Question

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36)Two pipes together can fill a tank in 6 hours 20 minutes. One tap takes
3hours more than the other to fill the tank separately, find the time in which
each tap can separately fill the tank.​

Answer 1

Given :

• Two pipes together can fill a tank in 6 hours *40 minutes.*[Time taken is 6 hrs 40 mins instead of 6 hrs 20 mins]
• One tap takes 3 hours more than the other to fill the tank separately.

To Find :

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

Solution :

Let the time taken by pipe A to fill the tank be x hours.

Case 1 :

It is given that pipe B takes 3 hrs more to fill the tank as compared to pipe A.

•°• Time taken by pipe B to fill the tank separately = (x + 3) hrs.

Case 2 :

The tank can be filled using both pipes together in *6 hrs 40 mins.*

The part of tank which will be filled by both pipes in 1 min,

$\longrightarrow$ $\sf{\dfrac{1}{6\:\:+\:\dfrac{40}{60}}}$

$\longrightarrow$ $\sf{\dfrac{1}{\dfrac{360+40}{60}}}$

$\longrightarrow$ $\sf{\dfrac{1}{\dfrac{400}{60}}}$

$\longrightarrow$ $\sf{\dfrac{1}{\dfrac{20}{3}}}$

$\longrightarrow$ $\sf{1\:\times\:\dfrac{3}{20}}$

$\longrightarrow$ $\sf{\dfrac{3}{20}}$

•°• Pipes A and B together in 1 min will fill 3/20 th part of the tank.

Pipe A can fill $\sf{\dfrac{1}{x}}$ part of tank in 1 min.

Pipe B can fill $\sf{\dfrac{1}{x+3}}$ part of tank in 1 min.

Pipe A + Pipe B :

$\longrightarrow$ $\sf{\dfrac{1}{x}\:+\:\dfrac{1}{x+3}}$ = $\sf{\dfrac{3}{20}}$

$\longrightarrow$ $\sf{\dfrac{x+x+3}{x(x+3)}\:=\:\dfrac{3}{20}}$

$\longrightarrow$ $\sf{\dfrac{2x+3}{x^2+3x}\:=\:\dfrac{3}{20}}$

$\longrightarrow$ $\sf{20(2x+3)=3(x^2+3x)}$

$\longrightarrow$ $\sf{40x+60=3x^2+9x}$

$\longrightarrow$ $\sf{3x^2+9x=40x+60}$

$\longrightarrow$ $\sf{3x^2+9x-40x-60=0}$

$\longrightarrow$ $\sf{3x^2-31x-60=0}$

$\longrightarrow$ $\sf{3x^2\:-\:36x+5x-60=0}$

$\longrightarrow$ $\sf{3x(x-12)\:+5(x-12)=0}$

$\longrightarrow$ $\sf{(x-12) \:\:or\:\:(3x+5)=0}$

$\longrightarrow$ $\sf{x-12=0\:\:or\:\:3x=-5}$

$\longrightarrow$ $\sf{x=12\:\:or\:\:x=\dfrac{-5}{3}}$

Time taken by pipe to fill the tank cannot be negative.

•°• x = -5/3 is neglected.

$\large{\boxed{\bold{Time\:Taken\:By\:Pipe\:A\:=\:x\:=\:12\:hrs}}}$

$\large{\boxed{\bold{Time\:Taken\:By\:Pipe\:B\:=\:x+3\:=\:12+3\:=\:15\:\:hrs}}}$