Question

Two taps running together can fill a tank in $3\frac{1}{13}$ hours. If one tap takes 3 hours more than the other to fill the tank then how much time will each tap take to fill the tank?

Answer 1

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