Question 46

A pipe, working at full speed, can fill an empty cistern in 1 hour. However, during the first hour it worked at one-twelfth of its capacity, during the second hour at one-ninth of its capacity, during the third hour at one-sixth of its usual capacity, during the fourth hour at one- fourth of its usual capacity and during the fifth hour it was only one-third as efficient as it was supposed to be. A second pipe also displayed similar performance,but if it worked at full speed would have filled the empty cistern in 2 hours. Together with a drain pipe that drained water out of the tank at a constant rate, the empty cistern could be filled in 5 hours,all the three pipes working concurrently. How many hours will it take the drain pipe to empty the filled cistern if no other pipe was functioning during the time?

Solution

Let,answer is x

In 1 hr=1/x

5hr=5/x

5hr,,,,

$$\frac{1}{12}+\frac{1}{9}+\frac{1}{6}+\frac{1}{4}+\frac{1}{3}$$

$$=\frac{(3+4+6+9+12)}{36}$$

$$=\frac{34}{36}$$

Now, for fill tank

5hr,,,,,

$$\frac{34}{36}×\frac{1}{2}=\frac{17}{36}$$

$$\frac{34}{36}+\frac{17}{36}-\frac{5}{x}=1$$

$$34x+17x-180=36x$$

15x=180

X=12

Answer = 12


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