Question

2. A can do a piece of work in 12 days, B can do the
same work in 15 days and A, B and C together
dan finish the work in 5 days. In how many days
will C alone complete the work?​

Answer 1

Given A can do the work in 10 days ,B can do the work in 12 days and C can do the work in 15 days.

Then A's one day's work = \frac{1}{10}

          B's one day's work = \frac{1}{12}

          C's one day's work = \frac{1}{15}

Then (A+B +C)'s one day's work = \frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{450}{1800} = \frac{1}{4}

(A+B+C)'s  two days' work = \frac{1}{4} ×2= \frac{1}{2}

But B leaves 3 days before the work gets finished, so C does the remaining  work alone

C's 3  days' work = \frac{1}{15} ×3= \frac{1}{5}

Then work done in 2+3 days = \frac{1}{2} + \frac{1}{5} = \frac{7}{10}

Work done by B+C together  =1− \frac{7}{10} = \frac{3}{10}

(B + C)'s one day work = \frac{1}{1} + \frac{1}{15} = \frac{3}{20}

So number of days worked by B and C together= \frac{3}{10} × \frac{20}{3} =2 days

Then total work done =2+3+2=7 days

Answer 2

$\huge \tt \blue {AnSWER}$

A+B = 12

B+C = 15

A+C = 20

Lcm of (12,15,20) = 60

Efficiency is

A+B = 5

B+C = 4

A+C = 3

Total efficiency

2 (A+B+C) = 12

A+B+C = 6

Individual efficiency

A = 6 - 4 = 2

B = 6 - 3 = 3

C = 6 - 5 = 1

Individual time taken

A = 60/2 = 30 days

B =60/3 = 20 days

C = 60/1 = 60 days