A, B and C working together completed a job in 12 days. However A only worked for the first four days when 36% of the job was done. If the work completed by B in 3 days equals to the work done by C in 1day, then how many days will the fastest worker took to complete the same work alone?

20

A + B + C = 12 days

Let, the total work = 100

ATQ,

⇒ 3B = 1C

⇒ B  :  C = 1  :  3  (efficiency)

Work done by (B + C) in 4 days = 4 × (1 + 3) = 16 days

Work done by A + B + C in 4 days = 36% of 100 = 36

Now,

4 (A + B + C) = 36

4A = 36 - 16 = 20

A's efficiency = 5

Efficiency of A is greater than B & C, thus A is the fastest worker.

⇒ A will take \(\frac{100}{5}\) = 20 days to complete the work.