Question

A and B can do a piece of work in $12$ days, B and C in $15$ days while A and C can do in $20$ days. In what time C alone can complete the work?

Answer 1

$A+B\left(1da{y}^{\prime }s\right)=\frac{1}{12}work$

$B+C\left(1da{y}^{\prime }s\right)=\frac{1}{15}work$

$A+C\left(1da{y}^{\prime }s\right)=\frac{1}{20}work$

Adding all the above,

$2(A+B+C)\left(1da{y}^{\prime }s\right)=\frac{1}{12}+\frac{1}{15}+\frac{1}{20}$

$=\frac{5+4+3}{60}=\frac{12}{60}=\frac{1}{5}work$

$\therefore A+B+C\left(1da{y}^{\prime }s\right)=\frac{1}{10}work$

$\therefore C\left(1da{y}^{\prime }s\right)=(A+B+C)\left(1da{y}^{\prime }s\right)-(A+B)\left(ada{y}^{\prime }s\right)$

$=\frac{1}{10}-\frac{1}{12}$

$=\frac{6-5}{60}=\frac{1}{60}$

Hence, C can complete the whole work in 60 days.