Question

A can complete a work in 10 days, B in 12 days and C in 15 days. All of them began the work together, but A
had to leave the work after 2 days of the start and B, 3 days before the completion of the work. How long did
the work last?​

Answer 1

A = ------ 10 days, B= -----12 days, C= ----15 days

L.C.M of 10,12,15 = 60unit = Total work

A's 1 day work= 6 unit

B's 1 day work= 5 unit

C's 1 day work = 4 unit

Working together (A+B+C) = 15 unit till 2 days = 15×2=30 unit

Remaining work = 60-30= 30 unit

Now B, left the work 3 days before i.e in the last 3 days only C will work

So, work of C in 3 days = 4×3=12unit

Now remaining work= 30-12=18 unit

When A left the work, (B+C) will work for some days

so (B+C)'s 1 day work = 5+4=9 unit while remaining work = 18 unit

then this 18 unit work will be done by (B+C)

So, time taken by (B+C) = 18/9 = 2 days

Now-

Total time taken to complete the whole work = 2+2+3= 7 days

Answer 2

A completes the work in 10 days

B completes the work in 12 days

C completes the work in 15 days

A's one day's work = 1/10

B's one day's work = 1/12

C's one day's work= 1/15

A,B and C's  one day's work = 1/10+1/12+1/15----------->15/60------->1/4

2(1/10) + x-3/12 + x/15 =1

12+5x-15+4x/60=1------------------------> -3+9x=60---------------------->x=63/9

x=7

The work will last for 7 days.