Pipes & Cistern Questions and Answers

Answer: D

Let the total time be \( t \). For the first half of it B works alone at \( \tfrac{1}{40} \); for the second half both work, at \( \tfrac{1}{60} + \tfrac{1}{40} \):

\[ \frac{t}{2}\cdot\frac{1}{40} + \frac{t}{2}\left(\frac{1}{60} + \frac{1}{40}\right) = 1 \]

\[ \frac{t}{80} + \frac{t}{2}\cdot\frac{5}{120} = 1 \quad\Longrightarrow\quad \frac{t}{80} + \frac{t}{48} = 1 \]

Over a denominator of 240:

\[ \frac{3t + 5t}{240} = 1 \quad\Longrightarrow\quad \frac{8t}{240} = 1 \quad\Longrightarrow\quad t = 30 \text{ minutes} \]

The phrase half the time means half of the unknown answer, which is why the equation has to be set up in \( t \) rather than solved forwards.

Answer: C

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

	1	+	1	=	1
	x		(x + 6)		4

	x + 6 + x	=	1
	x(x + 6)		4

 x2 - 2x - 24 = 0

 (x -6)(x + 4) = 0

 x = 6.     [neglecting the negative value of x]

Answer: C

Correct answer: 24 hours

Treat the emptying pipe as a negative rate.

\[\frac{1}{6} - \frac{1}{8} = \frac{4 - 3}{24} = \frac{1}{24}\]

The tank fills at one twenty-fourth per hour, so it takes 24 hours. Had B been the faster pipe the net rate would be negative and the tank would never fill.

Answer: B

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

As per question, we get

After solving this equation, we get

(x-15)(x+3) = 0,

As value can not be negative, so x = 15

Answer: C

The capacity of the tank is given by C = (p*q*r)/(r - p) litres
Where, p = time in which the leakage can empty the tank in hours = 5 hours
q = rate at which the inlet fills the tank in lph = 6*60 = 360 lph
r = time in which the tank gets emptied in hours = 8 hours
Hence, C = (5*360*8)/(8 - 5)
C = 4800 litres

Answer: D

We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak, as

Work done for 1 hour without leak = 1/3
Work done with leak =

3 1/2=7/2

Work done with leak in 1 hr = 2/7

Work done by leak in 1 hr = 1/3−2/7 = 1/21

So tank will be empty by the leak in 21 hours.

Answer: C

When we have question like one is filling the tank and other is empting it, then we subtraction as,

Filled in 1 hour = 1/4
Empties in 1 hour = 1/9

Net filled in 1 hour = 1/4 - 1/9
= 5/36

So cistern will be filled in 36/5 hours i.e. 7.2 hours

Answer: C

Substitute \( x = -3y \) into the first equation.

\[ -3y + 5y = 16 \quad\Longrightarrow\quad 2y = 16 \]

\[ y = 8 \]

Checking: \( x = -24 \), and \( -24 + 40 = 16 \). The \( -24 \) on offer is the value of \( x \), not \( y \).

Answer: B

Put both rates over a common unit of minutes. The filling tap does \( \tfrac{1}{48} \) of the tank each minute and the emptying tap removes \( \tfrac{1}{120} \), since 2 hours is 120 minutes:

\[ \frac{1}{48} - \frac{1}{120} = \frac{5 - 2}{240} = \frac{3}{240} = \frac{1}{80} \]

So the tank fills in 80 minutes, that is 1 hour 20 minutes.

\[ 11{:}40\text{ AM} + 1\text{ h }20\text{ min} = 1{:}00\text{ PM} \]

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).

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

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.