Question : When operated separately, pipe A takes 5 hours less than pipe B to fill a cistern, and when both pipes are operated together, the cistern gets filled in 6 hours. In how much time (in hours) will pipe B fill the cistern, if operated separately?
Option 1: 15
Option 2: 18
Option 3: 10
Option 4: 9
Correct Answer: 15
Solution :
Let the time it takes for pipe B to fill the cistern when operated separately as $t$.
Then, the time it takes for pipe A to fill the cistern when operated separately is $(t - 5)$.
The rates of pipes A and B are $\frac{1}{(t - 5)}$ and $\frac{1}{t}$, respectively.
When both pipes are operated together, their rates add up, and the cistern gets filled in 6 hours.
So, $\frac{1}{t - 5} + \frac{1}{t} = \frac{1}{6}$
Multiplying through by $6t(t - 5)$ to clear the fractions,
$⇒6t + 6(t - 5) = t(t - 5)$
$⇒t^2 - 17t + 30 = 0$
$⇒(t - 2)(t - 15) = 0$
$t = 2$ or $t = 15$
Since pipe A takes 5 hours less than pipe B to fill the cistern, then $t$ must be greater than 5.
$⇒t = 15$
So, pipe B will fill the cistern in 15 hours.
Hence, the correct answer is 15.
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