GMAT Math : Word Problems

Study concepts, example questions & explanations for GMAT Math

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Example Questions

Example Question #1 : Understanding Measurement

Restate 100 miles per hour in feet per second (rounded to the nearest whole number).

Possible Answers:

\displaystyle 74 \textrm{ ft/sec}

\displaystyle 49 \textrm{ ft/sec}

\displaystyle 147 \textrm{ ft/sec}

\displaystyle 68 \textrm{ ft/sec}

\displaystyle 136 \textrm{ ft/sec}

Correct answer:

\displaystyle 147 \textrm{ ft/sec}

Explanation:

One mile is equal to 5,280 feet, and one hour is equal to 3,600 seconds, The speed can be converted as follows:

\displaystyle \frac{100 \textrm{ mi}}{1\textrm{ hr}} = \frac{100 \textrm{ mi } \times 5,280 \textrm{ ft / mi } }{1\textrm{ hr} \times 3,600 \textrm{ sec / hr}}

\displaystyle = 146 \frac{2}{3} \textrm{ ft / sec }

 or, rounded, 147 feet per second.

Example Question #51 : Problem Solving Questions

Convert \displaystyle 45 ^{\circ }\textrm{ C} to Fahrenheit (to the nearest tenth).

\displaystyle F = \frac{9}{5} C + 32

Possible Answers:

\displaystyle 57^{\circ } \textrm{ F}

\displaystyle 113 ^{\circ } \textrm{ F}

\displaystyle 138.6 ^{\circ } \textrm{ F}

\displaystyle -7 ^{\circ } \textrm{ F}

\displaystyle 7.2 ^{\circ } \textrm{ F}

Correct answer:

\displaystyle 113 ^{\circ } \textrm{ F}

Explanation:

The Celsius-to-Fahrenheit conversion formula is:

\displaystyle F = \frac{9}{5} C + 32

Substitute 45 for \displaystyle C:

\displaystyle F = \frac{9}{5} \cdot 45+ 32 = 81 +32 = 113

The answer is \displaystyle 113 ^{\circ } \textrm{ F}.

Example Question #52 : Problem Solving Questions

Convert \displaystyle 80 ^{\circ } F to degrees Celsius (nearest whole degree).

\displaystyle F = \frac{9}{5} C + 32

Possible Answers:

\displaystyle 176^{\circ }C

\displaystyle 202^{\circ }C

\displaystyle 44^{\circ }C

\displaystyle 27 ^{\circ }C

\displaystyle 12^{\circ }C

Correct answer:

\displaystyle 27 ^{\circ }C

Explanation:

If \displaystyle F is the Fahrenheit temperature, then the equivalent Celsius temperature is 

\displaystyle C = \frac{5}{9} (F - 32).

Substitute \displaystyle F = 80 into the equation.

\displaystyle C = \frac{5}{9} (80 - 32)

\displaystyle C = \frac{5}{9} \cdot 48 = \frac{5}{9} \cdot \frac{48}{1} = \frac{5}{3} \cdot \frac{16}{1} = \frac{80}{3} =26 \frac{2}{3}

This rounds up to \displaystyle 27 ^{\circ }C.

Example Question #53 : Problem Solving Questions

Using the conversion factor 2.54 centimeters = 1 inch, rewrite 100 square inches in square centimeters (nearest tenth).

Possible Answers:

\displaystyle 155.0 \textrm{ cm}^{2}

\displaystyle 645.2 \textrm{ cm}^{2}

\displaystyle 1,550.0 \textrm{ cm}^{2}

\displaystyle 6,45 1.6 \textrm{ cm}^{2}

\displaystyle 64.5 \textrm{ cm}^{2}

Correct answer:

\displaystyle 645.2 \textrm{ cm}^{2}

Explanation:

1 inch = 2.54 centimeters, so 1 square inch = \displaystyle 2.54 ^{2} = 6.4516 square centimeters.

100 square inches are equal to \displaystyle 6.4516 \times 100 = 645.16 square centimeters, which rounds to 645.2 square centimeters.

Example Question #53 : Problem Solving Questions

Jason skips a rock across the surface of a pond.  The rock bounces three times over the pond's surface.  If the second bounce is half the distance of the first bounce, and the third bounce is one quarter of the second bounce, and the total distance between bounces is 65 inches, how many inches did the rock travel on the first bounce?

Possible Answers:

\displaystyle 24

\displaystyle 27

\displaystyle 40

\displaystyle 30

\displaystyle 42

Correct answer:

\displaystyle 40

Explanation:

We need to translate the words into a mathematical equation.  Since we need to solve for the distance the rock travels on the first bounce, let's use a variable to represent that distance.

Let "\displaystyle x" = distance traveled on the first bounce.

Notice that we can describe the distances traveled on the 2nd and 3rd bounces in terms of \displaystyle x:

Distance traveled on the 2nd bounce is half the distance of the 1st bounce, so 2nd bounce = \displaystyle \frac{1}{2}x

Distance traveled on the 3rd bounce is 1/4 of the 2nd bounce, so, 3rd bounce = \displaystyle \frac{1}{4}*\frac{1}{2}x=\frac{1}{8}x

The total distance, then, is:

\displaystyle x+\frac{1}{2}x+\frac{1}{8}x=65

\displaystyle \frac{13}{8}x=65

\displaystyle x=65\bigg(\frac{8}{13}\bigg)

\displaystyle x=40

 

Example Question #12 : Measurement Problems

 

 

Three years ago, Anna was four times as old as Jeff. Today Anna is three times as old as Jeff.

How old was Anna three years ago?

Possible Answers:

\displaystyle 6

\displaystyle 18

\displaystyle 21

\displaystyle 24

\displaystyle 27

Correct answer:

\displaystyle 24

Explanation:

First, we set up the variables and the equations. Let:

\displaystyle A= Anna's age three years ago

\displaystyle J = Jeff's age three years ago. So:

\displaystyle A+3 = Anna's age today

\displaystyle J+3 = Jeff's age today

 

Therefore: \displaystyle A = 4J and \displaystyle (A+3) = 3 (J+3). Then we solve for the system of equations:

\displaystyle A - 4J = 0

\displaystyle (A+3) - 3 (J+3) = 0 or \displaystyle A - 3J = 6

When we solve for J first we get \displaystyle J = 6.

Then we solve for A using the first equation of the system:

\displaystyle A = 4J = 4 (6) = 24

Therefore Anna was 24 years old three years ago.

Example Question #51 : Problem Solving Questions

\displaystyle 1 kilometer is approximately \displaystyle 3.9\times10^4 inches. How many kilometers are there in \displaystyle 1,170,000 inches?

Possible Answers:

\displaystyle \frac{10}{39}

\displaystyle \frac{117}{39}

\displaystyle \frac{171}{39}

\displaystyle \frac{17}{4}

\displaystyle 30

Correct answer:

\displaystyle 30

Explanation:

Since

\displaystyle 1 \text{ km}=3.9\times10^4 \text{ inches},

\displaystyle 1 \text{ inch}= \frac{1}{(3.9\times10^4)} \text{ km}

Therefore we multiply 1,170,000 by 1/(3.9*104) to obtain the number of kilometers in 1,170,000 inches.

\displaystyle \frac{1170000}{3.9\times 10^{4}}=\frac{1170}{39}=30

Example Question #13 : Measurement Problems

If one liter of gasoline costs \displaystyle \$1.50, how much would eighteen gallons of gasoline cost? (One gallon is approximately \displaystyle 3.8 liters.)

Possible Answers:

\displaystyle \$96.20

\displaystyle \$88.90

\displaystyle \$108.40

\displaystyle \$102.60

\displaystyle \$100.80

Correct answer:

\displaystyle \$102.60

Explanation:

One gallon of gasoline is about \displaystyle 3.8 liters of gasoline. The number of liters in eighteen gallons is:

\displaystyle 18\times 3.8=68.4

One liter costs \displaystyle \$1.50. Therefore, eighteen gallons (\displaystyle 68.4 liters) will cost:

\displaystyle 68.4\times 1.50=68.4\times 1 +\frac{68.4}{2}=68.4+34.2=102.6=\$102.60

Example Question #58 : Gmat Quantitative Reasoning

If an object travels at a speed of 10 feet per second, how many feet does it travel in half an hour?

Possible Answers:

\displaystyle 6000\:ft

\displaystyle 3000\:ft

\displaystyle 30,000\:ft

\displaystyle 18,000\:ft

\displaystyle 1800\:ft

Correct answer:

\displaystyle 18,000\:ft

Explanation:

If an object travels at 10 feet per second, then in half an hour it will travel 10 feet for every second there is in 30 minutes. We can set this multiplication up as follows:

\displaystyle 10\frac{feet}{sec}\cdot60\frac{sec}{min}\cdot30\:min=18,000\:ft

Here we can see that the units of seconds are cancelled after the first operation, and the units of minutes are cancelled after the second operation, leaving us with the unit of feet for our answer.

Example Question #51 : Word Problems

Fill in the blank:

Possible Answers:

\displaystyle 65

\displaystyle 39

\displaystyle 45

\displaystyle 78

\displaystyle 52

Correct answer:

\displaystyle 52

Explanation:

One pound is equal to 16 ounces. 

Multiply: 

\displaystyle 3 \frac{1}{4} \times 16 = 52

52 ounces is the correct choice.

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