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.
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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