Question

A tank is filled by 3 pipes with each pipe having uniform flow. The first two pipes operating simultaneously fill the tank in same time during in which a tank is filled by third pipe alone. The second pipe fills the tank 5 hour faster than first pipe and 4 hour slower than third pipe. Time required by first pipe to fill tank is.......

Answer 1

QUESTION DETAIL

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank ?

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

As per question, we get 

1x+1x−5=1x−9=>x−5+xx(x−5)=1x−9=>(2x−5)(x−9)=x(x−5)=>x2−18x+45=01x+1x−5=1x−9=>x−5+xx(x−5)=1x−9=>(2x−5)(x−9)=x(x−5)=>x2−18x+45=0

After solving this euation, we get 

(x-15)(x+3) = 0,

As value can not be negative, so x = 15

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Answer 2

Let the Capacity of the Tank to be filled be : T

Let the Time taken by the Third pipe to fill the Tank be : P hours

Given : Second pipe fills the tank 4 hours slower than Third pipe

$:\implies$  Time taken by Second pipe to fill the tank : (P + 4) hours

$\mathsf{:\implies Volume\;of\;Tank\;filled\;by\;Second\;Pipe\;in\;one\;hour = \dfrac{T}{P + 4}}$

Given : Second pipe fills the tank 5 hours faster than First pipe

$:\implies$  First pipe fills the tank 5 hours slower than Second pipe

$:\implies$  Time taken by 1st pipe to fill the tank : (P + 4 + 5) = (P + 9) hrs

$\mathsf{:\implies Volume\;of\;Tank\;filled\;by\;First\;Pipe\;in\;one\;hour = \dfrac{T}{P + 9}}$

Given : First two pipes are operating simultaneously

☯  Volume of Tank filled by First two pipes in one hour :

$\mathsf{\implies \dfrac{T}{P + 4} + \dfrac{T}{P + 9}}$

Given : First two pipes operating simultaneously fill the tank in same time during which the tank is filled by third pipe alone.

We assumed that : Time taken by the Third pipe to fill the Tank as P hours. It means First two pipes operating simultaneously fill the tank in P hours.

$\mathsf{\implies P\bigg[\dfrac{T}{P + 4} + \dfrac{T}{P + 9}\bigg] = T}$

$\mathsf{\implies \dfrac{P}{P + 4} + \dfrac{P}{P + 9} = 1}$

$\mathsf{\implies \dfrac{P(P + 9) + P(P +4)}{(P + 4)(P + 9)} = 1}$

$\mathsf{\implies P^2 + 9P + P^2 +4P = (P + 4)(P + 9)}$

$\mathsf{\implies 2P^2 + 13P = P^2 + 9P + 4P + 36}$

$\mathsf{\implies 2P^2 + 13P = P^2 + 13P + 36}$

$\mathsf{\implies P^2 = 36}$

$\mathsf{\implies P = \sqrt{36}}$

$\mathsf{\implies P = \sqrt{(\pm6)^2}}$

$\mathsf{\implies P = \pm6}$

$\mathsf{\implies P = 6\;\;(or)\;-6}$

$\mathsf{As\;Time\;cannot\;be\;Negative \implies P \neq -6}$

$\implies \textsf{Time taken by the Third pipe to fill the Tank = 6 hours}$

$\implies \textsf{Time taken by the First pipe to fill the Tank = (P + 9) hours}$

$\implies \textsf{Time taken by the First pipe to fill the Tank = (6 + 9) hours}$

$\textbf{Answer : }\textsf{Time taken by the First pipe to fill the Tank = \boxed{\textsf{15 hours}}}$