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RRB Group D 3-Oct-2018 Shift 3 Solved Paper

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Question : 14 of 100

Marks: +1, -0

A pipe, working at full speed, can fill an empty cistern in 2 hours. 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 4 hours. Together with a drain pipe that drained water out of the tank at a constant rate, half of the empty cistern could be filled in 5 hours, all the three pipes working concurrently. How many hours would it take the drain pipe to empty the filled cistern if no other pipe was functioning during the time?

32
20
30
24

Solution:

By pipe A, portion of the tank filled

in 1st hour

=\frac{1}{2\times 12}=\frac{1}{24}

in 2nd hour

=\frac{1}{2\times 9}=\frac{1}{18}

in 3rd hour

=\frac{1}{2\times 6}=\frac{1}{12}

in 4th hour

=\frac{1}{2\times 4}=\frac{1}{8}

in 5th hour

=\frac{1}{2\times 3}=\frac{1}{6}

Portion of tank filled by pipe A in 5 hours

=\frac{1}{24}+\frac{1}{18}+\frac{1}{12}

+\frac{1}{8}+\frac{1}{6}=\frac{17}{36}

Now, let empty pipe can empty the tank in

x

hours. So, portion of the tank empty in 5 hours

=\frac{5}{x}

According to the question,

\frac{17}{36}+\frac{17}{72}-\frac{5}{x}=\frac{1}{2}

\frac{51}{72}-\frac{5}{x}=\frac{1}{2}

x=\frac{72\times 5}{15}=24

Hence, drain pipe empty the full tank in 24 hours.

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