Free Full Length GRE Quantitative Reasoning Practice Test

The correct answer is Quantity B is greater.
\((x\ +\ y)^2=(x\ +\ y)\ (x\ +\ y)=x^2\ +\ 2\ x\ y\ +\ \ y^2 →\)Since \(y^2>0→x^2\ +\ 2\ x\ y\ +\ y^2> x^2\ +\ 2\ x\ y\)

The correct answer is Quantity B is greater.
Solve for \(p\) and \(q\) in the equation.
\((p,0): y=\frac{2}{3} \ x\ +\ 12→0=\frac{2}{3} \ p\ +\ 12\)
Solve for \(p\) in the equation.
\(0=\frac{2}{3} \ p\ +\ 12→\frac{2}{3}\ p=-\ 12→p=(-\ 12)\ ×\ (\frac{3}{2})=-\ 18\)
\((0,q): y=\frac{2}{3} \ x\ +\ 12→q=\frac{2}{3} \ (0)\ +\ 12→q=12\)
\(q>p\)

The correct answer is Quantity B is greater.
Since, \(x\) is a positive integer greater than \(1\), then the minimum value of \(\sqrt{x}\) is greater than \(1\).

The correct answer is Quantity B is greater.
Use factoring method to solve for x in the equation.
\(x^2\ -\ 2\ x\ -\ 15=0→(x\ -\ 5)\ (x\ +\ 3)=0\)
Then:
\((x\ -\ 5)=0→x=5\)
Or
\((x\ +\ 3)=0→x=-\ 3\)
Both values of \(x\) are less than \(6\). So, quantity B is greater

The correct answer is The relationship cannot be determined from the information given.
Let’s choose some values for \(x\) and \(y\).
\(x=1 ,\ y=0.5→(A=1.5)>(B=0.5)\) and if \(x=1\) and \(y=-\ 0.5 →B>A\)

The correct answer is \(3.5\)
The loan is \($15,000\) and its interest is \($1,050\). Since the interest is for \(2\) years. Therefore, the simple interest rate \((p)\) per year is \($525\). Then:
interest rate\(=\frac{interest\ amount}{loan}\ ×\ 100→p=\frac{525}{15000}\ ×\ 100=3.5\)

The correct answer is \(3 \ x\ + \ 2\)
Since, \(81=3^4\)
Then:
\((3^{6x} )\ (81)=3^{2y}→(3^{6x} )\ (3^4 )=3^{2y}\)
Use exponent “product rule”: x^n×x^m=x^(n+m)
\((3^{6x} )\ (3^4 )=3^{2y}→3^{6x+4}=3^{2y}\)
The bases are the same. Therefore, the powers must be equal.
\(6\ x\ +\ 4=2\ y\)
Divide both sides of the equation by \(2\):
\(6\ x\ +\ 4=2\ y →3\ x\ +\ 2=y\)

The correct answer is \(240\)
The ratio of boy to girls is \(2:3\). Therefore, there are 2 boys out of \(5\) students.
To find the answer, first divide the total number of students by \(5\), then multiply the result by \(2\).
\(600 \ ÷\ 5 = 120 ⇒ 120 \ ×\ 2 = 240\)

The correct answer is \(16\sqrt{3}\) cm\(^2\)
Area of the triangle is: \(\frac{1}{2}\) AD \(×\)BC and AD is perpendicular to BC.
Triangle ADC is a \(30^°-60^°- 90^°\) right triangle.
The relationship among all sides of right triangle \(30^°-60^°- 90^°\) is provided in the following triangle:
In this triangle, the opposite side of \(30^°\) angle is half of the hypotenuse. And the opposite side of \(60^°\) is opposite of \(30^° \ ×\ \sqrt{3}\)
CD \(= 4\), then AD \(= 4 \ ×\ \sqrt{3}\)
Area of the triangle ABC is: \(\frac{1}{2}\) AD\(×\)BC \(= \frac{1}{2 }\ 4\sqrt{3}\ ×\ 8=16\sqrt{3}\)

The correct answer is \(225\)
\(0.6\ x=(0.3)\ ×\ 20→x=10→(x+5)^2=(15)^2=225\)

The correct answer is \(60\) feet
The relationship among all sides of special right triangle
\(30^°\ -\ 60^°\ -\ 90^°\) is provided in this triangle:
In this triangle, the opposite side of \(30^°\) angle is half of the hypotenuse.
Draw the shape of this question:
The latter is the hypotenuse. Therefore, the latter is \(60\) ft.

The correct answer is \(87.5\)
The difference of \(94\) and \(69\) is \(25\).
average (mean) \(= \frac{sum \ of \ terms }{number\ of\ terms} ⇒ 88 = \frac{sum \ of \ terms }{50} ⇒ sum = 88 \ ×\ 50 = 4400\) Therefore, \(25\) should be subtracted from the sum.
\(4400 \ –\ 25 = 4375\), mean \(= \frac{sum \ of\ terms }{number \ of\ terms} ⇒\) mean \(= \frac{4375 }{50} = 87.5\)

The correct answer is \(\frac{1}{2}\)
To get a sum of \(6\) or \(9\) for two dice, we should get \(3\) and \(3\), or \(3\) and \(6\), or \(6\) and \(3\). Therefore, there are \(3\) options.
Since, we have \(6 \ ×\ 6 = 36\) total options, the probability of getting a sum of \(6\) and \(9\) is \(3\) out of \(36\) or \(\frac{1}{12}\).

The correct answer is \(10\) meters
The width of the rectangle is twice its length. Let \(x\) be the length. Then, width \(=2\ x\)
Perimeter of the rectangle is \(2\) (width \(+\) length) \(= 2\ (2\ x\ +\ x)=60 ⇒ 6\ x=60 ⇒ x=10\)
Length of the rectangle is \(10\) meters.

The correct answer is \(a=c\)
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
number of cities for each type of pollution: \(6, \ 3, \ 4, \ 9, \ 8\)
average (mean) \(=\frac{ sum \ of \ terms }{number \ of\ terms}⇒\)average\(=\frac{6\ +\ 3\ +\ 4\ +\ 9\ +\ 8}{5}=\frac{30}{5}=6\)
Median is the number in the middle. To find median, first list numbers in order from smallest to largest.
\(3, \ 4, \ 6, \ 8, \ 9\)
Median of the data is \(6\).
Mode is the number which appears most often in a set of numbers. Therefore, there is no mode in the set of numbers.
Median \(=\) Mean, then, \(a=c\)

The correct answer is \(60\% , 40\% , 90\%\)
percent of cities are in the type of pollution A : \(\frac{6}{10} \ ×\ 100= 60\%\)
percent of cities are in the type of pollution C : \(\frac{4}{10}\ ×\ 100= 40\%\)
percent of cities are in the type of pollution E : \(\frac{9}{10}\ ×\ 100= 90\%\)

The correct answer is \(y\ +\ 4\ x=Z\)
\(x\) and \(Z\) are colinear. \(y\) and \(5\ x\) are colinear. Therefore,
\(x\ +\ Z=y\ +\ 5\ x\), subtract x from bh sides,then, \(Z=y\ +\ 4\ x\)

The correct answer is \(16\)
average = \(\frac{\ sum\ \ of\ \ terms}{\ number\ \ of\ \ terms} ⇒\) (average of \(6\) numbers) \(12 = \frac{\ sum\ \ of\ \ numbers}{6} ⇒\)
sum of \(6\) numbers is \(12 \ ×\ 6 = 72\)
(average of \(4\) numbers) \(10 = \frac{\ sum\ \ of\ \ numbers }{4} ⇒\)sum of \(4\) numbers is \(10 \ ×\ 4 = 40\)
sum of \(6\) numbers \(–\) sum of \(4\) numbers \(=\) sum of \(2\) numbers \(72 \ –\ 40 = 32\)
average of \(2\) numbers \(=\frac {32 }{2}=16\)

The correct answer is \(-\ 2 \) 
Solving Systems of Equations by Elimination
Multiply the first equation by \(- \ 2\), then add it to the second equation.
\(\cfrac{\begin{align} - \ 2 \ (2 \ x \ + \ 5 \ y \ = \ 11 \\ 4 \ x \ - \ 2 \ y \ = - \ 14 \end{align}}{} \)
\(\cfrac{ \begin{align} - \ 4 \ x \ - \ 10 \ y \ = \ - \ 22 \\ 4 \ x \ - \ 2 \ y \ = - \ 14 \end{align} }{\begin{align} - \ 12\ y \ = - \ 36 \\ ⇒ y \ = \ 3 \end{align}} \)
Plug in the value of y into one of the equations and solve for \(x\).
\(2 \ x \ + \ 5 \ (3)= 11 ⇒\)
\(2 \ x \ + \ 15= 11 ⇒\)
\(2 \ x= \ - \ 4 ⇒ x= \ - \ 2\)

The correct answer is \(5\ x\ +\ 5\)
Five years ago, Amy was x times as old as Mike. Mike is \(10\) years now. Therefore, \(5\) years ago Mike was \(5\) years old.
Five years ago, Amy was: A\(=5\ ×\ x=5\ x\)
Now Amy is: A\(=5\ x\ +\ 5\)

The correct answer is \(5\ x\ +\ 5\)
Five years ago, Amy was x times as old as Mike. Mike is \(10\) years now. Therefore, \(5\) years ago Mike was \(5\) years old.
Five years ago, Amy was: A\(=5\ ×\ x=5\ x\)
Now Amy is: A\(=5\ x\ +\ 5\)

The correct answer is Quantity A is greater.
\(\frac{x\ +\ 5}{5}=\frac{x}{5}\ +\ 1\)
\(\frac{x^2\ -\ 36}{x^2\ -\ 6\ x}=\frac{(x\ -\ 6)\ (x\ +\ 6)}{(x\ (x\ -\ 6))}=\frac{(x\ +\ 6)}{x}=\frac{1\ +\ 6}{x}\)
Since,\(\frac{x}{5}>\frac{6}{x}\) for the values of \(6<\ x\ <9 →\) Quantity A \(>\) Quantity B

The correct answer is The two quantities are equal.
Use exponent “product rule”: \(x^n\ ×\ x^m=x^{n+m}\)
Quantity A: \((1.888)^4 (1.888)^8=(1.888)^{4+8}=(1.888)^{12}\)
Quantity B: \((1.88)^{12}\)
The two quantities are equal.

The correct answer is The two quantities are equal.
Area of a circle \(= π\ r^2→100=π\ r^2→r^2=\frac{100}{π}→r=\frac{10}{\sqrtπ}\)

The correct answer is the relationship cannot be determined from the information given
Choose different values for \(x\) and find the value of quantity A and quantity B.
\(x=1\), then:
Quantity A: \( x^{10}=1^{10}=1\)
Quantity B: \(x^{20}=1^{20}=1\)
The two quantities are equal.
\(x=2\), then: Quantity A: \(x^{10}=2^{10}\)
Quantity B: \(x^{20}=2^{20}\)
Quantity B is greater. 
Therefore, the relationship cannot be determined from the information given.

The correct answer is the Quantity A is greater.
Volume of a right cylinder \(= π\ r^2 \ h\to 50\ π=π\ r^2\  h=π\ (2)^2 \ h\to h=12.5\)
The height of the cylinder is \(12.5\) inches which is bigger than \(10\) inches.

The correct answer is \(\frac{22}{3}\)
Set A: {\(3, \ 5, \ 8, \ 10, \ x, \ y\)}
Set B: {\(4, \ 6, \ x, \ y\)}
The average of the \(4\) numbers in Set B is \(7\). Therefore:
\(\frac{4\ +\ 6\ +\ x\ += y}{4}=7\), multiply both sides of the equation by \(4. →10\ +\ x\ +\ y=28→x\ +\ y=18\)
Let’s find the average of the \(6\) numbers in Set A when the sum of \(x\) and \(y\) is \(18\).
\(\frac{3\ +\ 5\ +\ 8\ +\ 10\ +\ x\ +\ y}{6}=\frac{26\ +\ (x\ +\ y)}{6}=\frac{26\ +\ 18}{6}=\frac{44}{6}=\frac{22}{3}\)

The correct answer is \(27\)
average \(= \frac{\ sum\ \ of\ \ terms}{\ number\ \ of\ \ terms} ⇒24 = \frac{\ sum\ \ of\  5 \ numbers}{5} ⇒\)
sum of \(5\) numbers is \(24 \ ×\ 5 = 120\)
The sum of \(5\) numbers is \(120\). If a sixth number \(42\) is added, then the sum of \(6\) numbers is
\(120 \ +\ 42 = 162\)
average \(= \frac{\ sum\ \ of\ \ terms}{\ number\ \ of\ \ terms}=\frac {162 }{6}=27\)

The correct answer is \(0, 2, 3\)
Method 1: Plugin the options and check.
A. \( 0\)                               \((0^2\ +\ 0)\ (0\ -\ 6)=-\ 12\ (0)→0=0!\)                             It works!  
B. \(0, 3\ -\ \sqrt{3}, \sqrt{3}\ +\ 3\)                     
\(((3\ -\ \sqrt{3})^2\ +\ (\sqrt{3}\ -\ 3))\ (\sqrt{3}\ -\ 3\ -\ 6)=-\ 12\ (\sqrt{3}\ -\ 3)→ (12\ -\ \sqrt{3})\ (\sqrt{3}\ -\ 9)=-\ 12\sqrt{3}\ +\ 36\) \(21\sqrt{3}\ -\ 99≠-\ 12\sqrt{3}\ +\ 36\), 
not a solution!
C. \(0, -2, -3\)               \(((-\ 2)^2\ -\ 2)\ (-\ 2\ -\ 6)=-\ 12\ (-\ 2)→-16≠24!\)       not a solution!
D \(0, 2, 3\)                      \((2)^2\ +\ 2)\ (2-6)=-\ 12\ (2)→-\ 24=-\ 24!\),                 Bingo!
E. No solution             \(((3)^2\ +\ 2)\ (3\ -\ 6)=-\ 12\ (3)→-\ 36=-\ 36!\),             Bingo!
Method 2: Solve for \(x\).
\((x^2\ +\ x)\ (x\ -\ 6)=-\ 12\ x→x^3\ -\ 5\ x^2\ -\ 6\ x=-\ 12\ x→x(x^2\ -\ 5\ x\ +\ 6)=0
→x\ (x\ -\ 2)\ (x\ -\ 3)=0→x=0\) or \(x=2\) or \(x=3\)

The correct answer is \(0.49\)
\(a=b-0.3\ b=0.7\ b→\frac{a}{b}=\frac{0.7\ b}{b}=0.7→(\frac{a}{b})^2=(0.7)^2=0.49\)

The correct answer is \(38\%\)
The population is increased by \(15\%\) and \(20\%\). \(15\%\) increase changes the population to \(115\%\) of original population. 
For the second increase, multiply the result by \(120\%\).
\((1.15)\ ×\ (1.20)=1.38=138\%\)
\(38\) percent of the population is increased after two years.

The correct answer is \(\frac{2}{5}\)
Since the outcomes are mutually exclusive. Then, the sum of probabilities of all outcomes equals to \(1\).
Therefore: \(n\ +\ \frac{n}{2}\ +\ \frac{3\ n}{4}\ +\ \frac{n}{4}=1\)
Find a common denominator and solve for \(n\).
\(n\ +\ \frac{n}{2}\ +\ \frac{3\ n\ }{4} \ +\ \frac{n}{4}=1→\frac{4\ n}{4}\ +\ \frac{2\ n}{4}\ +\ \frac{3\ n}{4} \ +\ \frac{n}{4}=1→\frac{10\ n}{4}=1→10\ n=4→n=\frac{4}{10}=\frac{2}{5}\)

The correct answer is \(y\ +\ 4\ x=Z\)
\(x\) and \(Z\) are colinear. \(y\) and \(5\ x\) are colinear. Therefore,
\(x\ +\ Z=y\ +\ 5\ x\), subtract x from bh sides,then, \(Z=y\ +\ 4\ x\)

The correct answer is \(36\)
Let \(x\) be the smallest number. Then, these are the numbers: \(x, x\ +\ 1, x\ +\ 2, x\ +\ 3, x\ +\ 4\)
average \(= \frac{sum \ of\ terms }{number \ of\ terms} ⇒ 38 = \frac{(x\ +\ (x\ +\ 1)\ +\ (x\ +\ 2)\ +\ (x\ +\ 3)\ +\ (x\ +\ 4))}{5}⇒
38=\frac{5\ x\ +\ 10}{5} ⇒ 190 = 5\ x\ +\ 10⇒ 180 = 5\ x ⇒ x=36\)

The correct answer is \(600\) ml
\(4\%\) of the volume of the solution is alcohol.
Let \(x\) be the volume of the solution. Then: \(4\%\) of \(x = 24\) ml \(⇒ 0.04\ x = 24 ⇒ x = 24 \ ÷\ 0.04 = 600\)

The correct answer is \(0.97\)
ratio of women to men in cityA: \(\frac{570}{600}=0.95\)
ratio of women to men in city B: \(\frac{291}{300}=0.97\)
ratio of women to men in city C: \(\frac{665}{700}=0.95\)
ratio of women to men in city D: \(\frac{528}{550}=0.96\)

The correct answer is \(1.05\)
Percentage of men in city A \(= \frac{600}{1170}\ ×\ 100=51.28\%\)
Percentage of women in city C \(= \frac{665}{1365} \ ×\ 100=48.72\%\)
percentage of men in city A to percentage of women in C \(= \frac{51.28}{48.72}=1.05\)

The correct answer is \(132\)
\(\frac{528\ +\ x}{550}=1.2→528\ +\ x=660→x=132\)

The correct answer is \(  13 \ π\) cm
The rectangle is inscribed in a circle. Therefore, the diagonal of the rectangle is the diameter of the circle.
Use Pythagorean theorem to solve for the diagonal of the rectangle. \(a^2\ +\ b^2=c^2\) \(5^2\ +\ 12^2=c^2→25\ +\ 144=c^2→169=c^2→c=13\)
The diameter of the circle is \(13\). Therefore, the circumference of the circle is: \(C=π\ d=π\ ×\ 13=13\ π\)

The correct answer is \(𝑛^2\ +\ 2\ (𝑛\ −\ 1)\) , \(5 \ n \) , \(6\ (𝑛\ +\ 3)\)
\(n\) is even. Plug in an even number for n and check the options. Let’s choose \(2\) for \(n\). Then:
A.\(n\ +\ 13\)                           \(2 \ +\ 13 = 15\)                                          Odd
B.\(n^2\ +\ 2\ (n\ -\ 1)\)          \(2^2\ +\ 2 \ (2\ -\ 1)=4\ +\ 2\ (1)=6\)          Even
C.\(5\ n\)                                  \(5 \ ×\ 2 = 10\)                                             Even
D.\(3\ n^2\ +\ 5\ n\)                  \(3\ (2)^2\ +\ 5\ (2)=3\ ×\ 4\ +\ 10=17\)     Odd
E.\(n^3\ +\ 3\ n \ - \ 1\)             \(2^3\ +\ 3\ (2)\ -\ 1=8\ +\ 6\ -\ 1=13\)    Odd
F.\(6\ (n\ +\ 3)\)                      \(6\ (2\ +\ 3)=30\)                                       Even

The correct answer is \(80\) miles
The speed of car A is \(56\) mph and the speed of car B is \(64\) mph.
When both cars drive in a straight line toward each other, the distance between the cars decreases at the rate of \(120\) miles per hour: \(56 \ +\ 64 = 120\) 
\(40\) minutes is two third of an hour.
Therefore, they will be \(80\) miles apart \(40\) minutes before they meet. \(\frac{2}{3}\ ×\ 120=80\)