# Word Problems Algebra Worksheet

Word Problems Algebra Worksheet
• Page 1
1.
The table shows the perimeter and side lengths of three equilateral triangles.
Perimeter of the triangle

2.
A car travels at an average speed of 55 miles/hour for 180 minutes. Find the distance traveled by the car.
 a. 110 miles b. 165 miles c. 55 miles d. 220 miles

#### Solution:

Let x be the distance traveled by the car.

The time taken by the car to travel x miles = 180 minutes

180 minutes = 3 hours
[Convert minutes into hours.]

The average speed of the car = 55 miles/hour.

Distance traveled by the car, x = average speed x time taken

x = 55 x 3
[Replace the variables with the values, given.]

= 165
[Multiply.]

The car traveled a distance of 165 miles in 180 minutes.

3.
The formula for the volume of a rectangular pyramid is V = ($\frac{1}{3}$) $l$$w$$h$. Write a formula to find the height ($h$) of the pyramid.
 a. $\frac{3V}{l}$ units b. $\frac{3V}{lw}$ units c. $\frac{3V}{w}$ units d. $\frac{V}{lw}$ units

#### Solution:

V = (13) x lwh
[Formula.]

3V = (13) x lwh x (3)
[Multiply by 3 on each side.]

3V = lwh
[Simplify.]

3Vlw = lwhlw
[Divide each side by lw.]

3Vlw = h
[Simplify.]

So, the height of the rectangular pyramid is 3V / lw units.

4.
The perimeter of a rectangular wallpaper is 46 inches and its width is 10 inches. Find the length of the rectangular sheet of wallpaper.
 a. 12 inches b. 14 inches c. 13 inches d. 15 inches

#### Solution:

P = 2L + 2w
[Perimeter of a rectangle.]

46 = 2L + 2(10)
[Replace the variables with the values, given.]

46 = 2L + 20
[Multiply.]

46 - 20 = 2L + 20 - 20
[Subtract 20 from each side.]

26 = 2L
[Simplify.]

262 = 2L2
[Divide each side by 2.]

13 = L
[Simplify.]

So, the length of the rectangular wallpaper is 13 inches.

5.
An airplane flew 1182 miles at an average speed of 394 miles/hour. How long does it take to fly the entire distance?
 a. 4 hours b. 3 hours c. 6 hours d. 5 hours

#### Solution:

An airplane flew 1182 miles at an average speed of 394 miles/hour.

Time = DistanceSpeed
[Formula.]

Time = 1182394
[Substitute.]

= 3
[Simplify.]

The airplane took 3 hours to travel 1182 miles.

6.
Arthur left his house at 4:00 p.m. and drove directly to his aunt's house, which is 312 miles away. What was the average speed with which he drove, if he reached his aunt's house at 8:00 p.m.?
 a. 86 miles/hour b. 70 miles/hour c. 78 miles/hour d. 90 miles/hour

#### Solution:

The time taken by Arthur to reach his aunt's house = 8 - 4 = 4 hours

The distance from Arthur's house to his aunt's house = 312 miles

Speed = DistanceTime
[Formula.]

= 3124
[Replace the variables with the values, given.]

= 78
[Simplify.]

The average speed with which Arthur drove is 78 miles/hour.

7.
The average of 41 non-negative integers is 23.4. Find their sum.
 a. 918.4 b. 960.4 c. 959.4 d. 958.4

#### Solution:

The sum of the numbers = Average of the numbers x count of the numbers
[Formula.]

= 23.4 x 41
[Substitute the values.]

= 959.4
[Simplify.]

The sum of the numbers is 959.4.

8.
The formula F = ($\frac{9}{5}$) × C + 32 is used to convert temperature in Celsius (C) scale to that in Fahrenheit (F) scale. Which of the following best suits the conversion of temperature from Fahrenheit scale to Celsius scale?
 a. C = $\frac{5}{9}$ (F - 22) b. C = $\frac{5}{9}$ (F - 32) c. C = $\frac{5}{9}$ (F + 32) d. C = $\frac{5}{9}$ (F - 12)

#### Solution:

F = ( 95 ) x C + 32
[Formula.]

F - 32 = ( 95 ) x C + 32 - 32
[Subtract 32 from each side.]

F -32 = ( 95 ) x C
[Simplify.]

5(F - 32) = 5 x ( 95 ) x C
[Multiply by 5 on each side.]

5(F - 32) = 9C
[Simplify.]

59 (F - 32) = 9C9
[Divide by 9 on each side.]

59 (F - 32) = C
[Simplify.]

The formula C = 5 / 9 (F - 32) is used to convert temperature from Fahrenheit scale to Celsius scale.

9.
Find the celsius temperature that corresponds to a room temperature of 50oF by using the formula F = ( $\frac{9}{5}$ ) C + 32.
 a. 6o b. 4o c. 10o d. 8o

#### Solution:

F = ( 95 ) x C + 32
[Formula.]

50 = ( 95 ) x C + 32
[Replace the variable with the value, given.]

50 - 32 = ( 95 ) x C + 32 - 32
[Subtract 32 from each side.]

18 = ( 95 ) x C
[Simplify.]

(5)(18) = 5 x ( 95 ) x C
[Multiply by 5 on each side.]

90 = 9C
[Simplify.]

909 = 9C9
[Divide each side by 9.]

10 = C
[Simplify.]

The Celsius temperature corresponding to 50o F is 10oC.

10.
Convert 25oC into Fahrenheit scale by using the formula, F = ( $\frac{9}{5}$ ) C + 32.
 a. 77oF b. 82oF c. 72oF d. 85oF

#### Solution:

F = ( 95 ) x C + 32
[Original formula.]

F = ( 95 ) x 25 + 32
[Substitute 25 for C.]

= (9)(5) + 32
[Divide.]

= 45 + 32
[Multiply.]

= 77
[Simplify.]

25oC is equal to 77oF.