Question

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

Answer 1

$\text{A}$ and $\text{B}$ do a work in$=12$days

$\text{A}$ and $\text{B}$ do work in $1$ day $=\frac{1}{12}$

$\text{B}$ and $\text{C}$ do a work in $=15$ days

$\text{B}$ and $\text{C}$ do a work in $1$ day $=\frac{1}{15}$

$\text{A}$ and $\text{C}$ do a work in $=20$ days

$\text{A}$ and $\text{C}$ do a work in $1$ day $=\frac{1}{20}$

So,

$\text{A+B=}\frac{1}{12}\text{→}\overline{)1}$

$\text{B+C=}\frac{1}{15}\text{→}\overline{)2}$

$\text{A+C=}\frac{1}{20}\text{→}\overline{)3}$

By adding all these three equations we get,

$\text{A+B+B+C+A+C=}\frac{1}{12}\text{+}\frac{1}{15}\text{+}\frac{1}{20}$

$$\begin{gathered} \text{2(A+B+C)=}\frac{1}{12}\text{+}\frac{1}{15}\text{+}\frac{1}{20}\text{=}\frac{1\times 5+1\times 4+1\times 3}{60}\text{=}\frac{5+4+3}{60}\text{=}\frac{12}{60}\text{=}\frac{1}{5} \\ (\text{A+B+C)=}\frac{1}{2\times 5}\text{=}\frac{1}{10} \end{gathered}$$

Now $\text{C}$ work alone in one day

$$\begin{gathered} =(\text{A+B+C)-(A+B) (one day work)} \\ =\frac{1}{10}-\frac{1}{12}=\frac{1\times 6-1\times 5}{60}=\frac{6-5}{60}=\frac{1}{60} \end{gathered}$$

From here we say that $\text{C}$ completes the work in $60$ days.

Hence $\text{C}$ completes the whole work alone in $60$ days .