Question

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone. The second pipe fills the tank five hour faster than the first pipe and four hour slower than the third. The time required by the first pipe to fill the tank is:
$\left( a \right)$ 6 hrs.
$\left( b \right)$ 10 hrs.
$\left( c \right)$ 15 hrs.
$\left( d \right)$ 30 hrs.

Answer 1

Hint: In this particular question use the concept that assume any variable (say x) be the time required in hours to fill the tank by first pipe, so the time required to fill the tank by the second pipe is (x – 5) as second pipe fills the tank 5 hour faster than the first pipe, and accordingly for the third pipe so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
A tank is filled by three pipes with uniform flow.
Let the time required in hours to fill the tank by first pipe be x hrs.
Now it is given that the second pipe fills the tank 5 hours faster than the first pipe, so the time taken by the second pipe to fill the tank = x – 5 hours.
Now it is also given that the second pipe fills the tank 4 hours slower than the third i.e. we can say the third pipe fills the tank four hour earlier than the second pipe.
So the time taken by the third pipe to fill the tank – (x – 5) – 4 = x – 9 hours.
Now according to question we have,
The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone.
So the work done by the first pipe and the second pipe together is equal to the work done by the third pipe alone.
So, work done by first pipe + work done by second pipe = work done by third pipe.
Let the work be 1.
So the work done by the first pipe is the ratio of the work to the time taken by the first pipe, similarly for the other two pipes.
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{{x - 5}} = \dfrac{1}{{x - 9}}$
Now simplify we have,
$ \Rightarrow \dfrac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \dfrac{1}{{x - 9}}$
$ \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right)$
$ \Rightarrow 2{x^2} - 23x + 45 = {x^2} - 5x$
$ \Rightarrow {x^2} - 18x + 45 = 0$
Now factorize this we have,
$ \Rightarrow {x^2} - 3x - 15x + 45 = 0$
\[ \Rightarrow x\left( {x - 3} \right) - 15\left( {x - 3} \right) = 0\]
\[ \Rightarrow \left( {x - 3} \right)\left( {x - 15} \right) = 0\]
\[ \Rightarrow x = 3,15\]
So when x = 3,
Time taken by the second pipe = x – 5 = 3 – 5 = -2 hours which is not possible so we have to neglect x = 3.
So when x = 15,
Time taken by the second pipe = x – 5 = 15 – 5 = 10 hours
And the time taken by the third pipe = x – 9 = 15 – 9 = 6 hours.
So the time taken by the first pipe to fill the tank is 15 hours.
Hence option (c) is the correct answer.

So, the correct answer is “Option C”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the work done by first pipe is the ratio of the work to the time take by the first pipe, similarly for the other two pipes, so simplify substitute the values as above and simplify we will get the two values of x, so check which value of x is a valid case and which is not as above, valid case is the required answer.