Pipes & Cistern Questions and Answers

Answer: A

Let the slower pipe alone fill the tank in x minutes

then faster will fill in x/3 minutes.

Part filled by slower pipe in 1 minute = 1/x

Part filled by faster pipe in 1 minute = 3/x

Part filled by both in 1 minute =

x = 36 × 4 = 144 mins

Answer: D

Part filled by A in 1 hour = 1/5
Part filled by B in 1 hour = 1/10
Part filled by C in 1 hour = 1/30

Part filled by (A+B+C) in 1 hour =

1/5+1/10+1/30

=1/3

So all pipes will fill the tank in 3 hours.

Answer: B

Let B be turned off after x minutes. Then,

Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.

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).

\[ \frac{1}{X + 9} + \frac{1}{X + 16} = \frac{1}{X} \]

Clearing denominators:

\[ X(X + 16) + X(X + 9) = (X+9)(X+16) \]

\[ 2X^{2} + 25X = X^{2} + 25X + 144 \]

\[ X^{2} = 144 \quad\Longrightarrow\quad X = 12 \text{ minutes} \]

The result is the geometric mean of the two extra times: \( X = \sqrt{9 \times 16} = 12 \), a useful shortcut for this standard form.

Answer: C

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).

A runs for the whole 18 minutes:

\[ \frac{18}{24} = \frac{3}{4} \text{ of the tank} \]

B must supply the remaining \( \tfrac{1}{4} \) at \( \tfrac{1}{32} \) per minute:

\[ t = \frac{1}{4} \times 32 = 8 \text{ minutes} \]

So B is closed after 8 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).

\[ \frac{1}{38} + \frac{1}{19} - \frac{1}{133} \]

Over the common denominator 266:

\[ \frac{7 + 14 - 2}{266} = \frac{19}{266} = \frac{1}{14} \]

\[ t = 14 \text{ hours} \]

Answer: C

How we can solve this question ? First we will calculate the work done for 10 mins, then we will get the remaining work, then we will find answer with one tap work, As Part filled by Tap A in 1 min = 1/20 Part filled by Tap B in 1 min = 1/60 (A+B)'s 10 mins work =

Answer: B

Take the pool as 120 units, the LCM of the three times.

\[ M = 3, \qquad N = 2, \qquad P = -4 \text{ units per minute} \]

A full three-minute cycle nets \( 3 + 2 - 4 = 1 \) unit.

After 116 cycles (348 minutes) the pool holds 116 units. Then:

\[ \text{minute } 349 \ (M): \ 116 + 3 = 119 \]

\[ \text{minute } 350 \ (N): \text{ needs 1 more unit at 2 per minute} \rightarrow 0.5 \text{ min} \]

\[ \text{total} = 349.5 \text{ minutes} \]

The pool fills part-way through a minute, so the answer is not a whole number of cycles - checking the level after each of the three phases is what catches that.

Answer: C

Part filled without leak in 1 hour = 1/9

Part filled with leak in 1 hour = 1/10

Work done by leak in 1 hour

=1/9-1/10=190

We used subtraction as it is getting empty.

So total time to empty the cistern is 90 hours

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.