Question

A can do a piece of work in 12 days, B in 15 days and C in 18 days. All begin to work together. A leaves after 2 days B leaves before 3 days how long did the work last​

Answer 1

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

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

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

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

$= > \frac{450}{1800} = \frac{1}{4}$

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

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

$C's \:3  days' work  = \frac{1}{15} \times 3$

$=> \frac{1}{5}$

$Then \: work \: done \: in \: 2+3 \: days = \frac{1}{2} + \frac{1}{5}$

$=> \frac{7}{10}$

$work \: done \: by \: B+C \: together = 1 - \frac{7}{10}$

$=> \frac{3}{10}$

$(B + C)'s \: one \: day \: work = \frac{1}{12} + \frac{1}{15}$

$= > \frac{3}{20}$

So number of days worked by B and C

$together = \frac{3}{10} \times \frac{20}{3} = 2 \: days$

Then total work done =2+3+2=7 days

Answer 2

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

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

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

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

$= > \frac{450}{1800} = \frac{1}{4}$

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

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

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

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