Free GRE Quantitative Comparison Questions
with Solutions and Explanations Sample 1
Solutions and detailed explanations to questions and problems similar to the quantitative comparison GRE questions in sample 1.
When solving quantitative comparison questions, you asked to compare two quantities – Quantity A and Quantity B – and then determine which of the following statements describes the comparison:
(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.
Solution to Questions 1
\[
\begin{array}{|l|l|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
2(3 + 4)^2 + 10 & 2 + (3 + 4)^2 + 10 \\
\hline
\end{array}
\]
Evaluate expressions in A and B
A: \[ 2(3 + 4)^2 + 10 = 2(7)^2 + 10 = 2(49) + 10 = 108 \]
B: \[ 2 + (3 + 4)^2 + 10 = 2 + 49 + 10 = 61 \]
Quantity A is greater than quantity B
Answer (A)
Solution to Questions 2
\[
\begin{array}{|l|l|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
2 \times 3^{10} + 2 \times 3^{10} + 2 \times 3^{10} & 3 \times 3^{10} \\
\hline
\end{array}
\]
Simplify expressions in A and B
A: \[ \quad 2 \times 3^{10} + 2 \times 3^{10} + 2 \times 3^{10} = 2 \times \left( 3^{10} + 3^{10} + 3^{10} \right)
\]
\[
= 2 \times \left( 3 \times 3^{10} \right) = 2 \times 3^{11}
\]
B:
\[
\quad 3 \times 3^{10} = 3^{11}
\]
Quantity A is greater than quantity B
Answer (A)
Solution to Questions 3
\[
\begin{array}{|l|l|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
25\% \text{ of } 300 & \text{ three quarters of } 200 \\
\hline
\end{array}
\]
Evaluate expressions in A and B.
A: \[ 25\% \text{ of } 300 = \dfrac{25}{100} \times 300 = \dfrac{ 25 \times 300 }{100} = 75 \]
B: \[ \text{ three quarters of } 200 = \dfrac{3}{4} \times 200 = \dfrac{3 \times 200}{4} = 150 \]
Quantity B is greater than quantity A
Answer (B)
Solution to Questions 4
ABC is a triangle such that the measure of angle A is \( 45^{\circ} \). The measure of angle C is twice the measure of angle B.
\[
\begin{array}{|l|l|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
\text{The measure of angle } A & \text{The measure of angle } B \\
\hline
\end{array}
\]
Quantity A is known. Let us calculate quantity B. The measure of angle C is twice the measure of angle B is translated as follows.
\[ C = 2 B \]
The sum of the measures of the three angles A, B and C of the triangles is \( 180^{\circ} \).
\[ A + B + C = 180 \]
Substitute A by 45 and C by 2B in the above equation
\[ 45 + B + 2B = 180 \]
Solve for B
\[ 3 B = 135 \]
\[ B = 135 / 3 = 45^{\circ} \]
The measures of angle A and B are equal
Answer (C)
Solution to Questions 5
\(x \) is a variable that may take any real value.
\[
\begin{array}{|l|l|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
x^2 + 1 & 100x + 1 \\
\hline
\end{array}
\]
Evaluate expressions in A and B for different values of \( x \).
\[ \text{for} \; x = 0 \]
A: \[ x^2 + 1 = 0^2 + 1 = 1 \]
B: \[ 100x + 1 = 100(0) + 1 = 1 \]
\[ \text{for} \; x = 1 \]
A: \[ x^2 + 1 = 1^2 + 1 = 2 \]
B: \[ 100x + 1 = 100(1) + 1 = 101 \]
Conclusion: We have tried two values of \(x \): \( 0 \) and \( 1 \) and they gave different conclusions. Therefore the relationship between the quantities in A and B cannot be determined from the information given.
Answer (D)
Solution to Questions 6
\[
\begin{array}{|l|l|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
\text{Area of a rectangle with perimeter } 240 & \text{Area of a square with perimeter } 240 \\
\hline
\end{array}
\]
Given the perimeter of a rectangle, we cannot calcualte its area and therefore the relationship between the quantities in A and B cannot be determined from the information given.
Answer (D)
Solution to Questions 7
\[ x = - 10^{-2} \]
\[
\begin{array}{|c|c|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
x^3 & x^2 \\
\hline
\end{array}
\]
Evaluate expressions in A and B for the given value of x
A: \[ x^3 = (- 10^{-2})^3 = (-1)^3(10^{-2})^3 \]
\[ = - 10^{-6} \]
B: \[ x^2 = (- 10^{-2})^2 = (-1)^2(10^{-2})^2 \]
\[ = 10^{-4} \]
Quantity B is greater than quantity A
Answer (B)
Solution to Questions 8
\[
\begin{array}{|c|c|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
\text{The volume of a cylindrical tube of radius } 1\, \text{cm and length } 100\, \text{m} & \text{The volume of a cylindrical container of radius } 10\, \text{cm and length } 10\, \text{m} \\
\hline
\end{array}
\]
Before we calculate the volume, we need to convert all given units to one single unit, cm in this case.
A: radius 1 cm , length = \(100 \text{m} = 100 * 100 \text{cm} = 10^4 \) cm
B: radius 10 cm , length = \( 10 \text{m} = 10 \times 100 \text{cm} = 10^3 \) cm
We now calculate the volume using the formula for the volume of a cylindrical tube: \( \pi \times \text{radius}^2 \times \text{height} \).
A: \[ \pi \times 1^2 \times 10^4 = 10^4 \pi \]
B: \[ \pi \times 10^2 \times 10^3 = 10^5 \pi \]
The volume in B is greater than the volume in A.
Answer (B)
Solution to Questions 9
\[ 0 \lt x \lt 1 \]
\[
\begin{array}{|c|c|}
\hline
\textbf{Quantity A} & \textbf{Quantity B} \\
\hline
\dfrac{1}{x^2} & \dfrac{1}{x} \\
\hline
\end{array}
\]
Since x is positive, we can mutliply all terms of the given inequality \( 0 \lt x \lt 1 \) by x without changing the inequality symbols.
\[
0 \cdot x \lt x \cdot x \lt 1 \cdot x
\]
Which simplifies to
\[
0 \lt x^2 \lt x
\]
Again, since \( x \) is positive, the inequality \( x^2 \lt x \) gives an inequality of the reciprocals as follows:
\[
\dfrac{1}{x^2} '\gt \dfrac{1}{x}
\]
Quantity A is greater than quantity B
Answer (A)
Solution to Questions 10
A bag contains green, blue and yellow balls. The ratio of green to blue balls is \( 2:7 \). The ratio of green to yellow balls is \( 3:5 \).
\[
\begin{array}{|c|c|}
\hline
\text{Quantity A} & \text{Quantity B} \\
\hline
\text{Number of blue balls} & \text{Number of yellow balls} \\
\hline
\end{array}
\]
Let \( g, b \) and \( y \) be the numbers of green, blue and yellow balls respectively. Using the given ratios, we can write the following fractions
\[
\dfrac{g}{b} = \dfrac{2}{7} \quad \text{and} \quad \dfrac{g}{y} = \dfrac{3}{5}
\]
The first fraction gives:
\[
\dfrac{b}{g} = \dfrac{7}{2}
\]
We now evaluate the product of fractions \( \dfrac{g}{y} \) and \( \dfrac{b}{g} \) as follows:
\[
\dfrac{g}{y} \cdot \dfrac{b}{g} = \dfrac{3}{5} \cdot \dfrac{7}{2}
\]
\[
{\text{Note that } \dfrac{g}{y} \cdot \dfrac{b}{g} \text{ simplifies to } \dfrac{b}{y}, \text{ hence}}
\]
\[
\dfrac{b}{y} = \dfrac{21}{10}
\]
The above fraction indicates that the number of blue balls (A) is greater than the number of yellow balls (B).
Answer (A)
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