Question

Two pipes running together can fill a cistern in $2\frac{8}{11}$ minutes.If one pipe takes 1 minute more than the other to fill the cistern, find the time in which each pipe would fill the cistern alone.

Answer 1

Let the two pipe be A and B

Let time taken by A and B to fill the cistern individually be 'x' and 'x+1' minutes respectively.

Pipe A 1 minute work $=\frac{1}{x}$

Pipe B 1 minute work $=\frac{1}{x+1}$

Pipe A and B combined 1 minute work $=\frac{1}{x}+\frac{1}{x+1}$

They take $\frac{30}{11}$ minutes to fill the cistern together, there in 1 minute they will fill ${\left(\frac{11}{30}\right)}^{th}$ part of tank.

$\frac{1}{x}+\frac{1}{x+1}=\frac{11}{30}$

$\frac{x+1+x}{x(x+1)}=\frac{11}{30}$

$\frac{2x+1}{x^2+x}=\frac{11}{30}$

$30(2x+1)=11(x^2+x)$

$60x+30=11x^2+11x$

$11x^2-49x-30=0$

$(11x+6)(x-5)=0$

Since x cannot be negative

$\therefore x=5$

Hence pipe A will take 5 minutes and B will take 6 minutes.