Pipes & Cistern Questions and Answers

Answer: D

The tank is half filled in 1 hour, 40 minutes.

Pipe M fills \(\frac{1}{40}\) of the tank per minute and pipe N empties \(\frac{1}{50}\) per minute. Together:

\[ \begin{aligned} \frac{1}{40} - \frac{1}{50} &= \frac{5 - 4}{200} \\ &= \frac{1}{200} \end{aligned} \]

So the full tank would take 200 minutes, and half the tank takes 100 minutes, which is 1 hour 40 minutes.

3 hours 20 minutes (200 minutes) is the time to fill the whole tank, the most likely slip.

Answer: C

Pipe C alone fills the tank in 2 2/3 hours. In 3 hours A and B together fill \[3\left(\frac{1}{8} + \frac{1}{10}\right)\] \[= \frac{27}{40}\] of the tank, leaving 13/40. A and C together fill this in 39 minutes, which is 13/20 hour, so their combined rate is \[\frac{13/40}{13/20} = \frac{1}{2}\] tank per hour. Then C's rate is \[\frac{1}{2} - \frac{1}{8} = \frac{3}{8}\] and C alone takes 8/3 = 2 2/3 hours. The options 3, 3 1/2 and 4 1/2 give rates of 1/3, 2/7 and 2/9 for C, which do not fit.

Answer: B

The two inlet pipes fill 1/40 + 1/48 = 11/240 of the tank every hour, and the outlet empties 20 gallons per hour. All three together fill the tank in 40 hours, so in each hour they add 1/40 of the capacity C.

\[ \frac{C}{40} = \frac{11C}{240} - 20 \]

\[ 20 = \frac{5C}{240} \]

So C = 960 gallons, option (B). Check: 960 / 40 = 24, and 960 x 11 / 240 = 44, and 44 - 20 = 24. The values 840, 900 and 1000 gallons do not satisfy the equation.

Answer: C

Pipes A and C together can empty a full tank in 30 hours. In one hour, A fills 1/15 and B fills 1/18, which together make 11/90 of the tank. With C also open, 1/9 of the tank is filled in 5 hours, which is 1/45 per hour.

\[ \frac{11}{90} - c = \frac{1}{45} \]

\[ c = \frac{9}{90} = \frac{1}{10} \]

So C empties the tank at 1/10 per hour. A and C together then change the tank by 1/15 - 1/10 = -1/30 of it per hour, so they empty a full tank in 30 hours. The options 25, 28 and 35 hours do not follow from these rates.

Answer: D

Pipe C alone empties one-third of the tank in 6 hours. A and B together fill \(\frac{1}{20} + \frac{1}{30} = \frac{1}{12}\) of the tank per hour. With C also open, \(\frac{2}{9}\) of the tank fills in 8 hours, so the net rate is \[ \frac{2/9}{8} = \frac{1}{36} \] C's emptying rate is \[ \begin{aligned} \frac{1}{12} - \frac{1}{36} &= \frac{2}{36} \\ &= \frac{1}{18} \end{aligned} \] so C empties the whole tank in 18 hours and one-third of it in \(\frac{18}{3} = 6\) hours. The answer 18 hours would be for the whole tank; 3, 4 and 5 hours do not follow from the rate.

Answer: C

Capacity of the tank = (12*13.5) litres

= 162 litres

Capacity of each bucket = 9 litres.

So we can get answer by dividing total capacity of tank by total capacity of bucket.

Number of buckets needed = (162/9) = 18 buckets

Answer: C

Let the capacity of the tank be LCM (10, 20, 30) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 20 = 3 units / hour
=> Efficiency of pipe C = 60 / 30 = 2 units / hour
Now, all three work for some time, say ‘t’ hours.
So, B and C work for 2 more hours after ‘t’ hours and then, C works for another 14 hours.
=> Combined efficiency of pipe A, pipe B and pipe C = 11 units / hour
=> Combined efficiency of pipe B and pipe C = 5 units / hour
 
So, we have 11 x t + 5 x 2 + 14 x 2 = 60
=> 11 t + 10 + 28 = 60
=> 11 t = 60 - 38
=> 11 t = 22
=> t = 2
 
Therefore, A was closed after 2 hours.

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).

Without the leak the pump fills the tank at \( \tfrac{1}{2} \) per hour; with it, the effective rate is \( \tfrac{3}{7} \) per hour.

\[ \frac{1}{2} - \frac{1}{L} = \frac{1}{7/3} = \frac{3}{7} \]

\[ \frac{1}{L} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} \]

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

Answer: A

If the whole tank takes 6 hours, then in one hour the pipe fills

\[ \frac{1}{6} \text{ of the tank} \]

This reciprocal is the basic move in every pipes-and-cisterns problem: times cannot be added, but rates can.

Answer: A

It is given that Pipe A,B and C work simultaneously.

Pipe A and B are filling the tank and Pipe C is emptying it.

In 1 hour volume of tank filled by Pipe A: 1/5th

In 1 hour volume of tank filled by Pipe B: 1/6th

In 1 hour volume of tank emptied by Pipe C: 1/12th

Working together Pipe A B and C fill x volume of tank.

=>1/x = (1/5)+(1/6)-(1/12)

=>x=60/17 hours