Question

Two pipes A and B running together can filla tank in 3 1/3 minutes. If pipe B takes 5 minutes more than pipe A to fill the tank separately, then find the time in which each pipe would fill the tank separately.

Answer 1

Given :

• Pipes A and B running together can fill a tank in 3⅓ mins.
• Pipe B takes 5 mins more than pipe A to fill the tank.

To Find :

• Time taken by Pipes A and B to fill the tank separately.

Solution :

Let the time taken by Pipe A to fill the tank be x mins.

Case 1 :

It given that, Pipe B takes 5 mins more to fill the tank as compared to Pipe A.

•°• Time taken by Pipe B to fill the tank separately = (x + 5) mins.

Case 2 :

Tank is filled by both pipes in 3 ⅓ mins.

In 1 min, the pipes together will fill $\sf{\dfrac{1}{3\dfrac{1}{3}}}$ part of the tank.

In 1 min, pipe A will fill $\sf{\dfrac{1}{x}}$ part of the tank.

In 1 min, pipe B will fill, $\sf{\dfrac{1}{x+5}}$ part of the tank.

Pipe A + Pipe B :

$\longrightarrow$ $\sf{\dfrac{1}{x}\:+\:\dfrac{1}{x+5}}$ = $\sf{\dfrac{1}{3\:\dfrac{1}{3}}}$

$\longrightarrow$ $\sf{\dfrac{1}{x}\:+\:\dfrac{1}{x+5}}$ = $\sf{\dfrac{1}{3\:+\:\dfrac{1}{3}}}$

$\longrightarrow$ $\sf{\dfrac{1}{x}\:+\:\dfrac{1}{x+5}}$ = $\sf{\dfrac{1}{\frac{9+1}{3}}}$

$\longrightarrow$ $\sf{\dfrac{2x+5}{x^2+5x}}$ = $\sf{\dfrac{1}{\frac{10}{3}}}$

$\longrightarrow$ $\sf{\dfrac{2x+5}{x^2+5x}}$ = $\sf{1\:\times\:\dfrac{3}{10}}$

$\longrightarrow$$\sf{\dfrac{2x+5}{x^2+5x}\:=\:\dfrac{3}{10}}$

$\longrightarrow$ $\sf{10(2x+5)=3(x^2+5x)}$

$\longrightarrow$ $\sf{20x+50=3x^2+15x}$

$\longrightarrow$ $\sf{3x^2+15x=20x+50}$

$\longrightarrow$ $\sf{3x^2+15x-20x-50=0}$

$\longrightarrow$ $\sf{3x^2-5x-50=0}$

$\longrightarrow$ $\sf{3x^2-15x+10x-50=0}$

$\longrightarrow$ $\sf{3x(x-5) +10(x-5) =0}$

$\longrightarrow$ $\sf{(3x+10)\:\:\:(x-5) =0}$

$\longrightarrow$ $\sf{3x+10=0\:\:\:or\:\:x-5=0}$

$\longrightarrow$ $\sf{3x=-10\:\:or\:\:x=5}$

$\longrightarrow$ $\sf{x=\dfrac{-10}{3}\:\:or\:\:x=5}$

Time cannot be negative.

•°• x = -10/3 is neglected.

$\large{\boxed{\bold{Time\:Taken\:By\:Pipe\:A\:=\:x\:=\:5\:mins}}}$

$\large{\boxed{\bold{Time\:Taken\:By\:Pipe\:B\:=\:x+3\:=\:5+3\:=\:\:8\:mins}}}$