Question

1. A can do a piece of work in 120 days and B can do the work in 150 days. They work together for 20 days. Then A leaves and B continues the work. 12 days after that, C joins the work and the work is completed in 48 more days. In how many days C can do it alone? ​

Answer 1

Answer:

67 is your answer

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Answer 2

Answer:

I have tried the problem using LCM approach.

A→ 120

B→ 150

Total work = LCM (120,150) = 600

So,

A’s one day work will be 600/120 = 5

B’s one day work will be 600/150 = 4

Now, A+B work together for 20 days:

(A+B)’s total efficiency will be 5+4 = 9

Since they work together for 20 days, 9*20 = 180 which is the total work done.

Remaining work = 600–180 = 420.

B leaves, A alone continues the work for another 12 days:

A’s efficiency is 5 and he works for 12 days. So, 5*12 = 60 which is the work completed by A in 12 days.

Now, the remaining work will be 420 -60 = 360.

Then, A+C work for 48 days:

So, Total work / (efficiency of A+C) is 360/ 48 = 7.5 which is (A+C)’s one day work.

As we already know, A’s efficiency is 5,

C’s efficiency will be 7.5 - 5 = 2.5

Hence,

C alone complete the work in 600/ 2.5 = 240 days.

Hope this helps!

Cheers!