Question

A, B and C can complete a work in 10, 12 and 15 days respectively. All three of them starts together but after 2 days A leaves the job and B left the job 3 days before the work was completed. C completed the remaining work alone. In how many days was the total work completed?

A) 5 B) 6 C) 7 D) 8

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

→ LCM of 10,12 and 15 = 60units = Let Total work.

Than,

→ Efficiency of A = (Total work) / (Total Days) = (60/10) = 6 units/day.

→ Efficiency of B = (Total work) / (Total Days) = (60/12) = 5 units/day.

→ Efficiency of C = (Total work) / (Total Days) = (60/15) = 4 units/day.

Now, we have given that, all started work and after 2 days A leaves the job.

So,

We can say that, for first 2 days all three work together .

So,

→ in first 2 days all completed = 2(6 + 5 + 4) = 30 units of work.

____________

→ Left work now = 60 - 30 = 30 units.

Now, we have given that, B left the job 3 days before the work was completed.

So,

we can say that, As A already left, only C work alone for last three days.

Therefore,

→ C completed in last 3 days = 3 * 4 = 12 units .

_____________

→ Left work now = 30 - 12 = 18 units.

And, now, we can say that, As A already left, this was done by (B + C) together,

So,

→ No. of Days by (B+C) = 18/(5+4) = 18/9 = 2 Days.

______________

Hence,

→ Total Number of Days = 2(A + B + C) + 2(B + C) + 3(C) = 7 Days.(Ans.)

∴ The Total work was completed in 7 Days.