## Full Length GRE Quantitative Reasoning Practice Test

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GRE Quantitative Reasoning Practice Test 2

Section 1   20 questions Total time for this section: 35 Minutes You can use a basic calculator on this section.

1- $$x$$ and $$y$$ are positive numbers.
 Quantity A Quantity B $$x^2\ +\ 2\ x\ y$$ $$(x\ +\ y)^2$$
(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.
2- In the $$xy$$-plane, two points $$(p,0)$$ and $$(0,q)$$ are on a line with equation $$y=\frac{2}{3} \ x\ +\ 12$$.
 Quantity A Quantity B $$p$$ $$(q$$
(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.
3- $$x$$ is a positive integer greater than $$1$$.
 Quantity A Quantity B $$\sqrt{x\ +\ 1}$$ $$\sqrt{x\ +\ \sqrt{x}}$$
(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.
4- $$x^2 \ −\ 2\ x\ −\ 15 = 0$$.
 Quantity A Quantity B $$x$$ $$6$$
(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.
5- $$x > y$$ .
 Quantity A Quantity B $$|x^2\ +\ y|$$ $$|x^2\ - \ y|$$
(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.
6- Mr. Jones obtained a $$15,000$$ loan at a simple annual interest rate of p percent. After two years, he paid $$16,050$$ to repay the loan and its interest. What is the value of $$p$$?
(A) $$2.5$$
(B) $$3.5$$
(C) $$5.5$$
(D) $$6$$
(E) $$7$$
7- If $$(3^{6 x} ) \ (81)=3^{2 y}$$, where $$x$$ and $$y$$ are integers, what is the value of $$y$$ in terms of $$x$$?
(A) $$3\ x$$
(B) $$6\ x$$
(C) $$3\ x\ + \ 2$$
(D) $$3\ x\ + \ 81$$
(E) $$6\ x\ + \ 4$$
8- The ratio of boys to girls in a school is $$2:3$$. If there are $$600$$ students in a school, how many boys are in the school.
(A) $$540$$
(B) $$360$$
(C) $$300$$
(D) $$280$$
(E) $$240$$
9- What is the area of the following equilateral triangle if the side AB $$= 8$$ cm?
(A) $$16\sqrt{3}$$ cm$$^2$$
(B) $$8\sqrt{3}$$ cm$$^2$$
(C) $$\sqrt{3}$$ cm$$^2$$
(D) $$8$$ cm$$^2$$
(E) $$16$$ cm$$^2$$
10- If $$60\%$$ of $$x$$ equal to $$30\%$$ of $$20$$, then what is the value of $$(x\ +\ 5)^2$$?
(A) $$25.25$$
(B) $$26$$
(C) $$26.01$$
(D) $$2025$$
(E) $$225$$
11- A ladder leans against a wall forming a $$60^\circ$$ angle between the ground and the ladder. If the bottom of the ladder is $$30$$ feet away from the wall, how long is the ladder?
(A) $$30$$ feet
(B) $$40$$ feet
(C) $$50$$ feet
(D) $$60$$ feet
(E) $$120$$ feet
12- The mean of $$50$$ test scores was calculated as $$88$$. But, it turned out that one of the scores was misread as $$94$$ but it was $$69$$. What is the mean?
(A) $$85$$
(B) $$87$$
(C) $$87.5$$
(D) $$88.5$$
(E) $$90.5$$
13- Two dice are thrown simultaneously, what is the probability of getting a sum of $$6$$ or $$9$$?
(A) $$\frac{1}{3}$$
(B) $$\frac{1}{4}$$
(C) $$\frac{1}{6}$$
(D) $$\frac{1}{12}$$
(E) $$\frac{1}{36}$$
14- The perimeter of a rectangular yard is $$60$$ meters. What is its length if its width is twice its length?
(A) $$10$$ meters
(B) $$18$$ meters
(C) $$20$$ meters
(D) $$24$$ meters
(E) $$36$$ meters
15-

If a is the mean (average) of the number of cities in each pollution type category, b is the mode, and c is the median of the number of cities in each pollution type category, then which of the following must be true?

types of air pollution in $$10$$ cities of a cuntry

(A) $$π<π<π$$
(B) $$π<π<π$$
(C) $$π=π$$
(D) $$π<π=π$$
(E) $$π=π<π$$
16- What percent of cities are in the type of pollution A, C, and E respectively? types of air pollution in $$10$$ cities of a cuntry
(A) $$60\%, 40\%, 90\%$$
(B) $$30\%, 40\%, 90\%$$
(C) $$30\%, 40\%, 60\%$$
(D) $$40\%, 60\%, 90\%$$
(E) $$60\%, 30\%, 90\%$$
17- From the figure, which of the following must be true? (figure not drawn to scale)
(A) $$y = Z$$
(B) $$y =5\ x$$
(C) $$y \geq \ x$$
(D) $$y\ +\ 4\ x=Z$$
(E) $$y>x$$
18- The average of $$6$$ numbers is $$12$$. The average of $$4$$ of those numbers is $$10$$. What is the average of the other two numbers.
(A) $$10$$
(B) $$12$$
(C) $$14$$
(D) $$16$$
(E) $$24$$
19- What is the value of $$x$$ in the following system of equations?$$2\ x\ +\ 5\ y=11\\ 4\ x\ -\ 2\ y=-\ 14$$
(A) $$-\ 1$$
(B) $$1$$
(C) $$-\ 2$$
(D) $$4$$
(E) $$8$$
20- Five years ago, Amy was $$x$$ times as old as Mike was. If Mike is $$10$$ years old now, how old is Amy in terms of $$x$$?
(A) $$5\ x$$
(B) $$10 \ x$$
(C) $$5 \ x \ - \ 10$$
(D) $$5 \ x \ + \ 5$$
(E) $$5 \ x \ + \ 10$$

GRE Quantitative Reasoning Practice Test 2

Section 2   20 questions Total time for this section: 35 Minutes You can use a basic calculator on this section.

21- $$6<x<9$$
 Quantity A Quantity B $$\frac{x \ +\ 5}{5}$$ $$\frac{x^2\ -\ 36}{x^2\ -\ 6\ x}$$
(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.
22-
 Quantity A Quantity B $$(1.888)^4\ (1.888)^8$$ $$(1.88)^{12}$$
(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.
23-
 Quantity A Quantity B radius of a circle with the area of $$100$$ $$\frac{10}{\sqrt{π}}$$
(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.
24- $$x$$ is a positive number.
 Quantity A Quantity B $$x^{10}$$ $$x^{20}$$
(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.
25- A right cylinder with radius $$2$$ inches has volume $$50\ π$$ cubic inches.
 Quantity A Quantity B the height of the cylinder $$10$$ inches
(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.
26- If the average (arithmetic mean) of the $$4$$ numbers in Set B is $$7$$, what is the average of the $$6$$ numbers in Set A?
(A) 22/3
(B) 22/3
(C) 22/3
(D) 22/3
27- The average of five numbers is $$24$$. If a sixth number $$42$$ is added, then, what is the new average?
(A) $$25$$
(B) $$26$$
(C) $$27$$
(D) $$28$$
(E) $$36$$
28- What are the solutions of the following equation? $$(x^2\ +\ x)\ (x\ -\ 6)=-\ 12\ x$$
(A) $$0$$
(B) $$0, \ 3\ −\ \sqrt{3}, \sqrt{3}\ +\ 3$$
(C) $$0, -\ 2, -\ 3$$
(D) $$0, 2, 3$$
(E) No solution
29- If a and b are two positive natural numbers and a is $$30\%$$ less than b, what is the value of $$(\frac{a}{b})^2$$ ?
(A) $$0.7$$
(B) $$0.49$$
(C) $$\frac{10}{7}$$
(D) $$\frac{100}{49}$$
(E) $$1.69$$
30- In two successive years, the population of a town is increased by $$15\%$$ and $$20\%$$. What percent of the population is increased after two years?
(A) $$32\%$$
(B) $$35\%$$
(C) $$38\%$$
(D) $$68\%$$
(E) $$70\%$$
31- A certain experiment has $$4$$ possible mutually exclusive outcomes and have probabilities $$n,\frac{n}{2},\frac{3\ n}{4},\frac{n}{4}$$, respectively. What is the value of $$n$$?
(A) $$\frac{1}{10}$$
(B) $$\frac{3}{10}$$
(C) $$\frac{2}{5}$$
(D) $$\frac{1}{4}$$
(E) $$\frac{1}{2}$$
32- From the figure, which of the following must be true? (figure not drawn to scale)
(A) $$y = Z$$
(B) $$y =5\ x$$
(C) $$y ≥ \ x$$
(D) $$y \ + \ 4\ x = Z$$
(E) $$y \gt x$$
33- The average of five consecutive numbers is $$38$$. What is the smallest number?
(A) $$38$$
(B) $$36$$
(C) $$34$$
(D) $$12$$
(E) $$8$$
34- A chemical solution contains $$4\%$$ alcohol. If there is $$24$$ ml of alcohol, what is the volume of the solution?
(A) $$240$$ ml
(B) $$480$$ ml
(C) $$600$$ ml
(D) $$1,200$$ ml
(E) $$2,400$$ ml
35- What's the ratio of percentage of men in city to percentage of women in city ?
(A) $$0.98$$
(B) $$0.97$$
(C) $$0.96$$
(D) $$0.95$$
(E) $$0.94$$
36- What's the maximum ratio of number of women to number of men in the four cities?
(A) $$0.9$$
(B) $$0.95$$
(C) $$1$$
(D) $$1.05$$
(E) $$1.11$$
37- How many women should be added to city D until the ratio of the number of women to number of men will be $$1.2$$?
(A) $$120$$
(B) $$128$$
(C) $$132$$
(D) $$160$$
(E) $$162$$
38- A $$5$$ cm by $$12$$ cm rectangle is inscribed in a circle. What is the circumference of the circle?
(A) $$5 \ π$$ cm
(B) $$6.55 \ π$$ cm
(C) $$12 \ π$$ cm
(D) $$13 \ π$$ cm
(E) $$26 \ π$$ cm
39- If $$n$$ is even, which of the following cannot be odd? Select all that apply.
(A) $$π\ +\ 13$$
(B) $$π^2\ +\ 2\ (π\ −\ 1)$$
(C) $$5\ π$$
(D) $$3\ π^2\ +\ 5\ π$$
(E) $$π^3\ +\ 3\ π\ −\ 1$$
(F) $$6\ (π\ +\ 3)$$
40- Two cars are $$240$$ miles apart. They both drive in a straight line toward each other. If Car drives at $$56$$ mph and Car  drives at $$64$$ mph, then how many miles apart will they be exactly $$40$$ minutes before they meet?
(A) $$60$$ miles
(B) $$80$$ miles
(C) $$100$$ miles
(D) $$110$$ miles
(E) $$120$$ miles
 1- Choice B is correct The correct answer is Quantity B is greater.$$(x\ +\ y)^2=(x\ +\ y)\ (x\ +\ y)=x^2\ +\ 2\ x\ y\ +\ \ y^2 →$$Since $$y^2>0→x^2\ +\ 2\ x\ y\ +\ y^2> x^2\ +\ 2\ x\ y$$ 2- Choice B is correct 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.$$0=\frac{2}{3} \ p\ +\ 12→\frac{2}{3}\ p=-\ 12→p=(-\ 12)\ ×\ (\frac{3}{2})=-\ 18$$$$(0,q): y=\frac{2}{3} \ x\ +\ 12→q=\frac{2}{3} \ (0)\ +\ 12→q=12$$$$q>p$$ 3- Choice B is correct 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 $$1$$. 4- Choice B is correct 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 5- Choice D is correct 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 $$y=-\ 0.5 →B>A$$ 6- Choice B is correct 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$$=\frac{interest\ amount}{loan}\ ×\ 100→p=\frac{525}{15000}\ ×\ 100=3.5$$ 7- Choice C is correct 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 $$2$$:$$6\ x\ +\ 4=2\ y →3\ x\ +\ 2=y$$ 8- Choice E is correct 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 $$2$$. $$600 \ ÷\ 5 = 120 ⇒ 120 \ ×\ 2 = 240$$ 9- Choice A is correct 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 $$= \frac{1}{2 }\ 4\sqrt{3}\ ×\ 8=16\sqrt{3}$$ 10- Choice E is correct The correct answer is $$225$$$$0.6\ x=(0.3)\ ×\ 20→x=10→(x+5)^2=(15)^2=225$$ 11- Choice D is correct The correct answer is $$60$$ feetThe 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. 12- Choice C is correct 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 $$= \frac{4375 }{50} = 87.5$$ 13- Choice D is correct 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 $$\frac{1}{12}$$. 14- Choice A is correct The correct answer is $$10$$ metersThe 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. 15- Choice C is correct 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, $$a=c$$ 16- Choice A is correct 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 : $$\frac{9}{10}\ ×\ 100= 90\%$$ 17- Choice D is correct 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, $$Z=y\ +\ 4\ x$$ 18- Choice D is correct 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 $$=\frac {32 }{2}=16$$ 19- Choice C is correct The correct answer is $$-\ 2$$ Solving Systems of Equations by EliminationMultiply 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 $$x$$.$$2 \ x \ + \ 5 \ (3)= 11 ⇒$$$$2 \ x \ + \ 15= 11 ⇒$$$$2 \ x= \ - \ 4 ⇒ x= \ - \ 2$$ 20- Choice D is correct 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$$=5\ x\ +\ 5$$ 20- Choice D is correct 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$$=5\ x\ +\ 5$$ 21- Choice A is correct 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 22- Choice C is correct 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. 23- Choice C is correct The correct answer is The two quantities are equal.Area of a circle $$= π\ r^2→100=π\ r^2→r^2=\frac{100}{π}→r=\frac{10}{\sqrtπ}$$ 24- Choice D is correct The correct answer is the relationship cannot be determined from the information givenChoose 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. 25- Choice A is correct 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. 26- Choice D is correct 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 $$18$$.$$\frac{3\ +\ 5\ +\ 8\ +\ 10\ +\ x\ +\ y}{6}=\frac{26\ +\ (x\ +\ y)}{6}=\frac{26\ +\ 18}{6}=\frac{44}{6}=\frac{22}{3}$$ 27- Choice C is correct 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 $$= \frac{\ sum\ \ of\ \ terms}{\ number\ \ of\ \ terms}=\frac {162 }{6}=27$$ 28- Choice D is correct 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 $$x=3$$ 29- Choice B is correct The correct answer is $$0.49$$$$a=b-0.3\ b=0.7\ b→\frac{a}{b}=\frac{0.7\ b}{b}=0.7→(\frac{a}{b})^2=(0.7)^2=0.49$$ 30- Choice C is correct 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. 31- Choice C is correct 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 $$n$$.$$n\ +\ \frac{n}{2}\ +\ \frac{3\ n\ }{4} \ +\ \frac{n}{4}=1→\frac{4\ n}{4}\ +\ \frac{2\ n}{4}\ +\ \frac{3\ n}{4} \ +\ \frac{n}{4}=1→\frac{10\ n}{4}=1→10\ n=4→n=\frac{4}{10}=\frac{2}{5}$$ 32- Choice D is correct 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, $$Z=y\ +\ 4\ x$$ 33- Choice B is correct 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 $$= \frac{sum \ of\ terms }{number \ of\ terms} ⇒ 38 = \frac{(x\ +\ (x\ +\ 1)\ +\ (x\ +\ 2)\ +\ (x\ +\ 3)\ +\ (x\ +\ 4))}{5}⇒38=\frac{5\ x\ +\ 10}{5} ⇒ 190 = 5\ x\ +\ 10⇒ 180 = 5\ x ⇒ x=36$$ 34- Choice C is correct 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 $$⇒ 0.04\ x = 24 ⇒ x = 24 \ ÷\ 0.04 = 600$$ 35- Choice B is correct 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: $$\frac{528}{550}=0.96$$ 36- Choice D is correct 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 $$= \frac{51.28}{48.72}=1.05$$ 37- Choice C is correct The correct answer is $$132$$$$\frac{528\ +\ x}{550}=1.2→528\ +\ x=660→x=132$$ 38- Choice D is correct The correct answer is $$13 \ π$$ cmThe 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: $$C=π\ d=π\ ×\ 13=13\ π$$ 39- Choice F is correct The correct answer is $$π^2\ +\ 2\ (π\ −\ 1)$$ , $$5 \ n$$ , $$6\ (π\ +\ 3)$$$$n$$ 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.$$5\ n$$                                  $$5 \ ×\ 2 = 10$$                                             EvenD.$$3\ n^2\ +\ 5\ n$$                  $$3\ (2)^2\ +\ 5\ (2)=3\ ×\ 4\ +\ 10=17$$     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 40- Choice B is correct The correct answer is $$80$$ milesThe 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. $$\frac{2}{3}\ ×\ 120=80$$

### Practice Test 1

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### Practice Test 2

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