Question

A pump can fill a milk tank in 2 hours. Because of leakage problem, it took 20 minutes more to fill the tank. The plant supervisor wants to find that how much time would the leak take to empty a full tank.

Answer 1

Answer on Question #48444 – Math –Algebra

A pump can fill a milk tank in 2 hours. Because of leakage problem, it took 20 minutes more to fill the tank. The plant supervisor wants to find that how much time would the leak take to empty a full tank.

Solution:

$v$ – speed of filling;

$u$ – speed of draining;

$V$ – volume of the tank;

$t_1 = 2$ hours – time to fill tank in normal situation;

$t_2 = \frac{1}{3} h + 2 h = \frac{7}{3}$ hour – time to fill tank in leakage situation;

$t_3$ – time to empty the full tank;

Time to fill tank in normal situation:

$$t_1 = \frac{V}{v}$$ $$\frac{1}{t_1} = \frac{v}{V}$$

Time to fill tank in leakage situation:

$$t_2 = \frac{V}{v - u}$$ $$\frac{1}{t_2} = \frac{v - u}{V}$$ $$\frac{1}{t_2} = \frac{v}{V} - \frac{u}{V}$$

Time to empty full tank:

$$t_3 = \frac{V}{u}$$ $$\frac{1}{t_3} = \frac{u}{V}$$

Plug (3) and (1) in (2):

$$\frac{1}{t_2} = \frac{1}{t_1} - \frac{1}{t_3}$$ $$\frac{1}{t_3} = \frac{1}{t_1} - \frac{1}{t_2}$$ $$\frac{1}{t_3} = \frac{t_2 - t_1}{t_1 t_2}$$ $$t_3 = \frac{t_1 t_2}{t_2 - t_1} = \frac{2 h \cdot \frac{7}{3} h}{3 h - 2 h} = 14 \text{ hours}$$

Answer: 14 hours

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