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. $\frac{2}{3}\ ×\ 120=80$