This page presents a sample set of GMAT-style problem-solving questions. Each problem focuses on key quantitative concepts tested on the GMAT, complete with clear, step-by-step solutions integrated below.
Practice Questions with Solutions
Question 1: Triangle \(ABC\) has sides \(AB = 3\) and \(BC = 5\). Find the possible length of side \(AC\).
Options: (I) \(8\), (II) \(12\), (III) \(0.5\)
A) (I) only | B) (II) only | C) (III) only | D) (I) and (II) only | E) None
Solution & Explanation:
For any triangle, the length of the third side must satisfy:
\[ |a - b| < c < a + b \]Here, the given sides are \(3\) and \(5\):
\[ \text{Sum } = 3 + 5 = 8, \quad \text{Difference } = 5 - 3 = 2 \]Thus, the third side must satisfy \(2 < AC < 8\). None of the proposed values satisfy this condition.
Correct answer: E
Question 2: If \(\sqrt{x} = 3\), what is the value of \(x^4\)?
Options: A) \(6561\) | B) \(9\) | C) \(81\) | D) \(729\) | E) \(3\)
Solution & Explanation:
Given \(\sqrt{x} = 3\), square both sides:
\[ x = 9 \]Raise both sides to the fourth power:
\[ x^4 = 9^4 = 6561 \]Correct answer: A
Question 3: If \(n\) is an odd integer, which expressions are even? (I) \(n^2\), (II) \(n^2 + 1\), (III) \(3n^2 - 1\)
Options: A) (I) only | B) (II) only | C) (III) only | D) (I) and (II) only | E) (II) and (III) only
Solution & Explanation:
Since \(n\) is an odd integer, it can be written as \(n = 2k + 1\) (\(k \in \mathbb{Z}\)). Then:
\[ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 \quad \text{(Odd)} \]For expression (II):
\[ n^2 + 1 = 2(2k^2 + 2k + 1) \quad \text{(Even)} \]For expression (III):
\[ 3n^2 - 1 = 3[2(2k^2 + 2k) + 1] - 1 = 6(2k^2 + 2k) + 2 \quad \text{(Even)} \]Correct answer: E
Question 4: Evaluate: \(8 \times 2^{100} + 4 \times 2^{101}\)
Options: A) \(2^{101}\) | B) \(2^{102}\) | C) \(2^{103}\) | D) \(2^{104}\) | E) \(2^{105}\)
Solution & Explanation:
Rewrite constants as powers of \(2\): \(8 = 2^3\) and \(4 = 2^2\).
\[ 8 \cdot 2^{100} + 4 \cdot 2^{101} = 2^3 \cdot 2^{100} + 2^2 \cdot 2^{101} = 2^{103} + 2^{103} = 2 \cdot 2^{103} = 2^{104} \]Correct answer: D
Question 5: If \(3x + 5y = 5\) and \(x + 3y = 20\), find the value of \(2x + 4y\).
Options: A) \(25\) | B) \(\frac{25}{2}\) | C) \(\frac{25}{4}\) | D) \(15\) | E) \(\frac{15}{2}\)
Solution & Explanation:
Add the two equations together:
\[ (3x + 5y) + (x + 3y) = 5 + 20 \implies 4x + 8y = 25 \]Divide both sides by \(2\):
\[ 2x + 4y = \frac{25}{2} \]Correct answer: B
Question 6: If \(0 < n < 1\), which statements are true? (I) \(n^2 - n < 0\), (II) \(n^3 < n\), (III) \(n + 1 < 1\)
Options: A) (I) only | B) (II) only | C) (III) only | D) (I) and (II) only | E) (II) and (III) only
Solution & Explanation:
The condition \(0 < n < 1\) implies \(0 < n^2 < n\), which means \(n^2 - n < 0\) (Statement I is true).
Multiplying further by \(n\) gives \(0 < n^3 < n^2 < n\), so \(n^3 < n\) (Statement II is true).
Statement III is false because \(n + 1 > 1\).
Correct answer: D
Question 7: Which of the following has the greatest value?
Options: A) \(250\%\) | B) \(2 + \frac{1}{2}\) | C) \(5 \times 0.5\) | D) \(\frac{1}{0.1}\) | E) \(4\)
Solution & Explanation:
- \(250\% = 2.5\)
- \(2 + \frac{1}{2} = 2.5\)
- \(5 \times 0.5 = 2.5\)
- \(\frac{1}{0.1} = 10\)
- \(4 = 4\)
The greatest value is \(10\).
Correct answer: D
Question 8: Simplify: \(\dfrac{4x^2 - 4}{-3x + 3}\)
Options: A) \(-\frac{4}{3}(x + 1)\) | B) \(x + 1\) | C) \(\frac{4}{3}(x + 1)\) | D) \(4(x - 1)\) | E) \(\frac{4}{3}(x - 1)\)
Solution & Explanation:
Factor the numerator and denominator:
\[ 4x^2 - 4 = 4(x - 1)(x + 1) \] \[ -3x + 3 = -3(x - 1) \]Simplify the fraction:
\[ \frac{4(x - 1)(x + 1)}{-3(x - 1)} = -\frac{4}{3}(x + 1) \]Correct answer: A
Question 9: For which values of \(b\) and \(c\) does the equation \(x^2 + bx = c\) have solutions \(x = 2\) and \(x = -3\)?
Options: A) \(b = 2,\; c = -3\) | B) \(b = 1,\; c = 3\) | C) \(b = 1,\; c = -6\) | D) \(b = 1,\; c = 6\) | E) \(b = 2,\; c = 6\)
Solution & Explanation:
Substitute both solutions into the equation:
\[ 2^2 + 2b = c \implies 4 + 2b = c \] \[ (-3)^2 - 3b = c \implies 9 - 3b = c \]Equate and solve for \(b\):
\[ 4 + 2b = 9 - 3b \implies 5b = 5 \implies b = 1 \]Substitute \(b = 1\) back to find \(c\):
\[ c = 4 + 2(1) = 6 \]Correct answer: D
Question 10: Evaluate: \((\sqrt{12} - \sqrt{3})(-\sqrt{12} + \sqrt{3})\)
Options: A) \(9\) | B) \(-3\) | C) \(15\) | D) \(-15\) | E) \(18\)
Solution & Explanation:
\[ (\sqrt{12} - \sqrt{3})(-\sqrt{12} + \sqrt{3}) = -(\sqrt{12} - \sqrt{3})^2 \]Since \(\sqrt{12} = 2\sqrt{3}\):
\[ = -(2\sqrt{3} - \sqrt{3})^2 = -(\sqrt{3})^2 = -3 \]Correct answer: B