Question

If two pipes function simultaneously, a reservoir will be filled in $12$ hours. One pipe fills the reservoir $10$hours faster than the other. How many hours will be taken for the second pipe to fill the reservoir?

Answer 1

Compute the required hours:

Let us say, a faster pipe takes $x$ hours to fill the reservoir.

The portion of the reservoir that is filled by a faster pipe in one hour $=\frac{1}{x}$.

Time taken by slower pipe to fill reservoir $=(x+10)$ hours.

The part of the reservoir filled by the slower pipe is $=\frac{1}{x+10}$.

The reservoir is filled by both pipes in $12$hours.

The part of the reservoir filled by both pipes in one hour is $\frac{1}{12}$.

Now,

$$\begin{gathered} \Rightarrow \frac{1}{x}+\frac{1}{x+10}=\frac{1}{12} \\ \Rightarrow 12(2x+10)=x(x+10) \\ \Rightarrow x^2–14x–120=0 \\ \Rightarrow x^2–20x+6x–120=0 \\ \Rightarrow x(x–20)+6(x–20)=0 \\ \Rightarrow (x–20)(x+6)=0 \\ \Rightarrow x=20orx=-6 \end{gathered}$$

Therefore, $x=20,$ since, time cannot be negative.

So the faster pipe takes $20$hours and the slower pipes take $10+20=30$ hours, to fill the reservoir.

Hence, the time taken by second pipe is $30$ hours to fill the reservoir.