Free GRE Practice Questions with Solutions - Sample 1

For a number to be divisible by 6, it must be divisible by 2 and 3. All the given numbers above are divisible by 2. We need to select values of \( B \) such that the numbers are divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3. Hence,

\[ 3 + B + 3 + 2 + 4 = 12 + B \]

The values of \( B \) that make \( 12 + B \) divisible by 3 are:

\[ B = 0, 3, 6, 9 \]

Answer: A, D, G, J

A prime number is a number that has only two factors: 1 and itself. Let's list the factors:

• Factors of 27: 1, 3, 9, 27
• Factors of 13: 1, 13
• Factors of 43: 1, 43
• Factors of 49: 1, 7, 49
• Factors of 119: 1, 7, 17, 119
• Factors of 1111: 1, 11, 101, 1111

According to the definition, 27, 49, 119, and 1111 are not prime. The prime numbers are: 13 and 43.

Answer: B, C

The average of a set of unequal numbers is always greater than the smallest and less than the largest. So,

\[ a < \text{average of } a,b,c,d < d \]

Hence:

• Statement A: Always true
• Statement B: Always false
• Statement C: Always true
• Statement D: Not always true (Counterexample: Let \( a = 0, b = 1, c = 3, d = 100 \). Then the mean is \( m = \dfrac{0 + 1 + 3 + 100}{4} = 26 \), so \( 3 \times \text{mean} = 78 \), which is not greater than \( d = 100 \).)

Answer: A, C

The distance between \( x \) and -1 on the number line is:

\[ |x - (-1)| = |x + 1| \]

Solving:

\[ |x + 1| = 5 \implies x + 1 = \pm 5 \implies x = 4 \text{ or } -6 \]

Answer: A, C

Reduce \( \dfrac{6}{8} \):

\[ \dfrac{6}{8} = \dfrac{3}{4} \]

Check which fractions are equivalent to \( \dfrac{3}{4} \):

• \( \dfrac{30}{40} \) (divide numerator and denominator by 10) = \( \dfrac{3}{4} \)
• \( \dfrac{12}{16} \) (divide numerator and denominator by 4) = \( \dfrac{3}{4} \)
• \( \dfrac{99}{132} \) (divide numerator and denominator by 33) = \( \dfrac{3}{4} \)

Answer: A, C, D, E

In a right triangle, the sides \( a, b, c \) satisfy \( a^2 + b^2 = c^2 \). Checking each set:

• \( 1^2 + 2^2 = 5 \neq 3^2 \) (Not satisfied)
• \( 6^2 + 8^2 = 36 + 64 = 100 = 10^2 \) (Satisfied)
• \( 1.2^2 + 1.6^2 = 1.44 + 2.56 = 4.0 = 2.0^2 \) (Satisfied)
• \( 3^2 + 5^2 = 9 + 25 = 34 = (\sqrt{34})^2 \) (Satisfied)
• \( 10^2 + 12^2 = 244 \neq 20^2 \) (Not satisfied)

Answer: B, C, D

Triangle ABC has 2 equal sides, making it isosceles, so:

\[ \angle ABC = \angle ACB \]

Solving for interior angles:

\[ \angle ABC + \angle ACB + 30^\circ = 180^\circ \implies 2\angle ABC = 150^\circ \implies \angle ABC = 75^\circ \]

Evaluating statements:

• A: True
• B: False
• C: True
• D: Area = \( \dfrac{1}{2} ab \sin C = \dfrac{1}{2} \cdot 10 \cdot 10 \cdot \sin(30^\circ) = 50 \cdot \dfrac{1}{2} = 25 \) (True)

Answer: A, C, D

• If \( x = 0 \), \( x^2 = 0 \) and \( |x| = 0 \) (Statements A and B are false).
• Since \( x^2 \geq 0 \), then \( x^2 + 1 \geq 1 > 0 \) for all real \( x \) (Statement C is true).
• If \( x = -1 \), \( |x + 1| = 0 \) (Statement D is false).
• Since \( |x| \geq 0 \), then \( |x| + 1 \geq 1 > 0 \) for all real \( x \) (Statement E is true).

Answer: C, E

• \( |-x - y| = |-(x + y)| = |x + y| \) (Statement A is true)
• Counterexample for B: If \( x = 6, y = 9 \), then \( |6 - 9| = 3 \) while \( |6 + 9| = 15 \) (Statement B is false)
• \( |y - x| = |-(x - y)| = |x - y| \) (Statement C is true)
• \( |-x + y| = |-(x - y)| = |x - y| \) (Statement D is true)

Answer: A, C, D

• Given \( abc = 100 \), none of \( a, b, c \) can be zero (Statement A is false).
• Since \( ade = 0 \) and \( a \neq 0 \), it must be that \( de = 0 \), meaning \( d = 0 \) or \( e = 0 \). Similarly, \( bfh = 0 \) and \( b \neq 0 \implies fh = 0 \implies f = 0 \) or \( h = 0 \) (Statement B may be partially true, but D explicitly provides the correct independent condition for the second product).
• If \( a \neq 0 \) and \( ade = 0 \), then \( de = 0 \), so \( d \times e \neq 0 \) is false (Statement C is false).
• Since \( bfh = 0 \) and \( b \neq 0 \), either \( f = 0 \) or \( h = 0 \) (Statement D is true).

Answer: B, D