A can do a piece of work in 18 days. B and C do 15 units and 13 units of the work per day respectively. The work done by A and C together in 1 day is 3 units more than the work done by B and C together in 1 day. By what percent is the time taken by B more than the time taken by A?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 20%
C appears on both sides of the comparison, so it cancels: A does 3 units a day more than B. A's daily units and A's time give the total work, and then B's time.
A \(+\) C \(=\) B \(+\) C \(+3\), so A does \(15+3=18\) units a day
Total work \(=18\times18=324\) units
Time taken by B \(\displaystyle =\frac{324}{15}=21.6\) days
B takes \(21.6-18=3.6\) days more, which is \(\displaystyle \frac{3.6}{18}\times100=20\%\) more than A