Did you take the GRE Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve.

## GRE Math Practice Test Answers and Explanations

**Section 1**

**1- Choice A is correct**prime factoring of 77 is: 7 × 11, prime factoring of 136 is: 2 × 2 × 2 × 17, Quantity A = 7 and Quantity B = 2, Quantity A is greater.

**2- Choice A is correct**\(4\%\) of \(= 6\%\) of \(y → 0.04 x = 0.06 y→x=\frac{0.06}{0.04}y→x=\frac{6}{4} y=\frac{3}{2} y=1.5y\), therefore, \(x\) is bigger than \(y\).

**3- Choice A is correct.**Choose different values for \(x\) and find the value of quantity A, \(x=2\), then: Quantity A: \((\frac{1}{x})^2=(\frac{1}{2})^2=\frac{1}{4}=0.25\), Quantity B: \((\frac{1}{x})^3=(\frac{1}{2})^3=\frac{1}{8}=0.125\), Quantity A is greater.

**4- Choice C is correct**Choose different values for a and find the values of quantity A and quantity B.

\(a=2\), then: Quantity A: \(|a|=|2|=2\), Quantity B: \(|-a|=|-2|=2\), The two quantities are equal, \(a=-3\), then: Quantity A: \(|a|=|-3|=3\), Quantity B: \(|-a|=|-(-3)|=3\)

The two quantities are equal. Any other values of a and b provide the same answer.

**5- Choice D is correct**average \(=\frac{sum \space of \space terms}{number \space of \space terms}\), The sum of the weight of all girls is: 22 × 62 = 1364 kg

The sum of the weight of all boys is: 38 × 74 = 2812 kg, The sum of the weight of all students is: 1364 + 2812 = 4176 kg, average =\( \frac{4,176}{60}= 69.6\)

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**6- Choice C is correct**\(V_1=\frac{4π}{3}(\frac{6}{2})^3, V_2=\frac{4π}{3}(\frac{\sqrt{6}}{2})^3 → \frac{V_1}{V_2}=(\frac{6}{\sqrt{6}})^3=\frac{6^3}{6\sqrt{6}}=\frac{6^2}{\sqrt{6}}=6\sqrt{6}\)

**7- Choice C is correct**\(\begin{cases}\frac{x}{4}-\frac{y}{2}=2\\y+2x=1\end{cases}\) →multiply the top equation by 2 then: \(\begin{cases}\frac{x}{2}-y=4\\y+2x=1\end{cases}\) →add two equations

\(\frac{x}{2}+2x=5→\frac{5}{2} x=5→ x=\frac{2×5}{5}=2\) ,plug in the value of \(x\) into the second equation. → \(y=-3\)

**8- Choice E is correct**We know that, \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\) then: \(\sqrt{y}=\sqrt{3^6}=3^3=27→(3^6)^{\frac{27}{9}}=(3^6)^3=3^{18}\)

**9- Choice A is correct.**The profit of their business will be divided between Emma and Sophia in the ratio 5 to 7 respectively. Therefore, Emma receives \(\frac{5}{12}\) of the whole profile and Sophia receives \(\frac{7}{12}\) of the whole profile. Quantity A: The money Emma receives when the profit is $768 equals: \(\frac{5}{12}×768=\frac{3,840}{12}=320\), Quantity B: The money Sophia receives when the profit is $540 equals: \(\frac{7}{12}×540=\frac{3,780}{12}=315\), Quantity A is greater.

**10- Choice B is correct**The slop of line A is: \(m=\frac{y_2-y_1}{x_2-x_1}

=\frac{3-(-2)}{-4-6}=\frac{5}{-10}=-\frac{1}{2}\)

Parallel lines have the same slope and only choice D \((y=-\frac{1}{2} x)\) has slope of \(-\frac{1}{2}\).

**11- Choice D is correct**\((x-3)^4=256=4^4→x-3=4→x=7, →(x-5)(x-4)=(7-5)(7-4)=(2)(3)=6\)

**12- Choice D is correct**average \(= \frac{sum \space of \space terms}{number \space of \space terms}

→ \frac{x+y+4}{3}=6→x+y=14\)

\(\begin{cases}x+y=14\\x-y=-2\end{cases}\)

add both equations: \(2x=12→x=6→y=8→x×y=48\)

**13- Choice A is correct**Formula for the Surface area of a cylinder is:

\(SA=2πr^2+2πrh→216π=2πr^2+2πr(12)→r^2+12r-108=0\)

\((r+18)(r-6)=0→r=6\) or \(r= -18\) (unacceptable)

**14- Choice B is correct**Write the equation and solve for B: 0.40 A = 0.80 B, divide both sides by 0.80, then you will have \(\frac{0.40}{0.80}\)A = B, therefore: B = \(\frac{1}{2}\) A, and B it’s \(50\%\) of A.

**15- Choice C is correct**number of Chemistry book: 0.3 × 280 = 84, number of History book:

0.10 × 280 = 28

product of number of Chemistry and number of History book:

84 × 28 = 2,352

**16- Choice A is correct**The angle α is: \(0.15×360=54\circ\), The angle \(β\) is: \(0.20×360=72^\circ\)

**17- Choice D is correct**According to the chart, \(45\%\) of the books are in the Chemistry and Mathematics sections.

Therefore, there are 126 books in these two sections. \(0.45 × 280 = 126\)

\(γ+α=126\), and \(α=\frac{2}{5}γ\), Replace \(α\) by \(\frac{2}{5} γ\) in the first equation.

\(γ+α=126→γ+\frac{2}{5} γ=126→\frac{7}{5} γ=126\)→multiply both sides by \(\frac{5}{7}\)

\((\frac{5}{7}) \frac{7}{5} γ=126×(\frac{5}{7})→γ=\frac{126×5}{7}=90\), \(γ=90→α=\frac{2}{5} γ→α=\frac{2}{5}×90=36\)

There are 36 books in the Mathematics section.

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**18- Choice A is correct**Plug in the value of each option in the inequality.

\(0→ (0-3)^2+2>4(0)+3→11>3\) Bingo!

\(1 →(1-3)^2+2>4(1)+3→6>7\) No!

\(2 →(2-3)^2+2>4(2)+3→3>11\) No!

\(3 →(3-3)^2+2>4(3)+3→2>15\) No!

\(4 →(4-3)^2+2>4(4)+3→3>19\) No!

**19- Choice D is correct**This question is a combination problem. The formula for combination is: \(nCr = \frac{n!}{r!(n-r)!}\)

This formula is for the number of possible combinations of r objects from a set of n objects.

Using the information in the question: \(8C5 =\frac{8!}{5!(8-5)!} =\frac{8×7×6×5×4×3×2×1}{5×4×3×2×1×(3×2×1)}=56\)

**20- Choice D is correct**We have: \((x+3)(x+p)=x^2+(3+p)x+3p→3+p=9→p=6 , and r=3p=18\)

**Section 2**

**21- Choice B is correct**\(\frac{1}{5^n} <\frac{1}{625} → 5^{-n}<5^{-4}→-n<-4\), divide both side by \(-1→n>4\)

Quantities A: 4, Quantities B: n. The B quantities is greater.

**22- Choice B is correct**Quantity A is: \(\frac{7+5+x}{3}=7→12+x=21→x=21-12=9\), Quantity B is: \(\frac{2(9)+(9)+(9-3)}{3}=\frac{33}{3}=11\), The B quantities is greater.

**23- Choice D is correct.**Simply change the fractions to decimals. \(\frac{4}{5}=0.80, \frac{6}{7}=0.857\)…, \(\frac{5}{6}=0.8333\)…

As you can see, \(x\) lies between 0.80 and 0.857… and it can be 0.81 or 0.84. The first one is less than 0.833… and the second one is greater than 0.833… .

The relationship cannot be determined from the information given.

**24- Choice B is correct**Simplify both quantities. Quantity A: \((-3)^3=(-3)×(-3)×(-3)=-27\)

Quantity B: \(3^3=3×3×3=27\), The B quantities is greater.

**25- Choice C is correct.**First, find the number. Let \(x\) be the number. Write the equation and solve for \(x\).

\(160 \%\) of a number is 128, then: \(1.6×x=128 ⇒ x=128÷1.6=80\)

\(60 \%\) of \(80\) is: \(0.6 × 80 = 48\)

**26- Choice A is correct**Use PEMDAS (order of operation):

Quantity A \(=6-4×8+[2+30×6]÷2=6-32+[2+180]÷2=-26+[182]÷2=-26+91=65\)

Quantity B \(=[3+5×(-8)]+2-[6×4]÷8=[3-40]+2-[24]÷8=-37+2-3=-38, 65>-38\)

**27- Choice C is correct**I. \(|a|<1→-1<a<1\), Multiply all sides by b. Since, \(b>0→-b<ba<b\)

II. Since, \(-1<a<1\), and \(a<0→-a>a^2>a\) (plug in \(-\frac{1}{2}\), and check!)

III. \(-1<a<1\), multiply all sides by 3,then: \(-3<3a<3\), subtract 5 from all sides,the: \(-3-5<3a-5<3-5→-8<3a-5<-2\)

**28- Choice B is correct.**The ratio of boy to girls is 7 : 9. Therefore, there are 7 boys out of 16 students. To find the answer, first divide the total number of students by 16, then multiply the result by 7.

64 ÷ 16 = 4 ⇒ 4 × 7 = 28, There are 28 boys and 36 (64 – 28) girls. So, 8 more boys should be enrolled to make the ratio 1 : 1

**29- Choice A is correct.**Simplify quantity B. Quantity B: \((\frac{x}{7})^7=\frac{x^7}{7^7 }\)

Since, the two quantities have the same numerator (x^7) and the denominator in quantity B is bigger (\(7^7>7\)), then the quantity A is greater.

**30- Choice D is correct**Since line l is parallel to \(x\)-axis, therefore the slope of l is equal to 0 and the value of y is the same as the value of y in the point \((3, -2)\). Therefore, \(y\)-intercept is \(-2\).

**31- Choice D is correct**length of the rectangle is: \(\frac{7}{4}×12=21\), perimeter of rectangle is: \(2×(21+12)=66\)

**32- Choice D is correct**To find the discount, multiply the number by (\(100\%\) – rate of discount).

Therefore, for the first discount we get: \((D) (100\% – 30\%) = (D) (0.70) = 0.70 D\)

For increase of \(20 \%\): \((0.70 D) (100\% + 20\%) = (0.70 D) (1.20) = 0.84 D = 84\%\) of \(D\)

**33- Choice D is correct**Let \(x\) be the length of AB, then: \(42=\frac{x×7}{2}→x=12\)

The length of AC \(=\sqrt{12^2+16^2}=\sqrt{400}=20\)

The perimeter of ∆ABC \(=12+16+20=48\)

**34- Choice E is correct**Check each option provided:

A. \(3 → \frac{1+5+7+9+11}{5}=\frac{33}{5}=6.6\)

B. \(5 → \frac{1+3+7+9+11}{5}=\frac{31}{5}=6.2\)

C. \(7 → \frac{1+3+5+9+11}{5}=\frac{29}{5}=5.8\)

D. \(9 → \frac{1+3+5+7+11}{5}=\frac{27}{5}=5.4\)

E. \(11 → \frac{1+3+5+7+9}{5}=\frac{25}{5}=5\)

**35- Choice D is correct**First find the number of shoes sold in each month. January: 25, February: 35, March: 40, April: 20, May: 25, June: 35, Check each option provided.

January and February, \((\frac{25-35}{25})×100=\frac{10}{25}×100=40\%\)

February and March, \((\frac{35-40}{35})×100=\frac{5}{35}×100=14.28\%\)

March and April: there is a decrease from March and April.

April and May, \((\frac{20-25}{20})×100=\frac{5}{20}×100=25\%\)

May and Jun, \((\frac{25-35}{25})×100=\frac{10}{25}×100=40\%\)

**36- Choice A is correct**we order number sold of pants per month: \(65,70,85,88,90,110\)

median is: \(\frac{85+88}{2}= 86.5\), Put the number of shirts sold per month in order:

\(150,130,170,140,160,145\), mean is:

\(\frac{150+130+170+140+160+145}{6}=\frac{895}{6}= 149.16\)

**37- Choice D is correct**Let \(x\) be the number of pants need to be added in February. Then:

\(\frac{130}{88+x}=(\frac{2}{3})(\frac{150}{110})

→\frac{130}{88+x}=\frac{300}{330}→88+x=143→x=55\)

**38- Choice E is correct**Based on triangle similarity theorem: \(\frac{a}{a+b}=\frac{c}{3}→c=\frac{3a}{a+b}=\frac{3\sqrt{5}}{3\sqrt{5}}=1\)→ area of shaded region is: \((\frac{c+3}{2})(b)=\frac{4}{2}×2\sqrt{5}=4\sqrt{5}\)

**39- Choice E is correct**The area of ∆BED is 20, then: \(\frac{5×AB}{2} =20→5×AB=40→AB=8\)

The area of ∆BDF is 24, then: \(\frac{3×BC}{2}=24→3×BC=48→BC=16\)

The perimeter of the rectangle is \(= 2×(8+16)=48\)

**40- Choice A is correct**\(|2x+6-3x-7|=5→ |-x-1|=5 →-x-1=5 or-x-1=-5→x=-6\) or \(x=4\)

The product of all possible values of \(x = (-6)×4=-24\)

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