Three pipes A, B and C can fill a cistern in 10 h, 12 h and 15 h respectively. First A was opened, After 1 hour B was opened, and after 2 hours from the start of A, C also opened. Find the time in which the cistern was just fill.

3 hours 52 min.

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4 hours 52 min.

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5 hours 52 min.

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6 hours 52 min.

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Solution

The correct option is A 4 hours 52 min.
Let it take $x$ hours to fill the tank by all the three pipes.

Then work done by {A in x hours + B in (x - 1) hours + C in (x - 2) hours} = 1

$\Rightarrow \frac{x}{10} + \frac{(x - 1)}{12} + \frac{(x - 2)}{15} = 1$

$\Rightarrow 6 x + 5 (x - 1) + 4 (x - 2) = 60$

$\Rightarrow 15 x - 13 = 60 \Rightarrow 15 x = 73$

$\Rightarrow x = \frac{73}{15} = 4 \frac{13}{15} = 4 h o u r s + \frac{13}{15} \times 60 m i n$ $= 4 h o u r s 52 m i n$ .

The time in which the cistern was just filled is $= 4 h o u r s 52 m i n$ .

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