A can do a piece of work in 15 days and the efficiency of B and C is 12 unit/day and 10 unit/day respectively. Work done by A and C together in 1 day is 4 units more than B and C together. Then find the time taken by B is how much percent more than time taken by A?
Answer & explanation
Correct answer: option 2
B C
Efficiency 12 10
ATQ,
(A+C) - (B+C) = 4
A - B = 4 (A's eff. is 4 unit more than B's)
Hence,
A B
efficiency 16 12
Time taken by A & B 12 16
Total work = Efficiency × Number of days
So ratio of efficiency and time is inverse to each other
Now, more time is taken by B i.e 4R
Required % = \(\frac{4R}{12R}\) × 100 = 33\(\frac{1}{3}\)%