be equal: + +\[x + 1 = 6 \quad\Longrightarrow\quad x = 5\] + +So the answer is **5**. + +**The traps:** +- **D (6)** solves \(x + 1 = 6\) correctly but reports the exponent, forgetting to subtract \(1\). +- **B (4)** *adds* the exponents on the right (\(3^{3+2} = 3^{5}\)) instead of multiplying them, giving \(x+1 = 5\). +- **E (8)** reads \(27^{2}\) as a power tower \(3^{(3^{2})} = 3^{9}\), giving \(x+1 = 9\). +- **A (2)** ignores the square entirely, treating the right side as \(27 = 3^{3}\), giving \(x+1 = 3\). + +## Hints + +- Rewrite \(27\) as a power of \(3\) so both sides share a base, then equate the exponents. +- \(\left(3^{3}\right)^{2}\) means multiply the exponents — it is not \(3^{3+2}\) and not \(3^{3^{2}}\). 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..c249d92 --- /dev/null +++ b/content/questions/quant/quant-arithmetic-exponents-roots-0023.md @@ -0,0 +1,49 @@ +--- +id: quant-arithmetic-exponents-roots-0023 +section: quant +topic: arithmetic +subtopic: exponents-roots +type: problem-solving +difficulty: hard +tags: [roots, radicals, exponent-laws] +choices: + A: "2" + B: "10" + C: "18" + D: "92" + E: "-22" +answer: A +author: openmat +reviewers: [] +status: in-review +original: true +license: CC-BY-SA-4.0 +--- + +## Question + +What is the value of \(\dfrac{\sqrt{200}}{\sqrt{2}} - \sqrt{64}\)? + +## Explanation + +Simplify the quotient of roots before doing anything else. Dividing roots combines under a single +radical: + +\[\frac{\sqrt{200}}{\sqrt{2}} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10\] + +Then \(\sqrt{64} = 8\), so: + +\[10 - 8 = 2\] + +So the answer is **2**. + +**The traps:** +- **B (10)** stops after simplifying the fraction and forgets to subtract \(\sqrt{64}\). +- **C (18)** *adds* instead of subtracting: \(10 + 8\). +- **D (92)** drops the roots and computes \(200 \div 2 = 100\), then \(100 - 8\). +- **E (−22)** takes \(\sqrt{64}\) to be \(32\) (half of \(64\)) instead of \(8\): \(10 - 32\). + +## Hints + +- \(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\tfrac{a}{b}}\) — combine under one radical first. +- \(\sqrt{64}\) is the number whose square is \(64\); it is not \(64\div 2\). diff --git a/docs/data/content.json b/docs/data/content.json index 9ef981c..332c3c8 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 +301,214 @@ "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", + "exponent-laws" + ], + "choices": { + "A": "15", + "B": "25", + "C": "125", + "D": "625", + "E": "3125" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(\\dfrac{5^{6}}{5^{2}\\cdot 5}\\)?", + "explanation": "Combine the powers in the denominator first, then subtract exponents.\n\nThe denominator is \\(5^{2}\\cdot 5 = 5^{2}\\cdot 5^{1} = 5^{2+1} = 5^{3}\\). (The lone \\(5\\) is\n\\(5^{1}\\) — it's easy to skip.)\n\nNow divide, subtracting exponents:\n\n\\[\\frac{5^{6}}{5^{3}} = 5^{6-3} = 5^{3} = 125\\]\n\nSo the answer is **125**.\n\n**The traps:**\n- **E (3125 = \\(5^{5}\\))** comes from *adding* the lone \\(5\\) instead of subtracting it: \\(5^{6-2+1}\\).\n- **D (625 = \\(5^{4}\\))** comes from ignoring the lone \\(5\\) entirely: \\(5^{6-2}\\).\n- **B (25)** comes from *dividing* the exponents (\\(6\\div 3\\)) instead of subtracting them.\n- **A (15)** comes from reading \\(5^{3}\\) as \\(5\\times 3\\).", + "hints": [ + "Rewrite the lone \\(5\\) as \\(5^{1}\\) and combine the whole denominator into a single power.", + "Dividing powers of the same base means **subtracting** exponents, not dividing them." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0018", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "easy", + "tags": [ + "exponents", + "zero-exponent", + "negative-exponent" + ], + "choices": { + "A": "-3", + "B": "1/4", + "C": "5/4", + "D": "3/2", + "E": "1" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(3^{0} + 2^{-2}\\)?", + "explanation": "Handle each term with its definition — these are not things to \"multiply out.\"\n\nAny nonzero base raised to the zero power is \\(1\\):\n\n\\[3^{0} = 1\\]\n\nA negative exponent means the **reciprocal**, not a negative number:\n\n\\[2^{-2} = \\frac{1}{2^{2}} = \\frac{1}{4}\\]\n\nAdd them:\n\n\\[1 + \\frac{1}{4} = \\frac{5}{4}\\]\n\nSo the answer is **5/4**.\n\n**The traps:**\n- **B (1/4)** treats \\(3^{0}\\) as \\(0\\) instead of \\(1\\).\n- **A (−3)** treats the negative exponent as a negative value: \\(2^{-2} = -4\\), then \\(1 + (-4)\\).\n- **D (3/2)** takes \\(2^{-2} = \\tfrac{1}{2}\\), forgetting to square the base.\n- **E (1)** correctly gets \\(3^{0} = 1\\) but drops the \\(2^{-2}\\) term as if it were \\(0\\).", + "hints": [ + "\\(a^{0} = 1\\) for any nonzero \\(a\\); a zero exponent never gives \\(0\\).", + "A negative exponent flips the base into a fraction — it never makes the value negative." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0019", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "roots", + "radicals", + "simplifying" + ], + "choices": { + "A": "8√2", + "B": "15√2", + "C": "14√2", + "D": "2√17", + "E": "34√2" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is \\(\\sqrt{50} + \\sqrt{18}\\) in simplest form?", + "explanation": "You can only add square roots once they share the same radical, so simplify each one first by\npulling out perfect-square factors.\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\nSo the answer is **8√2**.\n\n**The traps:**\n- **D (2√17)** comes from adding *under* the root: \\(\\sqrt{50+18} = \\sqrt{68} = 2\\sqrt{17}\\). But \\(\\sqrt{a}+\\sqrt{b}\\neq\\sqrt{a+b}\\).\n- **B (15√2)** comes from *multiplying* the coefficients (\\(5\\times 3\\)) instead of adding them.\n- **C (14√2)** comes from simplifying \\(\\sqrt{18}\\) wrongly as \\(9\\sqrt{2}\\) (dividing the radicand by 2), giving \\(5\\sqrt{2}+9\\sqrt{2}\\).\n- **E (34√2)** comes from mis-extracting both roots by dividing the radicand by 2: \\(25\\sqrt{2}+9\\sqrt{2}\\).", + "hints": [ + "Factor each radicand into a perfect square times what's left, e.g. \\(50 = 25\\times 2\\).", + "Roots add like \"like terms\": only combine them when the part under the radical matches." + ] + }, + { + "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": "6", + "C": "9", + "D": "18", + "E": "27" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(27^{2/3}\\)?", + "explanation": "A fractional exponent packages a root and a power together: the denominator is the root, the\nnumerator is the power. Take the **root first** to keep the numbers small:\n\n\\[27^{2/3} = \\left(\\sqrt[3]{27}\\right)^{2}\\]\n\nThe cube root of \\(27\\) is \\(3\\), and then:\n\n\\[3^{2} = 9\\]\n\nSo the answer is **9**. (You can also do it the other way — \\(27^{2} = 729\\), then \\(\\sqrt[3]{729} = 9\\) — but the numbers are uglier.)\n\n**The traps:**\n- **A (3)** stops at the cube root and forgets to apply the power of \\(2\\).\n- **B (6)** takes the cube root (\\(3\\)) but then *multiplies* by \\(2\\) instead of squaring.\n- **D (18)** multiplies the base by the exponent: \\(27 \\times \\tfrac{2}{3}\\).\n- **E (27)** ignores the exponent altogether and leaves the base.", + "hints": [ + "Read \\(a^{m/n}\\) as \"take the \\(n\\)th root, then raise to the \\(m\\)th power.\"", + "The denominator of the fraction is the **root**; taking it first keeps the arithmetic easy." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0021", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "medium", + "tags": [ + "exponents", + "exponent-laws" + ], + "choices": { + "A": "35", + "B": "96", + "C": "128", + "D": "256", + "E": "512" + }, + "answer": "D", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(2^{x} = 32\\), what is the value of \\(2^{x+3}\\)?", + "explanation": "You don't even need to find \\(x\\) — split the power using the product rule.\n\n\\[2^{x+3} = 2^{x}\\cdot 2^{3}\\]\n\nYou're told \\(2^{x} = 32\\), and \\(2^{3} = 8\\), so:\n\n\\[2^{x+3} = 32 \\times 8 = 256\\]\n\nSo the answer is **256**. (As a check: \\(32 = 2^{5}\\), so \\(x = 5\\) and \\(2^{x+3} = 2^{8} = 256\\).)\n\n**The traps:**\n- **A (35)** *adds* the exponent as if it were a plain number: \\(32 + 3\\).\n- **B (96)** multiplies by \\(3\\) instead of by \\(2^{3} = 8\\): \\(32 \\times 3\\).\n- **C (128)** multiplies by only \\(2^{2} = 4\\) (an off-by-one on the exponent): \\(32 \\times 4\\).\n- **E (512)** multiplies by \\(2^{4} = 16\\), doubling one time too many.", + "hints": [ + "\\(2^{x+3}\\) is not \\(2^{x} + 3\\); a sum in the exponent means a **product** of powers.", + "Rewrite \\(2^{x+3}\\) as \\(2^{x}\\cdot 2^{3}\\) and substitute the value you're given." + ] + }, + { + "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": "2", + "B": "4", + "C": "5", + "D": "6", + "E": "8" + }, + "answer": "C", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "If \\(3^{\\,x+1} = 27^{2}\\), what is the value of \\(x\\)?", + "explanation": "The trick is to write both sides with the **same base** — here, base \\(3\\).\n\nSince \\(27 = 3^{3}\\), the right side is a power of a power, so **multiply** the exponents:\n\n\\[27^{2} = \\left(3^{3}\\right)^{2} = 3^{3\\times 2} = 3^{6}\\]\n\nNow both sides are powers of \\(3\\), so the exponents must be equal:\n\n\\[x + 1 = 6 \\quad\\Longrightarrow\\quad x = 5\\]\n\nSo the answer is **5**.\n\n**The traps:**\n- **D (6)** solves \\(x + 1 = 6\\) correctly but reports the exponent, forgetting to subtract \\(1\\).\n- **B (4)** *adds* the exponents on the right (\\(3^{3+2} = 3^{5}\\)) instead of multiplying them, giving \\(x+1 = 5\\).\n- **E (8)** reads \\(27^{2}\\) as a power tower \\(3^{(3^{2})} = 3^{9}\\), giving \\(x+1 = 9\\).\n- **A (2)** ignores the square entirely, treating the right side as \\(27 = 3^{3}\\), giving \\(x+1 = 3\\).", + "hints": [ + "Rewrite \\(27\\) as a power of \\(3\\) so both sides share a base, then equate the exponents.", + "\\(\\left(3^{3}\\right)^{2}\\) means multiply the exponents — it is not \\(3^{3+2}\\) and not \\(3^{3^{2}}\\)." + ] + }, + { + "id": "quant-arithmetic-exponents-roots-0023", + "section": "quant", + "topic": "arithmetic", + "subtopic": "exponents-roots", + "type": "problem-solving", + "difficulty": "hard", + "tags": [ + "roots", + "radicals", + "exponent-laws" + ], + "choices": { + "A": "2", + "B": "10", + "C": "18", + "D": "92", + "E": "-22" + }, + "answer": "A", + "author": "openmat", + "reviewers": [], + "status": "in-review", + "prompt": "What is the value of \\(\\dfrac{\\sqrt{200}}{\\sqrt{2}} - \\sqrt{64}\\)?", + "explanation": "Simplify the quotient of roots before doing anything else. Dividing roots combines under a single\nradical:\n\n\\[\\frac{\\sqrt{200}}{\\sqrt{2}} = \\sqrt{\\frac{200}{2}} = \\sqrt{100} = 10\\]\n\nThen \\(\\sqrt{64} = 8\\), so:\n\n\\[10 - 8 = 2\\]\n\nSo the answer is **2**.\n\n**The traps:**\n- **B (10)** stops after simplifying the fraction and forgets to subtract \\(\\sqrt{64}\\).\n- **C (18)** *adds* instead of subtracting: \\(10 + 8\\).\n- **D (92)** drops the roots and computes \\(200 \\div 2 = 100\\), then \\(100 - 8\\).\n- **E (−22)** takes \\(\\sqrt{64}\\) to be \\(32\\) (half of \\(64\\)) instead of \\(8\\): \\(10 - 32\\).", + "hints": [ + "\\(\\dfrac{\\sqrt{a}}{\\sqrt{b}} = \\sqrt{\\tfrac{a}{b}}\\) — combine under one radical first.", + "\\(\\sqrt{64}\\) is the number whose square is \\(64\\); it is not \\(64\\div 2\\)." + ] + }, { "id": "quant-arithmetic-fractions-decimals-0015", "section": "quant", @@ -968,6 +1176,23 @@ "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 all over GMAT Focus Quant — inside number-property questions, in\ncompound-growth word problems, and as fast-simplification steps. Almost every trap here comes\nfrom mixing up **adding**, **multiplying**, and **not touching** an exponent. Learn the handful of\nrules below cold and most \"scary-looking\" powers collapse to a small integer.\n\n## Core concepts\n\n**The exponent laws (same base).** These are the whole game. 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 \\left(a^{m}\\right)^{n} = a^{mn}\\]\n\nRead them out loud: **multiply → add exponents**, **divide → subtract exponents**, **power of a\npower → multiply exponents**. The last one is the single most common slip: \\(\\left(a^{m}\\right)^{n}\\)\nis \\(a^{mn}\\), *not* \\(a^{m+n}\\).\n\n**Zero and negative exponents.** These are definitions, not something to \"compute\":\n\n\\[a^{0} = 1 \\ (a\\neq 0) \\qquad a^{-n} = \\frac{1}{a^{n}}\\]\n\nA negative exponent means **reciprocal**, never a negative number. So \\(2^{-2} = \\tfrac{1}{4}\\),\nnot \\(-4\\).\n\n**Roots are fractional exponents.** A root is just an exponent with a fraction:\n\n\\[\\sqrt[n]{a} = a^{1/n} \\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** (numbers stay\nsmall), then apply the power.\n\n**Simplifying square roots.** Pull out perfect-square factors:\n\n\\[\\sqrt{50} = \\sqrt{25\\cdot 2} = 5\\sqrt{2}\\]\n\nYou can only **add or subtract** roots that already have the same radical: \\(5\\sqrt{2}+3\\sqrt{2} =\n8\\sqrt{2}\\). But \\(\\sqrt{a}+\\sqrt{b} \\neq \\sqrt{a+b}\\) — you cannot add under the root.\nMultiplication and division *do* combine: \\(\\sqrt{a}\\cdot\\sqrt{b} = \\sqrt{ab}\\) and\n\\(\\dfrac{\\sqrt{a}}{\\sqrt{b}} = \\sqrt{\\tfrac{a}{b}}\\).\n\n**Solving exponential equations.** Rewrite both sides with the **same base**, then set the\nexponents equal. If \\(9^{x} = 3^{x+4}\\), write \\(9 = 3^{2}\\) so the left side is \\(3^{2x}\\); then\n\\(2x = x+4\\), giving \\(x = 4\\).\n\n## Worked examples\n\n**Combining laws.** \\(\\dfrac{5^{6}}{5^{2}\\cdot 5} = \\dfrac{5^{6}}{5^{3}} = 5^{6-3} = 5^{3} = 125.\\)\nNote the lone \\(5\\) is \\(5^{1}\\) — don't forget to count it.\n\n**Fractional exponent.** \\(27^{2/3}\\): cube-root first, \\(\\sqrt[3]{27} = 3\\), then square,\n\\(3^{2} = 9\\). Doing it the other way (\\(27^{2} = 729\\), then \\(\\sqrt[3]{729} = 9\\)) gives the same\nanswer but with uglier numbers.\n\n**Factoring out a power.** \\(2^{x+2} + 2^{x} = 2^{x}\\left(2^{2}+1\\right) = 5\\cdot 2^{x}\\). If that\nequals \\(40\\), then \\(2^{x} = 8\\), so \\(x = 3\\). When you see a **sum** of powers, factor out the\nsmallest one — you cannot add the exponents of a sum.\n\n## Common traps\n\n- **Power of a power vs. product.** \\(\\left(a^{2}\\right)^{3} = a^{6}\\), but \\(a^{2}\\cdot a^{3} = a^{5}\\). Multiply vs. add.\n- **Negative exponent ≠ negative number.** \\(3^{-1} = \\tfrac{1}{3}\\), a positive fraction.\n- **\\(a^{0} = 1\\), not \\(0\\).** Any nonzero base to the zero power is 1.\n- **Adding under a root.** \\(\\sqrt{9+16} = \\sqrt{25} = 5\\), which is *not* \\(\\sqrt{9}+\\sqrt{16} = 3+4 = 7\\).\n- **Base times exponent.** \\(2^{5}\\) is \\(32\\), not \\(2\\times 5 = 10\\).\n- **Power towers.** \\(27^{2}\\) means \\(27\\) squared \\(\\big(=3^{6}\\big)\\); it is not \\(3^{(3^{2})} = 3^{9}\\).\n\n## Key takeaways\n\n- Same base: multiply → add exponents, divide → subtract, power of a power → multiply.\n- \\(a^{0}=1\\); \\(a^{-n}=\\tfrac{1}{a^{n}}\\) (reciprocal, never negative).\n- Roots are fractional exponents; take the root first to keep numbers small.\n- You can add/subtract roots only when the radical matches; you can never add under the root.\n- Solve exponential equations by rewriting to a common base and equating exponents." + }, { "id": "quant-arithmetic-percents-ratios", "section": "quant",