GMAT Math Practice Problems – Detailed Solutions (Sample 1)

This page provides clear, step-by-step solutions to the GMAT-style questions presented in Sample Test 1. Each explanation emphasizes logical reasoning and efficient problem-solving techniques.

Solution to Question 1

For any triangle, the length of the third side must be:

\[ |a - b| < c < a + b \]

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

Solution to Question 2

Given:

\[ \sqrt{x} = 3 \]

Square both sides:

\[ x = 9 \]

Raise both sides to the fourth power:

\[ x^4 = 9^4 = 6561 \]

Correct answer: A

Solution to Question 3

Since \(n\) is an odd integer, it can be written as:

\[ n = 2k + 1 \quad (k \in \mathbb{Z}) \]

Then:

\[ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 \]

So \(n^2\) is odd.

Now:

\[ n^2 + 1 = 2(2k^2 + 2k + 1) \]

which is even.

Finally:

\[ 3n^2 - 1 = 3[2(2k^2 + 2k) + 1] - 1 = 6(2k^2 + 2k) + 2 \]

which is also even.

Correct answer: E

Solution to Question 4

Rewrite constants as powers of 2:

\[ 8 = 2^3, \quad 4 = 2^2 \]

Then:

\[ 8 \cdot 2^{100} + 4 \cdot 2^{101} = 2^{103} + 2^{103} = 2 \cdot 2^{103} = 2^{104} \]

Correct answer: D

Solution to Question 5

Add the two equations:

\[ (3x + 5y) + (x + 3y) = 5 + 20 \] \[ 4x + 8y = 25 \]

Divide both sides by 2:

\[ 2x + 4y = \frac{25}{2} \]

Correct answer: B

Solution to Question 6

The condition \(0 < n < 1\) implies:

\[ 0 < n^2 < n \]

Thus:

\[ n^2 - n < 0 \]

Statement (I) is true.

Multiplying again by \(n\):

\[ 0 < n^3 < n^2 < n \]

Statement (II) is also true.

Statement (III) is false since \(n + 1 > 1\).

Correct answer: D

Solution to Question 7

Convert all expressions to decimals:

The greatest value is \(10\).

Correct answer: D

Solution to Question 8

Factor numerator and denominator:

\[ 4x^2 - 4 = 4(x - 1)(x + 1) \] \[ -3x + 3 = -3(x - 1) \]

Simplify:

\[ \frac{4(x - 1)(x + 1)}{-3(x - 1)} = -\frac{4}{3}(x + 1) \]

Correct answer: A

Solution to Question 9

Substitute the solutions into \(x^2 + bx = c\):

\[ 2^2 + 2b = c \Rightarrow 4 + 2b = c \] \[ (-3)^2 - 3b = c \Rightarrow 9 - 3b = c \]

Equate and solve:

\[ 4 + 2b = 9 - 3b \Rightarrow 5b = 5 \Rightarrow b = 1 \]

Then:

\[ c = 4 + 2(1) = 6 \]

Correct answer: D

Solution to Question 10

\[ (\sqrt{12} - \sqrt{3})(-\sqrt{12} + \sqrt{3}) = -(\sqrt{12} - \sqrt{3})^2 \] \[ = - (2\sqrt{3} - \sqrt{3})^2 = -(\sqrt{3})^2 = -3 \]

Correct answer: B

More GMAT & Math Practice Resources