Free GRE Numeric Entry Questions with Explanations
Sample 1
Solutions and detailed explanations to GRE numeric questions and problems in sample 1
Solution to Question 1
Pump A can fill the tank in 4 hours, therefore the quarter of the tank is filled in one hour, hence the rate of pump A in filling the tank is
\[
\dfrac{1}{4}
\]
If \( T \) is the number of hours for pump B to fill the tank, then its rate is
\[
\dfrac{1}{T}
\]
When working together for 3 hours both pumps are working at their rates to fill 1 tank. Hence
\[
3 \cdot \dfrac{1}{4} + 3 \cdot \dfrac{1}{T} = 1
\]
The term \( 3 \cdot \dfrac{1}{4} \) in the above equation is due to pump A working at its rate for 3 hours. The term \( 3 \cdot \dfrac{1}{T} \) is due to pump B and the "1" on the right of the equation corresponds to 1 tank. We now solve the above equation for \( T \):
\[
3 \cdot \dfrac{1}{T} = 1 - \dfrac{3}{4}
\]
\[
3 \cdot \dfrac{1}{T} = \dfrac{1}{4}
\]
\[
\dfrac{1}{T} = \dfrac{1}{12}
\]
\[
T = 12 \text{ hours}
\]
Solution to Question 2
"y = 45 when x = 3" is used to find the constant \( k \) by substituting \( y \) and \( x \) by their values in the equation \( y = \dfrac{k}{x} \).
\[
45 = \dfrac{k}{3}
\]
Solve for \( k \)
\[
k = 3 \cdot 45 = 135
\]
We now use the same equation with known value of \( k \) to find \( x \) when \( y = 180 \) as follows
\[
180 = \dfrac{135}{x}
\]
Solve for \( x \)
\[
x = \dfrac{135}{180} = \dfrac{3}{4}
\]
Solution to Question 3
Let \( L \), \( W \), and \( P \) be the length, width, and perimeter of the rectangle. Hence "a rectangle has a length that is one third of its perimeter" is translated as follows
\[
L = \dfrac{P}{3} = \dfrac{150}{3} = 50
\]
The perimeter \( P \) is given by the formula
\[
P = 2L + 2W
\]
Substitute \( P \) by 150 and \( L \) by 50 and solve for \( W \):
\[
150 = 2 \cdot 50 + 2W
\]
\[
150 = 100 + 2W
\]
\[
2W = 50
\]
\[
W = 25
\]
The area \( A \) of the rectangle is given by
\[
A = LW = 50 \cdot 25 = 1250
\]
Solution to Question 4
Let \( x \) and \( y \) be the two numbers. "The square of the sum of two numbers is 289" is translated as follows
\[
(x + y)^2 = 289
\]
Expand the left side of the above equation
\[
x^2 + y^2 + 2xy = 289
\]
\( xy \) is the product of the two numbers and is given and equal to \( 66 \) . Hence
\[
x^2 + y^2 + 2(66) = 289
\]
Which gives
\[
x^2 + y^2 = 157
\]
Hence the sum of the squares of \( x \) and \( y \) is 157
Solution to Question 5
\( 30\% \) of the money spent on energy is given by
\[ 30\% \times 30 = \dfrac{30}{100} \times 30 = \dfrac{30 \times 30}{100} = 9 \]
Solution to Question 6
Let \( x \) be the smallest of these numbers. \( x + 2 \) and \( x + 4 \) will be the next two odd integers. Hence
\[
x + (x + 2) + (x + 4) = 249
\]
Solve for \( x \) in the above equation:
\[
3x + 2 + 4 = 249
\]
\[
3x = 243
\]
\[
x = 81
\]
The largest of these numbers is \( x + 4 \) and its value is
\[
x + 4 = 81 + 4 = 85
\]
Solution to Question 7
Let \( x \) and \( y \) be the two numbers. Hence
\[
x + y = 3.6 \quad \text{and} \quad x - y = 1.2
\]
Solve the above system of equations by adding the left sides and right sides of the two equations
\[
(x + y) + (x - y) = 3.6 + 1.2
\]
\[
2x = 4.8
\]
\[
x = 2.4
\]
Use equation \( x + y = 3.6 \) to find \( y \)
\[
y = 3.6 - 2.4 = 1.2
\]
The largest of these numbers is 2.4
Solution to Question 8
Let \( x \) be the number. Hence
\[
20\% \cdot x = 125
\]
Solve for \( x \):
\[
\dfrac{20}{100} \cdot x = 125
\]
\[
x = \dfrac{125 \cdot 100}{20} = 625
\]
Solution to Question 9
Let \( x \) be the original price. The price after the first reduction of 10% is given by
\[
x - 10\% \, x = x - \dfrac{10}{100}x = x - 0.1x
\]
The price after the second reduction of 15% is given by
\[
(x - 0.1x) - 15\% (x - 0.1x) = x - 0.1x - \dfrac{15}{100}(x - 0.1x)
\]
\[
= x - 0.1x - 0.15(x - 0.1x)
\]
\[
= x - 0.1x - 0.15x + 0.015x
\]
\[
= 0.765x
\]
The final price is 22 dollars. Hence
\[
0.765x = 22
\]
Solve for \( x \):
\[
x = \dfrac{22}{0.765} = 28.7581
\]
Rounded to the nearest cent, the original price \( x \) is equal to
\[
28.76 \, \text{dollars}
\]
Solution to Question 10
The average of \( \dfrac{1}{2} \), \( \dfrac{1}{4} \), \( \dfrac{2}{3} \), and \( x \) is given by
\[
\dfrac{\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{2}{3} + x}{4}
\]
and is equal to \( \dfrac{3}{4} \). Hence
\[
\dfrac{\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{2}{3} + x}{4} = \dfrac{3}{4}
\]
Solve for \( x \). First, multiply both sides of the equation by 4 and simplify:
\[
\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{2}{3} + x = 3
\]
Now, solve for \( x \):
\[
x = 3 - \left( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{2}{3} \right)
\]
Set all fractions to a common denominator:
\[
x = \dfrac{36}{12} - \left( \dfrac{6}{12} + \dfrac{3}{12} + \dfrac{8}{12} \right) = \dfrac{36}{12} - \dfrac{17}{12}
\]
Finally,
\[
x = \dfrac{19}{12}
\]
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