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 the cistern, find the time in which each pipe would fill the cistern.

Answer 1

Let the volume of the cistern be $V$.

Together two pipes take $3\frac{1}{13}mins=40/13$

Rate of both the pipes together $=\frac{V}{\left(40/13\right)}$

Let pipes be $A$ and $B$,

Time taken by $A=tmins,$ So rate $=V/t$

Time taken by $B=t+3mins$, So rate $=V/(t+3)$

Combined rate $=V/t+V/(t+3)$

we already know that combined rate $=V/\left(40/13\right)$

Equating both,

$V/t+V/(t+3)=V/\left(40/13\right)$

$1/t+1/(t+3)=13/40$

$(t+3+t)/t(t+3)=13/40$

$(2t+3)/(t^2+3t)=13/40$

$80t+120=13t^2+39t$

$13t^2-41t-120=0$

On solving the quadratic equation we get $t=5$ and $t=-1.846,$

since time cannot be negative.

Therefore,

Time taken by pipes $A=5mins$

Time taken by pipes $B=5+3=8mins$