Question

A swimming pool is fitted with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool at the same time as that taken by the third pipe alone. The second pipe fills the pool 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the third pipe to fill the pool.
(A) 10 hours
(B) 6 hours
(C) 16 hours
(D) 5 hours

Answer 1

Hint: Assume that the time taken by the first pipe to fill the pool is x hours. It is given that the second pipe fills 5 hours faster and the third pipe fills 4 hours faster than the second pipe. So, the time taken by the second pipe and the third pipe should be \[\left( x-5 \right)\] hours and \[\left( x-9 \right)\] hours respectively. Now, calculate the part of the pool filled in a unit time by the first pipe, the second pipe, and the third pipe. The part of the pool filled in a unit time by the first pipe and the second pipe is equal to the part of the pool filled in a unit time by the third pipe. Now, solve it further and get the value of x. Take only those values of x for which the time taken by the second pipe and the third pipe cannot be equal to zero.

Complete step-by-step answer:
First of all, let us assume that the first pipe fills the swimming pool in x hours.
The time taken by the first pipe to fill the pool = x hours ………………………………..(1)
The part of the pool filled in a unit time by the first pipe = \[\dfrac{1}{x}\] ……………………………(2)
It is given that the second pipe fills the pool 5 hours faster than the first pipe.
From equation (1), we have the time taken by the first pipe to fill the pool.
The times taken by the second pipe to fill the pool = \[\left( x-5 \right)\] hours ……………………………..(3)
The part of the pool filled in a unit time by the second pipe = \[\dfrac{1}{\left( x-5 \right)}\] ……………………………(4)
It is given that the third pipe fills the pool 4 hours faster than the second pipe.
From equation (3), we have the time taken by the second pipe to fill the pool.
The times taken by the third pipe to fill the pool = \[\left( x-5-4 \right)=\left( x-9 \right)\] hours ……………………………..(5)
The part of the pool filled in a unit time by the third pipe = \[\dfrac{1}{\left( x-9 \right)}\] ……………………………(6)
It is also given that the first pipe and the second pipe operating simultaneously fill the pool at the same time as that taken by the third pipe alone. It means that the part of the pool filled in a unit time by the first pipe and the second pipe is equal to the part of the pool filled in a unit time by the third pipe.
From equation (2), equation (4), and equation (6), we have the part of the pool filled in a unit time by the first pipe, the second pipe, and the third pipe respectively. So,
\[\begin{align}
  & \Rightarrow \dfrac{1}{x}+\dfrac{1}{\left( x-5 \right)}=\dfrac{1}{\left( x-9 \right)} \\
 & \Rightarrow \dfrac{x-5+x}{x\left( x-5 \right)}=\dfrac{1}{\left( x-9 \right)} \\
 & \Rightarrow \left( x-9 \right)\left( 2x-5 \right)=x\left( x-5 \right) \\
 & \Rightarrow 2{{x}^{2}}-5x-18x+45={{x}^{2}}-5x \\
 & \Rightarrow 2{{x}^{2}}-{{x}^{2}}-18x+45=0 \\
 & \Rightarrow {{x}^{2}}-15x-3x+45=0 \\
 & \Rightarrow x\left( x-15 \right)-3\left( x-15 \right)=0 \\
 & \Rightarrow \left( x-3 \right)\left( x-15 \right)=0 \\
\end{align}\]
So, \[x=3\] and \[x=15\] .
From equation (3) and equation (5), we have the time taken by the first pipe and the second pipe to fill the pool.
If we take \[x=3\] , then the time taken by the second pipe and the third pipe will be negative. So, \[x=3\] is not possible.
Now, putting the value of x in equation (1), equation (3), and equation (5), we get
The time taken by the first pipe to fill the pool = 15 hours.
The time taken by the second pipe to fill the pool = \[\left( 15-5 \right)\] hours = 10 hours.
The time taken by the third pipe to fill the pool = \[\left( 15-9 \right)\] hours = 6 hours.
So, the correct answer is “Option B”.

Note: In this question, one might assume the time taken by the first pipe to fill the pool equal to x hours and then take the time taken by the second pipe and the third pipe equal to \[\left( x+5 \right)\] hours and \[\left( x+9 \right)\] hours respectively. This is wrong. Since the second pipe fills 5 hours faster and the third pipe fills 4 hours faster than the second pipe so, the time taken by the second pipe and the third pipe should be \[\left( x-5 \right)\] hours and \[\left( x-9 \right)\] hours respectively.