Two pipes A and B can fill a tank completely in 5 hours and 8 hours. respectively. Pipe C can empty the tank completely in 10 hours. If all the three pipes are opened simultaneously in an empty tank, then how much time will it take to fill the tank completely ?

Correct option is A

Given:

Pipe A fills the tank in 5 hours

Pipe B fills the tank in 8 hours

Pipe C empties the tank in 10 hours

Solution:

The rates of the pipes 

​$$\text{Rate of A} = \frac{1}{5} \text{ of the tank per hour.}$$

$$\text{Rate of B} = \frac{1}{8} \text{ of the tank per hour.}$$ 

​​$$\text{Rate of C} = -\frac{1}{10} \text{ of the tank per hour.}$$

​​When all three pipes are opened simultaneously, the total rate of filling or emptying the tank is the sum of the individual rates. So, 

​$$\text{Total Rate} = \text{Rate of A} + \text{Rate of B} + \text{Rate of C}$$ 

​$$\text{Total Rate} = \frac{1}{5} + \frac{1}{8} - \frac{1}{10}$$ 

$$\text{Total Rate} = \frac{(8 + 5 - 4)}{40} = \frac{9}{40}$$ 

the time to fill the tank completely is the reciprocal of this rate:

$$\text{Time} = \frac{1}{\text{Total Rate}} = \frac{1}{\frac{9}{40}} = \frac{40}{9} \text{ hours}$$ 

$$\text{Time} = 4 \frac{4}{9} \text{ hours}$$ 

thus time taken by all the three pipes filling the empty tank is $$4 \frac{4}{9} \text{ hours}$$ .​

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