GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #55 : Solving Equations

Solve for in the equation

$h = f+ \sqrt{3g+ 4}$

Possible Answers:

$y =\frac{ \sqrt{h - f} - 4}{3}$

$g= \frac{h^{2}-2hf+f^{2} - 12}{3}$

$g= \frac{h^{2}-2hf+f^{2} - 4}{3}$

$g= \frac{h^{2} +f^{2} - 4}{3}$

$y =\frac{ \sqrt{h - f} -12}{3}$

Correct answer:

$g= \frac{h^{2}-2hf+f^{2} - 4}{3}$

Explanation:

$h = f+ \sqrt{3g+ 4}$

$h- f = f+ \sqrt{3g+ 4} - f$

$\sqrt{3g+ 4} = h- f$

$\left (\sqrt{3g+ 4} \right )^{2}= \left (h- f \right )^{2}$

$3g+4= h^{2}-2hf+f^{2}$

$3g+4- 4= h^{2}-2hf+f^{2} - 4$

$3g = h^{2}-2hf+f^{2} - 4$

$\frac{3g }{3}= \frac{h^{2}-2hf+f^{2} - 4}{3}$

$g= \frac{h^{2}-2hf+f^{2} - 4}{3}$

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Example Question #56 : Solving Equations

is 44% of .

2K+3P is what percent of ?

Possible Answers:

$121 \frac{1}{3} \%$

$388 \%$

$135 \%$

$332\%$

$314\frac{2}{3} \%$

Correct answer:

$388 \%$

Explanation:

is 44% of , so is $2 \times 44 \% = 88 \%$ of .

Also, is 300% of .

Add these:

2K+3P is $88 \% + 300 \% = 388 \%$ of

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Example Question #361 : Algebra

Solve for :

-3x+18=7x-2

Possible Answers:

x=5

x=-2

x=-5

x=2

x=-4

Correct answer:

x=2

Explanation:

To solve the equation, we first group the terms on one side and the constants on the other side:

-3x+18=7x-2

-10x=-20

Now we can simply divide both sides by to solve for :

-10x=-20

x=2

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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 #1451 : Gmat Quantitative Reasoning

Solve the following equation for :

3x+5=26

Possible Answers:

x=26

x=7

x=0

x=6

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

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 #1451 : Problem Solving Questions

Solve for :

2x+5=23

Possible Answers:

x=16

x=9

x=14

x=2

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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All GMAT Math Resources

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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 #55 : Solving Equations

Example Question #56 : Solving Equations

Example Question #361 : Algebra

Example Question #371 : Algebra

Example Question #1451 : Gmat Quantitative Reasoning

Example Question #1451 : Problem Solving Questions

Example Question #61 : Solving Equations

Example Question #62 : Solving Equations

Example Question #372 : Algebra

Example Question #64 : Solving Equations

All GMAT Math Resources

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