Pipes & Cistern Questions and Answers

Answer: A

Flow through a pipe is proportional to its cross-sectional area, which varies as the square of the diameter.

\[ \frac{(2d)^{2}}{d^{2}} = 4 \]

Four times the flow means a quarter of the time:

\[ \frac{84}{4} = 21 \text{ minutes} \]

Halving the time to 42 minutes would assume flow proportional to diameter rather than to area.

Answer: C

 Let the capacity of the tank be LCM (10, 30) = 30 units
=> Efficiency of pipe A = 30 / 10 = 3 units / hour
=> Efficiency of pipe B = 30 / 30 = 1 units / hour
=> Combined efficiency of both pipes = 4 units / hour
Now, total time taken by A and B working together to fill the tank if there was no leak = 30 / 4 = 7.5 hours
=> Actual time taken = 7.5 + 2.5 = 10 hours
 
The tank filled by A and B in these 2.5 hours is the extra work done to compensate the wastage by the leak in 10 hours.
=> 2.5 hours work of A and B together = 10 hours work of the leak
=> 2.5 x 4 = 10 x E, where ‘E’ is the efficiency of the leak.
=> E = 1 unit / hour
 
Therefore, time taken by the leak alone to empty the full tank = 30 / 1 = 30 hours

Answer: A

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).

\[ \frac{1}{24} - \frac{1}{L} = \frac{1}{36} \]

\[ \frac{1}{L} = \frac{3 - 2}{72} = \frac{1}{72} \quad\Longrightarrow\quad L = 72 \text{ hours} \]

Answer: C

Work done by the waste pipe in 1 minute =	1	-		1	+	1
	15			20		24

=		1	-	11
		15		120

Volume of	1	part = 3 gallons.
	40

Volume of whole = (3 x 40) gallons = 120 gallons.

Answer: C

Let the slower pipe take \( x \) minutes; the faster one is three times as quick and so takes \( x/3 \). Their rates add:

\[ \frac{1}{x} + \frac{3}{x} = \frac{4}{x} = \frac{1}{36} \quad\Longrightarrow\quad x = 144 \text{ minutes} \]

Seen another way, the two pipes contribute in the ratio 1 : 3, so the slower one does a quarter of the work and needs four times the joint time: \( 4 \times 36 = 144 \). The faster pipe alone would take 48 minutes.

Answer: D

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).

For the first 30 minutes both taps run:

\[ \frac{1}{30} - \frac{1}{45} = \frac{3 - 2}{90} = \frac{1}{90} \text{ per minute} \]

\[ 30 \times \frac{1}{90} = \frac{1}{3} \text{ filled} \]

With Tap B closed, the remaining \( \tfrac{2}{3} \) is filled at \( \tfrac{1}{30} \) per minute:

\[ \frac{2}{3} \times 30 = 20 \text{ minutes} \]

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).

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

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

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

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

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.

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