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Pipes & Cistern Questions and Answers

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101.

Pipe P can fill a tank in 4 hours, pipe Q in 8 hours and pipe R in 24 hours. If all the pipes are open, in how many hours will the tank be filled?

Answer: B

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

\[ \frac{1}{4} + \frac{1}{8} + \frac{1}{24} = \frac{6 + 3 + 1}{24} = \frac{10}{24} = \frac{5}{12} \]

\[ t = \frac{12}{5} = 2.4 \text{ hours} \]

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102.

A tap can fill a tank in 6 hours. After half the tank is filled then 3 more similar taps are opened. What will be total time taken to fill the tank completely.

Answer: D

Half tank will be filled in 3 hours

Lets calculate remaining half,

Part filled by the four taps in 1 hour = 4 × (1/6) = 2/3

Remaining part after 1/2 filled = 1-1/2 = 1/2

capture Dup 5 -72357528

Total time = 3 hours + 45 mins = 3 hours 45 mins

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103.

Working alone, two pipes A and B require 9 hours and 6.25 hours more respectively to fill a pool than if they were working together. Find the total time taken to fill the pool if both were working together.

Answer: D

Let the time taken if both were working together be ‘n’ hours.
=> Time taken by A = n + 9
=> Time taken by B = n + 6.25
 
In such kind of problems, we apply the formula :
n2 = a x b, where ‘a’ and ‘b’ are the extra time taken if both work individually than if both work together.
Therefore, n2 = 9 x 6.25
=> n = 3 x 2.5 = 7.5
 
Thus, working together, pipes A and B require 7.5 hours.

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104.

To fill a cistern, pipes A, B and C take 20 minutes, 15 minute and 12 minutes respectively. The time in minutes that the three pipes together will take to fill the cistern is:

Answer: A

A can do a piece of work in x minutes
B can do a piece of work in y minutes
C can do a piece of work in z minutes
Minute’s work of each of the three is 1/x +1/y + 1/z
1 minutes work of each of the three pipes = 1/20 + 1/15 + 1/12
= 1/5 or = 5 minutes

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105.

Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in.

Answer: B

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

\[ \frac{1}{5} + \frac{1}{6} - \frac{1}{12} \]

Over the common denominator 60:

\[ \frac{12 + 10 - 5}{60} = \frac{17}{60} \text{ per hour} \]

\[ t = \frac{60}{17} = 3\tfrac{9}{17} \text{ hours} \]

The emptying pipe subtracts, so the tank still fills - but more slowly than the \( 2\tfrac{8}{11} \) hours A and B would manage alone.

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106.

Pipe R can empty a full tank in 30 hours. But two pipes P and Q can fill a tank in 15 hours and 10 hours respectively. Ram unknowingly opened all three taps. After 2 hours Shyam realized it and closed Pipe R. Due to this mistake how much time more would it take to fill the tank?

Answer: B

Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).

With all three open:

\[ \frac{1}{15} + \frac{1}{10} - \frac{1}{30} = \frac{2 + 3 - 1}{30} = \frac{2}{15} \text{ per hour} \]

\[ 2 \text{ hours} \rightarrow \frac{4}{15} \text{ filled} \]

After R is closed, P and Q fill at \( \tfrac{1}{15} + \tfrac{1}{10} = \tfrac{1}{6} \):

\[ \frac{11}{15} \div \frac{1}{6} = 4.4 \text{ hours} \]

\[ \text{total} = 2 + 4.4 = 6.4 \text{ h}, \qquad \text{without the mistake} = 6 \text{ h} \]

\[ \text{extra} = 0.4 \text{ h} = 24 \text{ minutes} \]

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