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Pipes & Cistern Questions and Answers

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81.

Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long it will take to fill the tank?

Answer: B

In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as

Part filled by A in 1 min = 1/20

Part filled by B in 1 min = 1/30

Part filled by (A+B) in 1 min = 1/20 + 1/30

= 1/12

So both pipes can fill the tank in 12 mins.

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82.

Three pipes A, B and C were opened to fill a cistern. Working alone, A, B and C require 12, 15 and 20 minutes respectively.After 4 minutes of working together, A got blocked and after another 1 minute, B also got blocked. C continued to work till the end and the cistern got completely filled. What is the total time taken to fill the cistern ?

Answer: C

Let the capacity of the cistern be LCM(12, 15, 20) = 60 units.
=> Efficiency of pipe A = 60 / 12 = 5 units / minute
=> Efficiency of pipe B = 60 / 15 = 4 units / minute
=> Efficiency of pipe C = 60 / 20 = 3 units / minute
=> Combined efficiency of pipe A, pipe B and pipe C = 12 units / minute
Now, the cistern is filled with the efficiency of 12 units / minute for 4 minutes.
=> Pool filled in 4 minutes = 48 units
=> Pool still empty = 60 - 48 = 12 units
Now, A stops working.
=> Combined efficiency of pipe B and pipe C = 7 units / minute
Now, the cistern is filled with the efficiency of 7 units / minute for 1 minute.
=> Pool filled in 1 minute = 7 units
=> Pool still empty = 12 - 7 = 5 units
Now, B also stops working.
These remaining 5 units are filled by C alone.
=> Time required to fill these 5 units = 5 / 3 = 1 minute 40 seconds
 
Therefore, total time required to fill the pool = 4 minutes + 1 minutes + 1 minute 40 seconds = 6 minutes 40 seconds

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83.

The product of two fractions is 3/4 and their quotient is 6/18. If the denominator of one is 1/3 of the other's numerator, then find the pair of fractions?

Answer: B

Let A = x/y, and B be two fractions

It is given that A×B = 3/4

Therefore x/y × B = 3/4

B = 3y/4x

Therefore, A = x/y and B = 3y/4x

It is given that A/B = (x/y) / (3y/4x) = 6/18

Therefore x/y × 4x/3y = 6 /18

Therefore 4x2/3y2 = 6/18

x2/y2 = 18/(4×18)

Therefore, x2/y2 = 1/4

x/y = 1/2

B = 3y/4x = 3/4 (1/(x/y)) = 3/4 [1/(1/2)] = 6/4

Therefore A = 1/2 and B = 3/2

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84.

Due to leakage in a tank, it empties in 6 hours. An inlet pipe fills the water at 4 litres per minute. If the tank is full and inlet is opened the tank empties in 8 hours because of the leakage. What is the capacity of the tank in litres?

Answer: C

Work done by the inlet in 1 hour =

1/6−1/8=1/24

Work done by inlet in 1 min=1/24∗1/60=1/1440

=>Volume of 1/1440 part

= 4 liters
Volume of whole = (1440 * 4) litres = 5760 litres.

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85.

3 pipes have diameters 2 cms, 3 cms and 4 cms. The ratio of water flowing through them is equal to ratio of the square of their diameters.The biggest pipe when open alone, can fill the entire pool in just 126 minutes. When all the pipes are opened together, the pool would be filled in how much time?

Answer: A

Flow varies as the square of the diameter, so the three pipes are in the ratio

\[ 2^{2} : 3^{2} : 4^{2} = 4 : 9 : 16 \]

The biggest carries 16 of the 29 parts and fills the pool in 126 minutes, so all three together work \( \tfrac{29}{16} \) times as fast:

\[ t = 126 \times \frac{16}{29} = \frac{2016}{29} \text{ minutes} \]

That is about 69.5 minutes - a little over half the 126, as expected when the flow nearly doubles.

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86.

Two pipes P and Q can fill a tank in 10 hours and 14 hours respectively. If both pipes are opened simultaneously, how much time will be taken 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).

\[ \frac{1}{10} + \frac{1}{14} = \frac{7 + 5}{70} = \frac{6}{35} \]

\[ t = \frac{35}{6} = 5.8333 \text{ hours} \]

\[ 0.8333 \times 60 = 50 \text{ minutes} \]

So the tank fills in 5 hours 50 minutes.

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87.

A water tank has two pipes to fill water in it. One of the pipes fills water at the rate of 100 litres per hour. The tank is filled by the other pipe in 6 hours and by both pipes together in 4.2 hours. What is the capacity of the tank in litres ?

Answer: D

Let the time taken by the 1st pipe to fill the tank be x hours.
When both the pipes are open, it is given that:
1/x + 1/6 = 1/(4.2) of tank is filled in 1 hour.
1/x + 1/6 = 5/21
21(6 + x) = 30 x
9x = 126
x = 14 hours
Therefore, capacity of the tank = 100 * 14 = 1400 litres.

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88.

Two pipes A and B are connected to drain out a water tank. A alone can drain out the tank in 20 hours and B can drain 20 liters per hour. Find the capacity of the water tank given that working together, they require 12 hours to completely drain out the tank.

Answer: A

Let the capacity of the tank be LCM (20, 12) = 60 units
=> Efficiency of A working alone = 60 / 20 = 3 units / hour
=> Efficiency of A and B working together = 60 / 12 = 5 units / hour
Therefore, Efficiency of B working alone = Efficiency of A and B working together - Efficiency of A working alone
=> Efficiency of B working alone = 5 - 3 = 2 units / hour
=> Time required by B alone to drain the tank = 60 / 2 = 30 hours
 
But we are given that B can drain the tank at the rate of 20 liters per hour.
Therefore, capacity of the water tank = 20 x 30 = 600 liters
 

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89.

4/7 of 2/3 of 5/6 of 5/8 of 1008 is

Answer: A

Work from the right, cancelling as you go.

\[ 1008 \times \frac{5}{8} = 630 \]

\[ 630 \times \frac{5}{6} = 525 \]

\[ 525 \times \frac{2}{3} = 350 \]

\[ 350 \times \frac{4}{7} = 200 \]

Multiplying all four fractions first gives \( \tfrac{200}{1008} \), and \( 1008 \times \tfrac{200}{1008} = 200 \) - the same answer.

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90.

A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) is

Answer: C

Let the capacity be \( C \) litres. The leak alone empties it in 6 hours, so it drains \( \tfrac{C}{6} \) litres an hour.

The inlet adds \( 4 \times 60 = 240 \) litres an hour, and the net effect empties the tank in 8 hours:

\[ \frac{C}{6} - 240 = \frac{C}{8} \]

\[ C\left(\frac{1}{6} - \frac{1}{8}\right) = 240 \quad\Longrightarrow\quad \frac{C}{24} = 240 \]

\[ C = 5760 \text{ litres} \]

The inlet rate must be converted from litres per minute to litres per hour to match the other terms.

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