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Work Problems Help (page 2)

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By — McGraw-Hill Professional
Updated on Sep 27, 2011

Examples

Example 1:

Tammy can wash a car in 40 minutes. When working with Jim, they can wash the same car in 15 minutes. How long would Jim need to wash the car by himself?

Let t represent the number of minutes Jim needs to wash the car alone.

Worker

Quantity

Rate

Time

Tammy

1

1/40

40

Jim

1

1/ t

t

Together

1

1/15

15

 

The equation to solve is 1/40 + 1/ t = 1/15. The LCD is 120 t .

Work Problems

Jim needs 24 minutes to wash the car alone.

Example 2:

Kellie can mow the campus yard in Work Problems hours. When Bobby helps, they can mow the yard in Work Problems hours. How long would Bobby need to mow the yard by himself?

Let t represent the number of hours Bobby needs to mow the yard himself. Kellie’s time is Work Problems or Work Problems . Then her rate is Work Problems

Work Problems

The together time is Work Problems or Work Problems , so the together rate is Work Problems

Work Problems

Worker

Quantity

Rate

Time

Kellie

1

2/5

Bobby

1

1/ t

t

Together

1

2/3

 

The equation to solve is 2/5 + 1/ t = 2/3. The LCD is 15 t .

Work Problems

Bobby needs Work Problems hours or 3 hours 45 minutes to mow the yard by himself.

Find practice problems and solutions at Work Problems Practice Problems - Set 1.

Two Workers, One Worker Joins or Finishes Early

Some work problems require part of the work being performed by one worker before the other worker joins in, or both start the job and one finishes the job. In these cases, the together quantity and one of the individual quantities will not be “1.” Take the time the one worker works alone divided by the time that worker requires to do the job alone, then subtract from 1. This is the proportion left over for both to work together.

Examples

Example 1:

Jerry needs 40 minutes to mow the lawn. Lou can mow the same lawn in 30 minutes. If Jerry works alone for 10 minutes then Lou joins in, how long will it take for them to finish the job?

Because Jerry worked for 10 minutes, he did Work Problems Examples of the job alone. So, there is Work Problems Examples of the job remaining when Lou started working. Let t represent the number of minutes they worked together—after Lou joins in. Even though Lou does not work the entire job, his rate is still 1/30.

Worker

Quantity

Rate

Time

Jerry

1

1/40

40

Lou

1

1/30

30

Together

3/4

Work Problems Examples

t

 

The equation to solve is 1/40 + 1/30 = 3/4 t . The LCD is 120 t .

Work Problems Examples

Together, they will work Work Problems Examples minutes.

Example 2:

A pipe can fill a reservoir in 6 hours. Another pipe can fill the same reservoir in 4 hours. If the second pipe is used alone for Work Problems Examples hours, then the first pipe joins the second to finish the job, how long will the first pipe be used?

The amount of time Pipe I is used is the same as the amount of time both pipes work together. Let t represent the number of hours both pipes are used. Alone, the second pipe performed Work Problems Examples parts of a 4-part job:

Work Problems Examples

Worker

Quantity

Rate

Time

Pipe

I

1

1/6

6

Pipe

II

1

1/4

4

Together

3/8

Work Problems Examples

t

 

The equation to solve is 1/6 + 1/4 = 3/8 t . The LCD is 24 t .

Work Problems Examples

Both pipes together will be used for Work Problems Examples hours or Work Problems Examples minutes. Hence, Pipe I will be used for 54 minutes.

Example 3:

Press A can print 100 fliers per minute. Press B can print 150 fliers per minute. The presses will be used to print 150,000 fliers.

(a) How long will it take for both presses to complete the run if they work together?

(b) If Press A works alone for 24 minutes then Press B joins in, how long will it take both presses to complete the job?

These problems are different from the previous work problems because the rates are given, not the times.

Before, we used Q = rt implies r = Q/t . Here, we will use Q = rt to fill in the Quantity boxes.

(a) Press A’s rate is 100, and Press B’s rate is 150. The together quantity is 150,000. Let t represent the number of minutes both presses work together; this is also how much time each individual press will run.

Press A’s quantity is 100 t , and Press B’s quantity is 150 t . The together rate is r = Q/t = 150,000/ t .

Worker

Quantity

Rate

Time

Press A

100t

100

t

Press B

150t

150

t

Together

150,000

150,000/ t

t

 

In this problem, the quantity produced by Press A plus the quantity produced by Press B will equal the quantity produced together. This gives the equation 100 t + 150 t = 150,000. (Another equation that works is 100 + 150 = 150,000/ t .)

Work Problems Examples

The presses will run for 600 minutes or 10 hours.

(b) Because Press A works alone for 24 minutes, it has run 24 × 100 = 2400 fliers. When Press B begins its run, there are 150,000 – 2400 = 147,600 fliers left to run. Let t represent the number of minutes both presses are running. This is also how much time Press B spends on the run. The boxes will represent work done together.

 

Worker

Quantity

Rate

Time

Press A

100t

100

t

Press B

150t

150

t

Together

147,600

147,600/ t

t

 

The equation to solve is 100 t + 150 t = 147,600. (Another equation that works is 100 + 150 = 147,600/ t .)

Work Problems Examples

The presses will work together for 590.4 minutes or 9 hours 50 minutes 24 seconds. (This is 590 minutes and 0.4(60) = 24 seconds.)

Find practice problems and solutions at Work Problems Practice Problems - Set 2.

More practice problems for this concept can be found at: Algebra Word Problems Practice Test.

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