# Work-related Problems

**Case 1: Workers have different rates**

Work rate = (1 job done) / (Time to finish the job)

Time of doing the job = (1 job done) / (Work rate)

For example

Albert can finish a job in A days

Bryan can finish the same job in B days

Carlo can undo the job in C days

1/A = rate of Albert

1/B = rate of Bryan

1/C = rate of Carlo

Albert and Bryan work together until the job is done: (1/A + 1/B)t = 1

Albert is doing the job while Carlo is undoing it until the job is done: (1/A - 1/C)t = 1

Problem

Lejon can finish a job in 6 hours while Romel can do the same job in 3 hours. Working together, how many hours can they finish the job?

Solution

Rate of Romel = 1/3

$(1/6 + 1/3)t = 1$

$\frac{1}{2}t = 1$

$t = 2 \, \text{ hours}$ *answer*

**Case 2: Workers have equal rates**

Work done = no. of workers × time of doing the job

To finish the job

If a job can be done by 10 workers in 5 hours, the work load is 10(5) = 50 man-hours. If 4 workers is doing the job for 6 hours, the work done is 4(6) = 24 man-hours. A remaining of 50 - 24 = 26 man-hours of work still needs to be done.

Problem

Eleven men could finish the job in 15 days. Five men started the job and four men were added at the beginning of the sixth day. How many days will it take them to finish the job?

Solution

Work done in 5 days = 5(5) = 25 man-days

Let $x$ = no. of days for them to finish the job

$25 + (5 + 4)(x - 5) = 165$

$25 + 9(x - 5) = 165$

$9x = 185$

$x = 20.56 \, \text{ days}$ *answer*