Question

Four pipes can fill a reservoir in 15, 20, 30 and 60 hours respectively. The first one was opened at 6 AM, second at 7 AM, third at 8 AM and the fourth at 9 AM. When will the reservoir be filled?

Answer 1

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Task. Four pipes can fill a reservoir in 15, 20, 30 and 60 hours respectively. The first one was opened at 6 AM, second at 7 AM, third at 8 AM and the fourth at 9 AM. When will the reservoir be filled?

Solution. The first pipe can fill reservoir in 15 hours. This means that it fills $\frac{1}{15}$ of reservoir in one hour.

Similarly, the second, third and fourth pipes fill $\frac{1}{20}$, $\frac{1}{30}$ and $\frac{1}{60}$ of reservoir in one hour respectively.

Let $t$ be the number of hours spent by first pipe. The second pipe opened at 7 AM, i.e. one hour later, therefore it worked $t-1$ hours.

Similarly, the third and fourth pipes worked $t-2$ and $t-3$ hours respectively.

Together all pipes fill the whole reservoir, so we obtain the following equation:

$\frac{1}{15}\ t+\frac{1}{20}\ (t-1)+\frac{1}{30}\ (t-2)+\frac{1}{60}\ (t-3)=1$

Let us solve it.

$\frac{t}{15}+\frac{t-1}{20}+\frac{t-2}{30}+\frac{t-3}{60}=1,$

$\frac{4t}{60}+\frac{3(t-1)}{60}+\frac{2(t-2)}{60}+\frac{t-3}{60}=1$

$\frac{4t+3(t-1)+2(t-2)+t-3}{60}=1$

$4t+3t-3+2t-4+t-3=60$

$10t-10=60$

$10t=60+10=70$

$t=\frac{70}{10}=7\ \text{hours}$

Answer. 7 hours