Question

A can do a piece of work in 40 days. He works at it for 8 days and then B finishes the remaining work in 16 days. How long will they take to complete the work if they do it together?

Answer 1

A can do the work in 40 days but,
According to the question-
A works at it on 8 days
So,work done by A in 1 day=1/8
B can do the remaining work in 16 days
So,work done by B in 1 day=1/16
Total no. of days they need to complete the work together=1/8+1/16
=L.C.M of 8 and 16=16
=2+1/16=3/16 i.e.16/3 days Ans.

Answer 2

Answer:

40/3 DAYS

SOLUTION :-)

Given,

A can do a piece of work in =40 days

So, A’s 1 day's work = 1 / 40

So, A’s 8 day's work = 8 / 40 = 1 / 5

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∴  Remaining work = 1 − 1 / 5 = 4 / 5  

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Now, B can do 4 / 5 piece of work in =16 days

So, B’s 1 day's work =  (4 / 5) / 16 = 4 / 5×16 = 1 / 20

Therefore, (A+B)’s 1 day's work = 1 / 40 + 1 / 20 =  1 + 2 / 40 = 3 / 40

∴   Both (A+B) can do a piece of work in = 1 / (3 / 40) days = 40 / 3 days

=13 3 / 1 days