## How to Solve Work Problems Part 3

This is the third part of the Solving Work Problems Series. The first part of this series discussed in detail the concept behind work problems and the second part discussed the basic work problems and their solutions.

In this post, we discuss two more work problems. The first problem is about two persons who started to work together and after a while, the other person stopped. The second problem is about filling a pool whose outlet pipe is left open.

Sample Problem 4

Jack can dig a ditch alone in 5 days, while John alone can do it in 8 days. The two of them started working together, but after two days, Jack left the job. How many more days do John need to work to finish the job alone?

Solution

Jack can finish the job in 5 days, so he can finish 1/5 of the job in 1 day.

John can finish the job in 8 days, so he can finish 1/8 of the job in 1 day.

In 1 day, both of them can finish 1/5 + 1/8 of the job.

Since both of them worked for two days, they have worked 2(1/5) + 2(1/8) before Jack left.
Now, let x be the number of days John need to finish the job, so he needs to work 1/8(x) more days. The whole job is equal to 1, so we form the following equation. $2(\frac{1}{5}) + 2(\frac{1}{8}) + \frac{1}{8}(x) = 1$ $\frac{2}{5} + \frac{2}{8} + \frac{1}{8}x = 1$

Multiplying both sides by 40, the least common denominator of the three fractions, we have $40(\frac{2}{5}) + 40(\frac{2}{8}) + 40(\frac{1}{8}x) = 40(1)$ $\frac{80}{5} +\frac{80}{8} + \frac{40}{8}x = 40$ $16 + 10 + 5x = 40$ $26 + 5x = 40$ $5x = 40 - 26$ $5x = 14$ $x = \frac{14}{5}$.

So, John still needs $\frac{14}{5}$ days or about $2 \frac{4}{5}$.

Sample Problem 5

A swimming pool can be filled with water using an inlet pipe in 6 hours. It can be emptied using an outlet pipe in 8 hours. One day, after emptying the pool, the owner opened the inlet pipe but forgot to close the outlet pipe. How many hours will it take to fill the pool with both pipes open?

Solution

The inlet pipe can fill the pool in 6 hours, so it can fill 1/6 of the pool in 1 hour.

The outlet pipe can empty the pool in 8 hours, so it can empty 1/8 of the pool in 1 hour.

If both the inlet and outlet pipes are opened, then in 1 hour the pool is 1/6 filled but emptied 1/8 of water. So, the remaining water is 1/6 – 1/8.

Therefore, if we let x be the number of hours, then in x hours, the pool is filled with $\frac{1}{6}x - \frac{1}{8}x$.

So, to fill the entire pool, we equate the preceding equation with 1 (can you see why?). $\frac {1}{6}x - \frac{1}{8}x = 1$.

Multiplying both sides of the equation by 24 which is the least common denominator of 1/6 and 1/8, we have $\frac{24}{6}x - \frac{24}{8}x = 24(1)$ $4x - 3x = 24$ $x = 24$.

Therefore, with both pipes open, the pool can be filled in 24 hours.