GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #15 : Equations

The period of a pendulum - that is, the time it takes for the pendulum to swing once and back - varies directly as the square root of its length.

The pendulum of a giant clock is 18 meters long and has period 8.5 seconds. If the pendulum is lengthened to 21 meters, what will its period be, to the nearest tenth of a second?

Possible Answers:

$11.2 \textrm{ sec}$

$10.4 \textrm{ sec}$

$7.9 \textrm{ sec}$

$9.2 \textrm{ sec}$

$11.6 \textrm{ sec}$

Correct answer:

$9.2 \textrm{ sec}$

Explanation:

The variation equation for this situation is

$\frac{P_{1}}{\sqrt{L_{1}}} = \frac{P_{2}}{\sqrt{L}_{2}}$

Set $P_{1} = 8.5 , L_{1} = 18, {L}_{2}= 21$ , and solve for $P_{2}$ ;

$\frac{8.5}{\sqrt{18}} = \frac{P_{2}}{\sqrt{21}}$

$\frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} = \frac{P_{2}}{\sqrt{21}} \cdot{\sqrt{21}}$

$P_{2} = \frac{8.5}{\sqrt{18}} \cdot{\sqrt{21}} \approx 9.2 \textrm{ sec}$

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Example Question #327 : Algebra

$x = 8^{N}$

Which of these expressions is equal to $\log_{32} x$ ?

Possible Answers:

$\ln \frac{5}{3}N$

$\frac{3}{5} N$

$N\sqrt[3]{N^{2}}$

$\frac{5}{3} N$

$\sqrt[5]{N^{3}}$

Correct answer:

$\frac{3}{5} N$

Explanation:

$2^{5} = 32$

$2 = \sqrt[5]{32}$

$8 = 2^{3} = \left ( \sqrt[5]{32} \right ) ^{3} = 32^{ \frac{3}{5} }$

$x = 8^{N}= \left (32^{ \frac{3}{5} } \right \)^ {N } =32^{ \frac{3}{5} N }$

$\log_{32} x =\log_{32}32^{ \frac{3}{5} N } = \frac{3}{5} N$

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Example Question #16 : Equations

Solve for :

| 4x - 18 | - 18 = 57

Possible Answers:

$x= -23.25 \textrm{ or }x=23.25$

x=23.25

$x= -14.25 \textrm{ or }x=14.25$

$x= -14.25 \textrm{ or }x=23.25$

x= -14.25

Correct answer:

$x= -14.25 \textrm{ or }x=23.25$

Explanation:

First, isolate the absolute value expression on one side:

| 4x - 18 | - 18 = 57

| 4x - 18 | -18+ 18 = 57+18

| 4x - 18 | = 75

Rewrite as a compound sentence:

$4x - 18 = - 75 \textrm{ or } 4x - 18 = 75$

Solve each separately:

4x - 18 = - 75

4x - 18+18 = -75+18

4x= - 57

$4x \div 4= - 57 \div 4$

x= - 14.25

4x - 18 = 75

4x - 18+18 = 75+18

4x= 93

$4x \div 4=93 \div 4$

x = 23.25

The solution set is $\left \{ -14.25, 23.25 \right \}$

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Example Question #17 : Equations

Solve for :

| 4x - 18 | + 18 = 57

Possible Answers:

x=-14.25

$x=-5.25 \textrm{ or }x=14.25$

$x=-5.25 \textrm{ or }x=5.25$

$x=-14.25\textrm{ or }x=14.25$

x=14.25

Correct answer:

$x=-5.25 \textrm{ or }x=14.25$

Explanation:

First, isolate the absolute value expression on one side:

| 4x - 18 | + 18 = 57

| 4x - 18 | + 18 -18 = 57-18

| 4x - 18 | = 39

Rewrite as a compound sentence:

$4x - 18 = - 39 \textrm{ or } 4x - 18 = 39$

Solve each separately:

4x - 18 = - 39

4x - 18+18 = - 39+18

4x= - 21

$4x \div 4= - 21 \div 4$

x= - 5.25

4x - 18 = 39

4x - 18+18 = 39+18

4x= 57

$4x \div 4=57 \div 4$

x = 14.25

The solution set is $\left \{ -5.25, 14.25\right \}$ .

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Example Question #1415 : Gmat Quantitative Reasoning

Give the solution set of the equation:

$\left | x ^{2} - 2x \right | = 3x -6$

Possible Answers:

$x \in \left \{ -3, 3\right \}$

The equation has no solution.

$x \in \left \{ -3, 2\right \}$

$x \in \left \{ 2, 3\right \}$

$x \in \left \{ -3, 2, 3\right \}$

Correct answer:

$x \in \left \{ 2, 3\right \}$

Explanation:

Write this as a compound equation and solve each separately.

$\left | x ^{2} - 2x \right | = 3x -6$

$x ^{2} - 2x = 3x -6 \textrm{ or }x ^{2} - 2x = -\left ( 3x -6 \right )$

$x ^{2} - 2x = 3x -6$

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

$x ^{2} - 5x + 6 =0$

(x-3) (x-2) =0

$x - 3 = 0 \Rightarrow x = 3$

$x - 2 = 0 \Rightarrow x =2$

$x ^{2} - 2x = -\left ( 3x -6 \right )$

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

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

$x ^{2} +x - 6= 0$

(x + 3) (x - 2)= 0

$x + 3 = 0 \Rightarrow x = -3$

$x - 2 = 0 \Rightarrow x =2$

This gives us three possible solutions - $\left \{ -3, 2, 3\right \}$ . We check all three.

x = -3

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | \left ( -3\right ) ^{2} - 2 (-3) \right | = 3 (-3) -6$

$\left | \left9 - (-6) \right | = -9-6$

$\left |15| = -15$

15 = -15

This is a false statement so we can eliminate as a false "solution".

x = 2

$\left | x ^{2} - 2x \right | = 3x -6$

$\left |2^{2} - 2 \cdot 2 \right | = 2 \cdot 3 -6$

$\left | \left 4-4 \right | = 6-6$

|0| = 0

0=0

2 is a solution.

x = 3

$\left | x ^{2} - 2x \right | = 3x -6$

$\left | 3 ^{2} - 2 \cdot 3 \right | = 3 \cdot 3 -6$

$\left | \left 9 - 6\right | = 9-6$

$\left |3 | = 3$

3= 3

3 is a solution.

The solution set is $x \in \left \{ 2, 3\right \}$ .

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Example Question #13 : Equations

Solve for :

$4 ^{5x - 7} = 8 ^{2x -1}$

Possible Answers:

The equation has no solution

x = -1.0625

x = 0.6875

x = 2.75

x = 4.25

Correct answer:

x = 2.75

Explanation:

Since $2 ^{2} = 4$ and $2^{3} = 8$ , replace, and use the exponent rules:

$4 ^{5x - 7} = 8 ^{2x -1}$

$\left ( 2 ^{2} \right ) ^{5x - 7} = \left ( 2^{3} \right ) ^{2x -1}$

$2 ^{2 \left ( 5x - 7 \right )}=2 ^{3 \left ( 2x - 1 \right )}$

Set the exponents equal to each other and solve for :

$2 \left ( 5x - 7 \right )=3 \left ( 2x - 1 \right )$

10x -14=6x - 3

10x-6x -14+14=6x -6x - 3+14

4x = 11

$4x \div 4 = 11 \div 4$

x = 2.75

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Example Question #1411 : Problem Solving Questions

Give the solution set of the equation:

$2x + \left | x\right | = 60$

Possible Answers:

$x= -60 \textrm{ or } x = 20$

$x = 20 \textrm{ or }x= 60$

x = 60

x = -60

x = 20

Correct answer:

x = 20

Explanation:

Case 1: $x \geq 0$

Then $\left | x\right | = x$ , and the equation becomes

2x + x= 60

3x= 60

$3x \div 3 = 60 \div 3$

x = 20

Case 2: x < 0

Then $\left | x\right | = -x$ , and the equation becomes

$2x +\left ( - x \right ) = 60$

x = 60

However, this conflicts with the fact that x < 0 , so this is a false solution.

The only solution is therefore x = 20 .

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Example Question #21 : Solving Equations

Give the solution set of the equation:

$x^{2} =|x| + 6$

Possible Answers:

$x \in\left \{ 2,3\right \}$

$x \in \left \{ -3,3\right \}$

$x \in\left \{ -2,2\right \}$

$x \in\left \{ -3,-2,2,3\right \}$

$x \in\left \{ -3,-2\right \}$

Correct answer:

$x \in \left \{ -3,3\right \}$

Explanation:

Case 1: is nonnegative - that is, |x| = x .

The equation becomes

$x^{2} =x+ 6$

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

$x^{2} -x - 6=0$

(x-3) (x+2 )=0

Either x-3 = 0 , in which case x = 3 ,

or x + 2 = 0 , in which case x = -2 , which is impossible, as we are assuming that is nonnegative.

case 2: is negative - that is, |x| = -x

The equation becomes

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

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

$x^{2} + x - 6=0$

(x-2) (x+3 )=0

Either x-2= 0 , in which case x = 2 , which is impossible since we are assuming that is negative,

or x + 3 = 0 , in which case x = -3 .

x = -3 and x = 3 can both be confirmed as solutions.

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Example Question #21 : Solving Equations

What is the equation of the line that goes through (1,2) and is parallel to y=3x+4 ?

Possible Answers:

y=3x+1

y=3x-1

y=3x+3

y=3x-4

Correct answer:

y=3x-1

Explanation:

First, we have to find the slope from y=3x+4 using the general form y=mx+b. Therefore m=3

Parallel lines have the same slope, so the new line has a slope of and point (1,2) .

Use the point-slope equation: $y-y_{1}=m(x-x_{1})$

y-2=3(x-1)

y-2=3x-3

y=3x-1

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Example Question #22 : Solving Equations

What is the -intercept of 5y+2x=4 ?

Possible Answers:

(2,0)

(4,0)

(0,2)

(0,4)

Correct answer:

(2,0)

Explanation:

To solve for the -intercept, you set the to zero and solve for :

5y+2x=4

5(0)+2x=4

2x=4

x=2

-intercept: (2,0)

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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 #15 : Equations

Example Question #327 : Algebra

Example Question #16 : Equations

Example Question #17 : Equations

Example Question #1415 : Gmat Quantitative Reasoning

Example Question #13 : Equations

Example Question #1411 : Problem Solving Questions

Example Question #21 : Solving Equations

Example Question #21 : Solving Equations

Example Question #22 : Solving Equations

All GMAT Math Resources

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