GMAT Math : Problem-Solving Questions

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Example Questions

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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Example Question #381 : Algebra

D = E - 12

$E = \sqrt{F}$

F = 2 G

Express in terms of .

Possible Answers:

$G = \sqrt{2D}- 12$

$G = 4D^{2} - 96D + 576$

$G = \sqrt{2D-24}$

$G = \frac{1}{2} D^{2} -12D +72$

$G = \frac{1}{2} D^{2} +12D +72$

Correct answer:

$G = \frac{1}{2} D^{2} +12D +72$

Explanation:

F = 2 G

$F\div 2 = 2 G \div 2$

$\frac{1}{2} F = G$

$E = \sqrt{F}$

$E^{2} = (\sqrt{F})^{2}$

$E^{2} =F$

D = E - 12

D + 12 = E - 12 + 12

D + 12 = E

By two substitutions:

$G = \frac{1}{2} F = \frac{1}{2} E^{2} = \frac{1}{2} (D+12)^{2}$

$G = \frac{1}{2} (D^{2} +2 \cdot 12 \cdot D +12^{2})$

$G = \frac{1}{2} (D^{2} +24 D +144)$

$G = \frac{1}{2} D^{2} +12D +72$

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Example Question #71 : Solving Equations

Which of the following is the solution set of the equation:

$x-3 = \sqrt{2x+9}$

Possible Answers:

$\left \{ 0,8\right \}$

$\left \{ 0\right \}$

$\left \{ -8, 0\right \}$

$\left \{ -8\right \}$

$\left \{ 8\right \}$

Correct answer:

$\left \{ 8\right \}$

Explanation:

Square both sides of the equation, then solve the resulting quadratic equation by factoring:

$x-3 = \sqrt{2x+9}$

$\left (x-3 \right ) ^{2}=\left ( \sqrt{2x+9 \right )}^{2}$

$x^{2}-6x+9=2x+9$

$x^{2}-8x=0$

x(x-8)=0

Either x=0

x-8=0

x=8

Checking both solutions, however:

$0-3 = \sqrt{2\cdot 0+9}$

-3 = 3

This eliminates 0 as a solution.

$8-3 = \sqrt{2\cdot 8+9}$

5=5

8 turns out to be the only solution.

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Example Question #1 : Solving Quadratic Equations

Solve for .

x^2+4x+9=15

Possible Answers:

$x=\pm 3$

$x=-2\pm \sqrt{10}$

$x=-4\pm2\sqrt{10}$

$x=4\pm4\sqrt{10}$

Correct answer:

$x=-2\pm \sqrt{10}$

Explanation:

Solve using the quadratic formula:

$ax^2+bx+c=0;\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

x^2+4x+9=15

$x=\frac{-4\pm\sqrt{4^2-(4)(1)(-6)}}{(2)(1)}$

$x=\frac{-4\pm\sqrt{16+24}}{2}$

$x=\frac{-4\pm\sqrt{40}}{2}$

$x=\frac{-4\pm2\sqrt10}{2}$

$x=-2\pm\sqrt{10}$

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Example Question #2 : Solving Quadratic Equations

Multiply: $(x^{2} - 5x + 3) (x - 2)$

Possible Answers:

$x^{3} - 7 x^{2}+7x - 6$

$x^{3} +3 x^{2}+13x - 6$

$x^{3} - 7 x^{2}+13x - 6$

$x^{3} +7 x^{2}+10x - 6$

$x^{3} +3 x^{2}+7x - 6$

Correct answer:

$x^{3} - 7 x^{2}+13x - 6$

Explanation:

Distribute:

$(x^{2} - 5x + 3) (x - 2)$

$=x^{2}(x - 2) - 5x(x - 2) + 3(x - 2)$

$=x^{2}\cdot x - x^{2}\cdot 2- 5x \cdot x +5x \cdot 2 + 3 \cdot x - 3 \cdot 2$

$=x^{3} - 2 x^{2}- 5 x^{2} +10x + 3 x - 6$

$=x^{3} - 7 x^{2}+13x - 6$

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All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

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

Example Question #381 : Algebra

Example Question #71 : Solving Equations

Example Question #1 : Solving Quadratic Equations

Example Question #2 : Solving Quadratic Equations

All GMAT Math Resources

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