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 the second pipe take to fill the reservoir?

Answer

Two pipes can fill the, reservoir in = 12 hours
Let first pipe can fill the reservoir in = x hrs
Then second pipe will fill it in = (x - 10) hours
Now according to the condition,
$\frac{1}{\text{x}}+\frac{1}{\text{x}-10}=\frac{1}{12}$
$\Rightarrow\frac{\text{x}-10+\text{x}}{\text{x}(\text{x}-10)}=\frac{1}{12}$
$\Rightarrow\frac{2\text{x}-10}{\text{x}^2-10\text{x}}=\frac{1}{12}$
$\Rightarrow x^2 - 10x = 24x - 120$
$\Rightarrow x^2 - 10x - 24x + 120 = 0$
$\Rightarrow x^2 - 34x + 120 = 0$
$\Rightarrow x^2 - 30x - 4x +120 = 0$
$\Rightarrow x(x - 30) - 4(x - 30) = 0$
$\Rightarrow (x - 30)(x - 4) = 0$
Either x - 30 = 0, then x = 30
Or x - 4 = 0 but it is not possible as it is < 10
The second pipe will fill the reservoir in = x - 10 = 30 - 10 = 20 hours.

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