Free Solutions and Explanations to GRE Quantitative MCQ (Multiple Answers) to Sample 1

Free Solutions and Explanations to Practice GRE Quantitative MCQ (Multiple Answers) to sample 1.

Solution to Question 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: \[ B = 0, 3, 6, 9 \] The answer to the above question is: A, D, G, J

Solution to Question 2

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

The answer is: B, C

Solution to Question 3

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 \[ m = \dfrac{0 + 1 + 3 + 100}{4} = \dfrac{104}{4} = 26 \] \[ 3 \times \text{mean} = 78,\ \text{which is not greater than } d = 100 \] Answer: A, C

Solution to Question 4

The distance between \( x \) and -1 on the number line is: \[ |x - (-1)| = |x + 1| \] Solving: \[ |x + 1| = 5 \Rightarrow x + 1 = \pm 5 \Rightarrow x = 4 \text{ or } -6 \] Answer: A, C

Solution to Question 5

Reduce \( \dfrac{6}{8} \): \[ \dfrac{6}{8} = \dfrac{3}{4} \] Check which fractions are equivalent to \( \dfrac{3}{4} \):

\( \text{divide numerator and denominator by 10} : \dfrac{30 \div 10}{40\div10} = \dfrac{3}{4} \)

\( \text{divide numerator and denominator by 4} : \dfrac{12\div4}{16\div4} = \dfrac{3}{4} \)

\( \text{divide numerator and denominator by 33} : \dfrac{99\div33}{132\div33} = \dfrac{3}{4} \)

Answer: A, C, D, E

Solution to Question 6

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

\( 1^2 + 2^2 = 5 \neq 9 = 3^2 \quad \text{(Not satisfied)} \)

\(6^2 + 8^2 = 36 + 64 = 100 = 10^2 \quad \text{(Satisfied)} \)

\(1.2^2 + 1.6^2 = 1.44 + 2.56 = 4.0 = 2.0^2 \quad \text{(Satisfied)} \)

\(3^2 + 5^2 = 9 + 25 = 34 = (\sqrt{34})^2 \quad \text{(Satisfied)} \)

\(10^2 + 12^2 = 244 \neq 400 = 20^2 \quad \text{(Not satisfied)} \)

Answer: B, C, D

Solution to Question 7

Triangle ABC has 2 equal sides → Isosceles, so: \[ \angle ABC = \angle ACB \] Let’s solve: \[ \angle ABC + \angle ACB + 30^\circ = 180^\circ \Rightarrow 2\angle ABC + 30^\circ = 180^\circ \Rightarrow \angle ABC = 75^\circ \] Statements: A: True

B: False

C: True

D: Area using formula: \[ \text{Area} = \dfrac{1}{2} ab \sin C = \dfrac{1}{2} \cdot 10 \cdot 10 \cdot \sin(30^\circ) = \dfrac{1}{2} \cdot 100 \cdot \dfrac{1}{2} = 25 \] D: True Answer: A, C, D

Solution to Question 8

If \( x = 0 \): \[ x^2 = 0,\quad |x| = 0 \Rightarrow \text{Statements A and B are false} \] Since \( x^2 \geq 0 \), then \( x^2 + 1 > 0 \Rightarrow \) Statement C is true

If \( x = -1 \): \[ |x + 1| = 0 \Rightarrow \text{Statement D is false} \] Since \( |x| \geq 0 \Rightarrow |x| + 1 > 0 \Rightarrow \) Statement E is true

Answer: C, E

Solution to Question 9

\[ |-x - y| = |-(x + y)| = |x + y| \Rightarrow \text{Statement A is true} \] If \( x = 6, y = 9 \): \[ |x - y| = |6 - 9| = 3,\quad |x + y| = |6 + 9| = 15 \Rightarrow \text{Statement B is false} \] \[ |y - x| = |-(x - y)| = |x - y| \Rightarrow \text{Statement C is true} \] \[ |-x + y| = |-(x - y)| = |x - y| \Rightarrow \text{Statement D is true} \] Answer: A, C, D

Solution to Question 10

Given \( abc = 100 \), none of \( a, b \) or \( c \) is equal to zero ⇒ Statement A is false

If \( a \times d \times e = 0 \) and \( b \times f \times h = 0 \), then:

At least one factor from each must be 0, but we do not know which ⇒ Statement B may not be true

If \( a \neq 0 \) and \( ade = 0 \), then \( de = 0 \) ⇒ Statement C is false

If \( b \neq 0 \) and \( bfh = 0 \), then \( f = 0 \), \( h = 0 \), or both ⇒ Statement D is true

Answer: B, D