Pipes & Cistern Questions and Answers

Answer: C

Convert everything to decimals and compare.

\[ \frac{2}{3} = 0.667, \qquad \frac{4}{5} = 0.800 \]

• \( \tfrac{1}{2} = 0.500 \) - too small
• \( \tfrac{9}{10} = 0.900 \) - too large
• \( \tfrac{5}{6} = 0.833 \) - too large
• \( \tfrac{3}{4} = 0.750 \) - in range ✓

So \( \tfrac{3}{4} \) is the only one strictly between the two bounds.

Answer: C

Let the fraction be \( \tfrac{n}{d} \). Decreasing the denominator by 80% leaves \( 0.2d \); increasing the numerator by 300% makes it \( 4n \).

\[ \frac{4n}{0.2d} = \frac{20n}{d} = \frac{2}{9} \]

\[ \frac{n}{d} = \frac{2}{9 \times 20} = \frac{2}{180} = \frac{1}{90} \]

An increase of 300% makes a quantity four times as large, not three - the step that decides the answer.

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.

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

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

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.

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.

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.

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.

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