Question

A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously, fill the pool at the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.

Answer 1

Hint: First, we will assume the volume of the pool to be V. Assuming the second pipe to be as random variable as y. So, as the data given in question, the first pipe will take $\left( y+5 \right)$ hours and the third pipe will take $\left( y-4 \right)$ hours. Also, we will assume time in one hour as ‘t’. So, we will get equation as $\left( \dfrac{V}{y+5}+\dfrac{V}{y} \right)t=\left( \dfrac{V}{y-4} \right)t$ because it is said that time taken by first and second pipe simultaneously is same as that of third pipe. Thus, on solving the equation we will get the answer of y.

Complete step-by-step answer:
Here, we will assume the volume of the pool as V. The number of hours required by the second pipe alone to fill the pool alone let say is y.
Now, it is given as the second pipe fills the pool five hours faster than the first pipe. So, the first pipe takes $\left( y+5 \right)$ hours. Also told that the second pipe takes four hours slower than the third pipe. So, we can write the third pipe takes $\left( y-4 \right)$ hours. So, the parts of the pool filled by the first, second and third pipes in one hour are $\dfrac{V}{y+5},\dfrac{V}{y},\dfrac{V}{y-4}$ respectively.
Now, we assume that time taken by the first and second pipe to fill the fill simultaneously be ‘t’ hours. Also, it is said that the third pipe takes the same time as that of the first and second pipe simultaneously. So, in mathematical form we can write it as
$\left( \dfrac{V}{y+5}+\dfrac{V}{y} \right)t=\left( \dfrac{V}{y-4} \right)t$ ……………………………(1)
Now, we will cancel the same terms on the RHS and LHS side. So, we will get as
$\dfrac{1}{y+5}+\dfrac{1}{y}=\dfrac{1}{y-4}$
Now, we will take LCM on the LHS side. We will get as
$\dfrac{y+y+5}{y\left( y+5 \right)}=\dfrac{1}{y-4}$
On further solving, we can write it as
$\left( y+y+5 \right)\left( y-4 \right)=y\left( y+5 \right)$
$\left( 2y+5 \right)\left( y-4 \right)={{y}^{2}}+5y$
$2{{y}^{2}}-8y+5y-20={{y}^{2}}+5y$
Taking all the terms on LHS side, we will get quadratic equation as
$2{{y}^{2}}-{{y}^{2}}+5y-5y-8y-20=0$
${{y}^{2}}-8y-20=0$
Now, here we will split the middle term in order to get factors of this quadratic equation. So, we will get as
${{y}^{2}}-10y+2y-20=0$
$y\left( y-10 \right)+2\left( y-10 \right)=0$
$\left( y+2 \right)\left( y-10 \right)=0$
Thus, we get two factors i.e. $y=-2,y=10$ .
Time cannot be negative so, here the answer will be 10 hours.
So, second pipe will get 10 hours, first pipe will take $y+5=10+5=15hours$ and third pipe will take $y-4=10-4=6hours$ .

Note: Sometimes students make mistake in understanding the statement i.e. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. In this case they assume the first pipe to be say y so, the second pipe will take $\left( y+5 \right)$ hours. Also, the second pipe has $\left( y-4 \right)$ hours when the third pipe is let say y. Now, here the second pipe is filled in $\left( y+5 \right)$ and $\left( y-4 \right)$ hours. So, here students get confused about how to apply the formula and end up getting the incorrect answer. So, understand the statement clearly and then assume the data carefully to avoid mistakes.