Question

A can do a piece of work in 10 days, B in 12 days and C in 15 days. All begin together but A leaves the work after 2 days and B leaves 3 days before the work is finished. 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 $=\frac{1}{15}$
A, B and C's one day's work $=\frac{1}{10}+\frac{1}{12}+\frac{1}{15}$
$=\frac{6+5+4}{60}=\frac{15}{60}=\frac{1}{4}$
Let the work completed in x days.
$\therefore$ A's 2 days work + B's (x - 3) days work + C's x days work = 1
$\Rightarrow 2\times \frac{1}{10}+(x-3)\times \frac{1}{12}+x\times \frac{1}{15}=1$
$\Rightarrow \frac{1}{5}+\frac{x-3}{12}+\frac{x}{15}=1$
$=\frac{12+5x-15+4x=60}{60}$
(L.C.M. of 5, 12, 15 = 60)
$\Rightarrow 9x=60-12+15\Rightarrow 9x=63$
$\Rightarrow x=\frac{63}{9}=7$
$\therefore$ Work will last for 7 days.