Q) A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?

a) 30 days              b) 40 days             c) 60 days                d) 70 days

Ans:

Method 1:

Let’s consider A completes the work in A days and B completes the work in B days

∴ A’s 1 day’s work = 1/A

and B’s 1 day’s work = 1/B

∴ (A + B)’s 1 day’s work = 1/A + 1/B = (A + B) / AB

Given that A & B together take 30 days

∴ (A + B)/AB = 1/30

∴ 30 (A + B) = AB ……….. (i)

A’s 16 days’ work = 16 x 1/A = 16/A

Balance work = 1 – 16/A = (A – 16)/A

This balance work is completed by B

B completed 1/B work in 1 day

B will complete (A – 16)/A work in = [(A – 16)/A] /(1/B) = (A – 16)B/A days

(A – 16)B/A = 44 (given)

A B – 16 B = 44 A …………(ii)

By substituting value of AB in equation (ii), we get:

30 (A + B) – 16 B = 44 A

∴ 14 A = 14 B

∴ A = B

From equation (i), we get: 30 (2 B) = B x B

∴ B = 60 days

Method 2:

A’s 1 day’s work = A

B’s 1 day’s work = B

A + B = 1/30

∴ 30 A + 30 B = 1 ….(i)

Given that A for 16 days and B worked for 44 days to complete the work

∴ 16 A + 44 B = 1 …. (ii)

From equation (i) & (ii), we get:

14 A – 14B = 0 => A = B

From equation (i), we get:

30 A  + 30 B = 1

∴ 60 B = 1

∴ B = 1/60

Since Time taken by B to finish 1/60 work = 1 day

∴ Time taken by B to finish whole work = 1 /(1/60) = 60 days