Question

If two pipes function simultaneously a resirvoir will be filled in 12hours. One pipe fills the resirvor 10hours faster than the other.how many hours will the second pipe fill the resirvor

Answer 1

Suppose the slower pipe alone can fill the reservoir in $x$ hours,
faster pipe alone can fill the reservoir in $(x-10)$ hours.

Part filled by slower pipe in 1 hr $=\dfrac{1}{x}$
Part filled by faster pipe in 1 hr $=\dfrac{1}{x-10}$

If pipes functions simultaneously, the reservoir will be filled in 12 hours.
=> Part filled by both the popes in 1hr = $\dfrac{1}{12}$

Thus we have,
$\dfrac{1}{x}+\dfrac{1}{x-10}=\dfrac{1}{12}\\
12(x-10)+12x=x(x-10)\\
12x-120+12x=x^2-10x\\
x^2-34x+120=0\\
(x-30)(x-4)=0\\
x=30\text{ or } 4$

$x$ cannot be 4 because $(x-10)$ will be negative.
Therefore, $x=30$

Part filled by second pipe (slower pipe) in 1 hr = $\dfrac{1}{30}$
=> Second pipe (slower pipe) alone can fill the reservoir in 30 hours.