GMAT Math : Equations

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Example Questions

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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Example Question #68 : Solving Equations

AB = 100,000

AC = 25

BD = 40

Which of the following is equal to ?

Possible Answers:

160,000

0.01

0.00000625

100

62,500

Correct answer:

0.01

Explanation:

AC = 25 and BD = 40 , so

$AC \cdot BD = 25 \cdot 40 = 1,000$

$AB \cdot CD = 1,000$

Since AB= 100,000 , it follows that

$100,000 \cdot CD = 1,000$

$CD = 1,000 \div 100,000 = 0.01$

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Example Question #1461 : Problem Solving Questions

True or false: N = 5 .

Statement 1: $(N+2)^{2} = 2N + 19$

Statement 2: $1 + \frac{10}{N-5} = 0$

Possible Answers:

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.x

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

Correct answer:

EITHER STATEMENT ALONE provides sufficient information to answer the question.

Explanation:

Substitute 5 for in both statements, and it becomes immediately apparent that it is a solution of neither statement:

$(N+2)^{2} = 2N + 19$

$(5+2)^{2} = 2 \cdot 5 + 19$

$7^{2} = 10+19$

49= 29 , a false statement.

$1 + \frac{10}{N-5} = 0$

$1 + \frac{10}{5-5} = 0$

$1 + \frac{10}{0} = 0$ , a false statement since the left quantity, having a zero denominator, is undefined.

Therefore, it follows from either statement alone that N = 5 is false.

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Example Question #70 : Solving Equations

is a rational number. True or false: x = 5

Statement 1: $x^{2} - 3x - 10 = 0$

Statement 2: $1 - \frac{7}{x+2} = 0$

Possible Answers:

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

Correct answer:

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

Explanation:

Assume Statement 1 alone and solve for :

$x^{2} - 3x - 10 = 0$

(x-5)(x+2) = 0

Either:

x - 5 = 0 , in which case x = 5 ,

x+2 = 0 , in which case x = -2 .

Therefore, Statement 1 alone is inconclusive.

Assume Statement 2 alone and solve for :

$1 - \frac{7}{x+2} = 0$

$1 =\frac{7}{x+2}$

$1 \cdot (x+ 2) =\frac{7}{x+2} \cdot (x+ 2)$

x+ 2 = 7

x = 5

The only solution to this equation is x = 5 , so Statement 2 alone answers the question.

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GMAT Math : Equations

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

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

Example Question #68 : Solving Equations

Example Question #1461 : Problem Solving Questions

Example Question #70 : Solving Equations

All GMAT Math Resources

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