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Question:

Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately in 5 hours, 10 hours and 15 hours respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at 3/4th of their normal rates for the first hour after which they supplied water at the normal rate. Pipe C supplied water at 2/3rd of its normal rate for first 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled.

Explanation:

Let the pipes work for n hours.

By the given conditions,

$\frac{1}{5 \times \frac{4}{3}}$ + $\frac{1}{10 \times \frac{4}{3}}$ + (n - 1) $[\frac{1}{5} + \frac{1}{10}]$ + $\frac{2}{15 \times \frac{3}{2}}$ + $\frac{n - 2}{15}$ = 1

∴ $\frac{3}{20}$ + $\frac{3}{40}$ + (n - 1) $\frac{3}{10}$ + $\frac{4}{45}$ + $\frac{n - 2}{15}$ = 1

∴ $\frac{9}{40}$ + $\frac{3 n - 3}{10}$ + $\frac{4}{45}$ + $\frac{n - 2}{15}$ = 1

∴ $\frac{9 n - 9}{30}$ + $\frac{2 n - 4}{30}$ = 1 - $\frac{9}{40}$ - $\frac{4}{45}$

∴ $\frac{11 n - 13}{30}$ = $\frac{360 - 81 - 32}{360}$

∴ 11n - 13 = $\frac{247 \times 30}{360}$

∴ n ≈ 3.05

Hence, option (c).

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