Question

Two pipes running together can fill a cistern in $3\frac{1}{13}$ minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Answer 1

$$\begin{gathered} \text{Let one pipe fills the cistern in}x\text{mins.} \\ \mathrm{Therefore},\mathrm{the}\mathrm{o}\text{ther pipe will fill the cistern in (}x+3)\text{mins.} \\ \text{Time taken by both, running together, to fill the cistern = 3}\frac{1}{13}\text{mins =}\frac{40}{13}\text{mins} \\ \text{Part filled by one pipe in 1 min =}\frac{1}{x} \\ \text{Part filled by the other pipe in 1 min =}\frac{1}{x+3} \\ \text{Part filled by both pipes, running together, in 1 min =}\frac{1}{x}+\frac{1}{x+3} \\ \therefore \frac{1}{x}+\frac{1}{x+3}=\frac{1}{{\displaystyle \raisebox{1ex}{$40$}\!\left/ \!\raisebox{-1ex}{$13$}\right.}} \\ \Rightarrow \frac{(x+3)+x}{x(x+3)}=\frac{13}{40} \\ \Rightarrow \frac{2x+3}{x^2+3x}=\frac{13}{40} \\ \Rightarrow 13x^2+39x=80x+120 \\ \Rightarrow 13x^2-41x-120=0 \\ \Rightarrow 13x^2-(65-24)x-120=0 \\ \Rightarrow 13x^2-65x+24x-120=0 \\ \Rightarrow 13x(x-5)+24(x-5)=0 \\ \Rightarrow (x-5)(13x+24)=0 \\ \Rightarrow x-5=0\text{or 13}x\text{+ 24 = 0} \\ \text{⇒}x\text{= 5 or}x\text{=}\frac{-24}{13} \\ \Rightarrow x=5(\because \text{Speed cannot be a negative fraction}) \\ \mathrm{Thus},\mathrm{o}\text{ne pipe will take 5 mins and the other will take {}(5+3)=8\}\text{mins to fill the cistern}\text{.} \end{gathered}$$