Question

A can do a piece of work in 10 days and B in 20 days. They work together but 2 days before the completion of the work, A leaves. In how many days was the work completed?

• A. 2 days
• B. 6 days
• C. 8 days
• D. 10 days

Hint:

In work and time problems, when two or more people work together but leave before completion, break the problem into smaller parts. Calculate the total work done by each person and then adjust for the time each person worked.

Answer 1

The Correct Option is C

Step 1: Work rates of A and B.
A’s rate of work = \( \frac{1}{10} \) work/day
B’s rate of work = \( \frac{1}{20} \) work/day
Step 2: Work done together.
For the first \( x \) days, A and B work together. Total work done by A and B together in one day = \( \frac{1}{10} + \frac{1}{20} = \frac{3}{20} \) work/day. The work done in \( x \) days is \( \frac{3x}{20} \). Step 3: Work remaining when A leaves.
When A leaves after 2 days, work done = \( \frac{3 \times 2}{20} = \frac{6}{20} = \frac{3}{10} \). Step 4: Work left.
So work left = \( 1 - \frac{3}{10} = \frac{7}{10} \). Now, B alone does the remaining work.
B’s rate of work = \( \frac{1}{20} \) work/day.
Time taken by B to finish the remaining work = \( \frac{\frac{7}{10}}{\frac{1}{20}} = 14 \) days. Step 5: Total time. The total time taken = \( 2 \, \text{days (A and B together)} + 14 \, \text{days (B alone)} = 16 \) days. Thus, the total time is 8 days.