Question

Two pipes A and B can fill a tank in 15 hours and 20 hours respectively, while a third pipe C can empty the full tank in 25 hours. All the three pipes are opened in the beginning. After 10 hours, C is closed in how much time will the tank be full?
(a) 12 hours
(b) 10 hours
(c) 9 hours
(d) 8 hours.

Answer 1

Hint: We start solving the problem by assigning a value for the total capacity of the tank. We then find the portion of the tank filled if only pipe A and only pipe B are opened in one hour. We then find the portion of the tank emptied if only pipe C is opened for one hour. We then find the portion of the tank filled if all the pipes are opened for one hour. We then multiply this with 10 hours and find the remaining portion present in the tank. We then find the portion of the tank filled in one hour if both pipes A and B are opened. We use this value to find the hours required to fill the whole tank.

Complete step-by-step solution:
According to the problem, we have two pipes A and B which can fill a tank in 15 hours and 20 hours, and the third pipe C can empty the full tank in 25 hours.
Let us assume the total capacity of the tank be x.
Let us find the part of the tank that pipe A can fill in 1 hour. Let us assume that the pipe A fills y part of the tank in 1 hour. So, we get $15\times y=x$.
$\Rightarrow y=\dfrac{x}{15}$ -----------(1).
Now, let us find the part of the tank that the pipe B can fill in 1 hour. Let us assume that the pipe A fills z part of the tank in 1 hour. So, we get $20\times z=x$.
$\Rightarrow z=\dfrac{x}{20}$ ------------(2).
Now, let us find the part of the tank that the pipe C can empty in 1 hour. Let us assume that the pipe C fills p part of the tank in 1 hour. So, we get $25\times p=x$.
$\Rightarrow p=\dfrac{x}{25}$ ------------(3).
Let us find the part of the tank that is filled in one hour if all the pipes are opened. So, we get $\dfrac{x}{20}+\dfrac{x}{15}-\dfrac{x}{25}$.
$\Rightarrow \dfrac{x}{20}+\dfrac{x}{15}-\dfrac{x}{25}=\dfrac{15x+20x-12x}{300}$.
$\Rightarrow \dfrac{x}{20}+\dfrac{x}{15}-\dfrac{x}{25}=\dfrac{23x}{300}$.
So, the tank is filled $\dfrac{23x}{300}$ in one hour if all the pipes are opened. Let us find the part of the tank that is filled in 10 hours. So, we multiply $\dfrac{23x}{300}$ with 10.
We have $\dfrac{23x}{300}\times 10=\dfrac{23x}{30}$.
We have found that the $\dfrac{23x}{30}$ of the tank is filled. Let us find the remaining space present in the tank.
So, we have $x-\dfrac{23x}{30}=\dfrac{30x-23x}{30}$.
$\Rightarrow x-\dfrac{23x}{30}=\dfrac{7x}{30}$--------(4).
Let us find the part of the tank that can be filled if pipes A and B are opened. So, we have $\dfrac{x}{20}+\dfrac{x}{15}$.
$\Rightarrow \dfrac{x}{20}+\dfrac{x}{15}=\dfrac{3x+4x}{60}$.
$\Rightarrow \dfrac{x}{20}+\dfrac{x}{15}=\dfrac{7x}{60}$ -------(5).
Let us find the total time required to fill the remaining $\dfrac{7x}{30}$ of the tank if the pipes A and B are open.
So, we get $\dfrac{\dfrac{7x}{30}}{\dfrac{7x}{60}}=2$.
We need two hours to fill the remaining portion of the tank after 10 hours.
So, the total time required is $\left( 10+2 \right)=12$hours.
We have found that 12 hours is required to fill the whole tank.
$\therefore$The total time required to fill the whole tank is 12 hours.
The correct option for the given problem is (a).

Note: We can also solve this problem by assuming the total capacity of the tank as 1 unit. But it will create unnecessary confusion while making calculations. Whenever we get this type of problem, it is preferable to assume the variables for every unknown present or required in the problem. We can also expect problems to find the total time required if only one of the pipes is opened for filling and the pipe for emptying is opened.