Pipes & Cistern Questions and Answers

Answer: C

Run the hours in pairs, since A and B alternate with A first. Each pair of hours fills

\[ \frac{1}{6} + \frac{1}{4} = \frac{2 + 3}{12} = \frac{5}{12} \]

After two full pairs, that is 4 hours:

\[ 2 \times \frac{5}{12} = \frac{5}{6} \]

leaving \( \tfrac{1}{6} \). The fifth hour belongs to A, which fills exactly \( \tfrac{1}{6} \) in an hour, so the tank is full at the end of it:

\[ \text{total} = 5 \text{ hours} \]

The remainder happening to match A's hourly output is what makes the answer a whole number here.

Answer: B

Let the capacity of the tank be LCM (10, 15, 20) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 15 = 4 units / hour
=> Efficiency of pipe C = 60 / 20 = 3 units / hour
=> Combined efficiency of all three pipes = 13 units / hour
 
Till 9 AM, A works for 2 hours and B work for 1 hour.
=> Tank filled in 2 hours by A = 12 units
=> Tank filled in 1 hour by B = 4 units
=> Tank filled till 9 AM = 16 units
=> Tank still empty = 60 - 16 = 44 units
 
Now, all three pipes work for 3 hours with the efficiency of 13 units / hour.
=> Tank filled in 3 more hours = 39 units
=> Tank filled till 12 PM = 16 + 39 units = 55 units
=> Tank empty = 60 - 55 = 5 units
 
Now, C is closed for 1 hour and these remaining 5 units would be filled by A and B working together with the efficiency 10 units / hour.
=> Time taken to fill these remaining 5 units = 5 / 10 = 0.5 hours
 
Therefore, time at which the tank will be completely filled = 12 PM + 0.5 hours = 12 : 30 PM

Answer: B

Let the capacity of the pool be LCM (10, 20, 15) = 60 units.
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 20 = 3 units / hour
=> Efficiency of pipe C = 60 / 15 = 4 units / hour
 
=> Efficiency of pipe A and pipe C working together = 10 units / hour
=> Efficiency of pipe B and pipe C working together = 7 units / hour
 
=> Pool filled in first hour = 10 units
=> Pool filled in second hour = 7 units
=> Pool filled in 2 hours = 10 + 7 = 17 units
 
We will have 3 cycles of 2 hours each such that A and C, and, B and C work alternatively.
=> Pool filled in 6 hours = 17 x 3 = 51 units
=> Pool empty = 60 - 51 = 9 units
Now, these 9 units would be filled by A and C working together with the efficiency of 10 units / hour.
=> Time required to fill these 9 units = 9/10 hour = 0.9 hours = 54 minutes
 
Therefore, total time required to fill the pool = 6 hours 54 minutes
 

Answer: B

(A+B)'s 2 hour's work when opened =

Now, its A turn in 5th hour

1/6 work will be done by A in 1 hour

Total time = 4 + 1 = 5 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).

\[ \frac{1}{10} + \frac{1}{15} - \frac{1}{w} = \frac{1}{18} \]

\[ \frac{3 + 2}{30} - \frac{1}{w} = \frac{1}{18} \quad\Longrightarrow\quad \frac{1}{6} - \frac{1}{w} = \frac{1}{18} \]

\[ \frac{1}{w} = \frac{3 - 1}{18} = \frac{1}{9} \quad\Longrightarrow\quad w = 9 \text{ minutes} \]

Answer: C

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.

(2x - 5)(x - 9) = x(x - 5)

x2 - 18x + 45 = 0

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

x = 15. [neglecting x = 3]

Answer: A

There are two important points to learn in this type of question,

First, if both will open then tank will be empty first.

Second most important thing is,

If we are calculating filling of tank then we will subtract as (filling-empting)

If we are calculating empting of thank then we will subtract as (empting-filling)

So lets come on the question now,

Part to emptied 2/5

Part emptied in 1 minute =

1/6 - 1/10 = 1/15 => 1/15 : 25 :: 1: x =>25 × 15 = 6mins

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).

\[ \text{A's rate} = \frac{1}{2.4} - \frac{1}{4} = \frac{5 - 3}{12} = \frac{1}{6} \]

So A alone would fill the cistern in 6 hours. At 100 litres an hour:

\[ \text{capacity} = 6 \times 100 = 600 \text{ litres} \]

Pipe B's own time is only needed to isolate A's rate; the capacity then follows from A's known flow.

Answer: D

We can get the answer by subtracting 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 =

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

Answer: A

Let there be 100 chocolates

215th of 100 40 chocolates eaten on first day

314th of the remaining 314th of (100-40) 45 chocolates

Chocolates left 100-40-45 15 chocolates

15 chocolates are left when 100 are there

75 chocolates are left when ? chocolates are there? = 500 Chocolates