• The time it takes to complete a given task, if two objects or people are working on it.

• The time it takes to empty (fill) an object, if the fill (empty) time, and

combined fill/empty times are known.

• The time a second participant must work, if the total time to complete the job, and the time of the other participant is given.

The quantity in these problems is ONE, i.e. to build one house , to fill one order, to empty one pool, etc.

The rates are a fraction or a part of the task (quantity) that can be completed in one unit of time. If a roofer can roof a house in 12 hrs, then he can roof1/12 th of the house per hour. If Ford motor co. takes X hours to build a car, then they produce 1/X th of a car per hour. If a carpenter takes 23 days to do a remodeling job, then she can do 1/23 rd of a remodeling job per day.

Examples:

1) How long does it take to fill a ditch with dirt by Joe and Sam, if Joe can fill it by himself in 3 hrs and Sam can fill it by himself in 5 hrs.

1. Fill in known items.

2. Use 2 variables for unknown values.

3. Write 2 equations:

a. Q = R•T: 1 = XY

b. Sum of rates = total rate: 1/3 + 1/5 = X; X = 8/15

4. Solve for Y:

1 = XY 1/3 + 1/5 = X

X = 1/Y

Substitute 1/Y for X in 2^nd equation: 1/3 + 1/5 = 1/Y

Multiply both sides of equation by LCM of 15Y: 5Y + 3Y = 15

Let the total number of hours that it takes both of them to complete the job be represented by X. Since work = rate · time, the amount of the job each can complete is equal to their respective rates (1/3 & 1/5) times the total time X. Thus Joe can complete X/3 of the job and Sam can complete X/5 of the job. The sum of how much each can complete is the whole job or 1.

X/3 + X/5 = 1

15(X/3 + X/5) = 15(1)

5X + 3X = 15

8X = 15

X = 1.875 hours = hours

2) How long will it take to fill a tank that was leaking, if a pipe could fill it in 10 hrs if it were not leaking, and the leak could drain it in 12 hrs (with the fill shut off).

1. Fill in known items.

2. Use 2 variables for unknown values.

3. Write 2 equations:

a. Q = R•T: 1 = XY

b. Sum of rates = total rate: 1/10 - 1/12 = X; X = 1/60 Note: The rates are subtracted because they are working against one another.

4. Solve for Y:

1 = XY

1/X = Y

1/(1/60) = Y; Y = 60 hrs <------

Let X represent the total time needed to fill the tank. Since work = rate · time, the amount of work done by the "fill" is X/10, and the "leak" is -X/12 (its’ sign is negative since it is working against the fill). Again, the total work is one job or 1.

X/10 - X/12 = 1

60(X/10 - X/12) = 60(1)

6X - 5X = 60

X = 60 hours

3) A warehouse person can fill an order in 6 hrs. With the assistance of a part-time worker the order can be filled in 4 hrs. How long would it take the part time worker to fill the order alone?

1. Fill in known items.

2. Use 2 variables for unknown values.

3. Write 2 equations:

a. Q = R•T: 1 = Y4; Y = 1/4

b. Sum of rates = total rate: 1/6 + 1/X = Y

4. Solve for X: substitute 1/4 for Y in the 2^nd equation

1/6 + 1/X = 1/4 This equation relates how much of the job is done in 1 hour. This is how it would be solved with a single variable.

1/6 + 1/X = 1/4

2X + 12 = 3X

12= X; X = 12 hrs <------