GMAT Math : Algebra

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Example Questions

Example Question #371 : Algebra

Solve the following equation:

2x-5+3x=18-6x+10

Possible Answers:

x=3

x=5

x=1

x=6

x=11

Correct answer:

x=3

Explanation:

To solve the equation, we must simplify it by adding together the like terms. We can group the terms on the left side of the equation and the constants on the right side of the equation:

2x-5+3x=18-6x+10

2x+3x+6x=18+10+5

11x=33

$x=\frac{33}{11}$

x=3

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Example Question #56 : Equations

Solve the following equation for :

3x+5=26

Possible Answers:

x=7

x=6

x=26

x=0

$x=\frac{31}{3}$

Correct answer:

x=7

Explanation:

To solve, we must isolate . First, subtract from both sides and then divide both sides by .

3x+5-5=26-5

3x=21

$\frac{3x}{3}=\frac{21}{3}$

x=7

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Example Question #371 : Algebra

Solve for :

2x+5=23

Possible Answers:

x=16

x=2

x=14

x=9

Correct answer:

x=9

Explanation:

To solve, isolate .

2x+5=23

2x=18

x=9

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Example Question #61 : Solving Equations

and are the volume and the radius of the same sphere.

$r = n ^{3}$

$V = \sqrt[3]{t}$

Which of the following is a true statement?

Possible Answers:

varies directly as the twenty-seventh power of

varies directly as the cube of .

varies directly as the ninth power of .

varies directly as the eighteenth power of .

varies directly as .

Correct answer:

varies directly as the twenty-seventh power of

Explanation:

The volume and radius of a sphere are related as follows:

$V = \frac{4}{3} \pi r^{3}$ .

By substitution,

$\sqrt[3]{t}= \frac{4}{3} \pi \left ( n^{3}\right ) ^{3}$

$\sqrt[3]{t}= \frac{4}{3} \pi n^{9}$

$\left (\sqrt[3]{t} \right )^{3}= \left (\frac{4}{3} \pi n^{9} \right )^{3}$

$t= \left (\frac{4}{3} \pi \right )^{3} \cdot \left ( n^{9} \right )^{3}$

$t= \left (\frac{4}{3} \pi \right )^{3} \cdot n^{27}$

If $K =\left ( \frac{4}{3} \pi \right )^{3}$ , then

$t=K n^{27}$ ,

and varies directly as the twenty-seventh power of .

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Example Question #62 : Solving Equations

and are the surface area and the radius of the same sphere.

$A = d^{2}$

$r = \sqrt[4]{H}$ .

Which of the following is a true statement?

Possible Answers:

$d\textup{ varies directly as the square root of }H$

$d\textup{ varies directly as the fourth power of } H$

$d\textup{ varies directly as the square of } H$

$d\textup{ varies directly as }H$

$d\textup{ varies directly as the fourth root of } H$

Correct answer:

$d\textup{ varies directly as the fourth root of } H$

Explanation:

The surface area and radius of a sphere are related as follows:

$A = 4 \pi r^{2}$

Substituting:

$d^{2} = 4 \pi (\sqrt[4]{H})^{2}$

$\sqrt{d^{2} }=\sqrt{ 4 \pi (\sqrt[4]{H})^{2}}$

$d=\sqrt{ 4 \pi } \cdot \sqrt{ ( \sqrt[4]{H})^{2}}$

$d=\sqrt{ 4 \pi } \cdot \sqrt[4]{H}$

If $K=\sqrt{ 4 \pi }$ , then

$d=K \sqrt[4]{H}$ ,

and varies directly as the fourth root of .

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Example Question #372 : Algebra

Solve the following equation for .

x^2-7=42

Possible Answers:

$x=\pm \sqrt{35}$

$x=\pm \sqrt{42}$

$x=\pm 7$

x=7

Correct answer:

$x=\pm 7$

Explanation:

To solve, simply isolate .

x^2-7=42

x^2=49

$x=\pm \sqrt{49}$

$x=\pm 7$

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Example Question #64 : Solving Equations

$H = J^{2}$

$J = K ^{3}$

$K= L^{4}$

Express in terms of . You may assume all variables assume positive values.

Possible Answers:

$L = \sqrt[9]{H}$

$L = H^{9}$

$L = \sqrt[24]{H}$

$L = H^{3}$

$L = H^{24}$

Correct answer:

$L = \sqrt[24]{H}$

Explanation:

$H = J^{2}$ , $J = K ^{3}$ , and $K= L^{4}$ , so, by two substitutions,

$H = (K ^{3})^{2} = K ^{3 \cdot 2} = K ^{6} = ( L ^{4})^{6} = L ^{4 \cdot 6} = L ^{24}$

$H = L ^{24}$ , so

$L = \sqrt[24]{H}$

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Example Question #65 : Solving Equations

3D + 5E = 17

D - 3F = 19

$\frac{1}{2}E - \frac{1}{3}G = 8$

Express in terms of .

Possible Answers:

$F =- \frac{10}{27}G- \frac{40}{3}$

$F =- \frac{5}{6}G- \frac{160}{9}$

F= 9G + 99

$F =- \frac{10}{3}G- 82$

$F= \frac{1}{9}G + \frac{11}{9}$

Correct answer:

$F =- \frac{10}{27}G- \frac{40}{3}$

Explanation:

3D + 5E = 17

D - 3F = 19 , so

D - 3F + 3F = 19 + 3F

D = 3F + 19

$\frac{1}{2}E - \frac{1}{3}G = 8$ , so

$2 \cdot \left (\frac{1}{2}E - \frac{1}{3}G \right )= 2\cdot 8$

$E - \frac{2}{3}G =16$

$E - \frac{2}{3}G + \frac{2}{3}G =16 + \frac{2}{3}G$

$E = \frac{2}{3}G + 16$

Now substitute twice, and solve for :

3D + 5E = 17

$3 (3F + 19) + 5 \left ( \frac{2}{3}G + 16 \right ) = 17$

$9F +57 + \frac{10}{3}G + 80 = 17$

$9F + \frac{10}{3}G + 137= 17$

$9F + \frac{10}{3}G + 137- \frac{10}{3}G - 137 = 17 - \frac{10}{3}G - 137$

$9F = - \frac{10}{3}G - 120$

$\frac{1}{9} \cdot 9F = \frac{1}{9} \cdot \left ( - \frac{10}{3}G - 120 \right )$

$F =- \frac{10}{27}G- \frac{40}{3}$

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Example Question #373 : Algebra

P = 3Q

Q = R + 7

$R = S \div 5$

Express in terms of .

Possible Answers:

S =15P+21

$S = \frac{3}{5}P+21$

$S = \frac{5}{3}P - \frac{5}{3}$

$S = \frac{5}{3}P - 35$

$S = \frac{3}{5}P - \frac{7}{5}$

Correct answer:

$S = \frac{5}{3}P - 35$

Explanation:

$R = S \div 5$

$R \cdot 5 = S \div 5 \cdot 5$

S = 5R

Q = R + 7

Q - 7 = R + 7 - 7

R = Q - 7

P = 3Q

$\frac{P }{3} = \frac{3Q}{3}$

$Q = \frac{P }{3} = \frac{1 }{3} P$

Substituting twice:

$S = 5R = 5(Q-7) = 5Q - 35 = 5 \cdot \frac{1}{3}P - 35 = \frac{5}{3}P - 35$

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Example Question #67 : Solving Equations

A = 3B

$B = C^{3}$

C = - 4D

Express in terms of .

Possible Answers:

$D = -4 \sqrt[3 ]{\frac{1}{3}A}$

$D = -\frac{1}{4} \sqrt[3 ]{\frac{1}{3}A}$

$D =-12 \sqrt[3 ]{A}$

$D = -\frac{4}{3} \sqrt[3 ]{A}$

$D = -\frac{1}{12} \sqrt[3 ]{A}$

Correct answer:

$D = -\frac{1}{4} \sqrt[3 ]{\frac{1}{3}A}$

Explanation:

A = 3B

$A \div 3 = 3B \div 3$

$B = \frac{1}{3}A$

$B = C^{3}$

$\sqrt[3 ]{B} =\sqrt[3 ]{ C^{3}}$

$C = \sqrt[3 ]{B}$

C = - 4D

$C \div (-4) = - 4D \div (-4)$

$D = -\frac{1}{4} C$

Substitute twice:

$D = -\frac{1}{4} C = -\frac{1}{4} \sqrt[3 ]{B}= -\frac{1}{4} \sqrt[3 ]{\frac{1}{3}A}$

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #371 : Algebra

Example Question #56 : Equations

Example Question #371 : Algebra

Example Question #61 : Solving Equations

Example Question #62 : Solving Equations

Example Question #372 : Algebra

Example Question #64 : Solving Equations

Example Question #65 : Solving Equations

Example Question #373 : Algebra

Example Question #67 : Solving Equations

All GMAT Math Resources

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