Aptitude - Pipes and Cistern - Discussion

@Mouni.

Please check this one:

Let us take Cistern is filled by pipe A and leak is due to pipe B (In this way it is easy to understand).

First point : Cistern filled by pipe A in 8 hrs.

Second point : Cistern filled by pipe A and pipe B (it is actually leak) is 10hrs (8+2).

Now take LCM of these two LCM of 8, 10 is 40. This implies to fill Cistern, total 40 units of work needed.

Then if only pipe A can fill the Cistern in 8 hrs that means 5 units of work for min by pipe A. Both pipe A and pipe B can fill the Cistern in 10 hrs that means 4 units of work by pipe A and pipe B.

So if you see, in second case 4 units of work per min where as in first case it is 5 units of work. This means if you include leak it reduces the work units by 1 unit ---> leakage per hr is 1 unit.

So to drain all the water in cistern, leak has to do 40 units of work. So if it is doing 1 unit of work to drain the tank. For 40 units it has to 40 hrs of work. So answer : 40 hrs.

Let total units to be filled be 1 unit.

Then work done by the pump in 1h=1/2 now let x be the time taken by leakage to empty the tank i.e. 1 unit. The work done by leakage in one is 1/x units.

Now, as it takes 2 1/3 (or 7/3) to fill tank then work done in one hour with leakage will be =1/ (7/3) =3/7.

Therefore.

Work is done by pump without leakage per hour-work done by leakage in one-hour =work done by pumy with leakage.

i.e 1/2-1/x = 3/7.
i.e x = 14hours.

It takes 2 hours to fill the tank.

It takes 2 1/3 (7/3 hrs)hours to fill the tank including the leak.
Now,
Tank fill in 1 hour=1/2,
Tank fill in 1 hour including leak = 3/7,
Tank leak for 1 hour= tank fill in 1 hour- tank fill including tank leak in 1 hour = 1/2 - 3/7=1/14,
So, 1/14 of the tank will be empty in 1 hour.
Now, 1/14*14 of the tank will be empty in 1*14 hour. (Multiply 14 on both sides),
The Complete tank will be empty by leak in 14 hours.

I'm confused, please check my logic
1. It takes 2 hours for the pump to fill the tank without the leak.
2. The leak adds 1/3hrs onto that time time, which is an increase in time of 1/6th.
3. Then the amount leaked = 1/6 of the amount pumped.
4. When the pumped is turned off the leak will drain the tank at 1/6 of the speed at which the pump fills it
5. If it takes the pump 2hrs to fill it then it should take 12 hours for the leak to drain the tank

Lets us simplify by doing it minute basis:.

Let the Volume be V without leak tank is filled in 120 min.

So, in 1 min = V/120.

With leak, = V/140.

Leak is withdrawing water per min = V/120 - V/140 = V/840.

i.e 840th part is filled in 1 min.

So the tank is filled in 840 min i.e. 14 hours.

----------------------------------------------------------------Hour Basis,

In 1 hour V/2 with leak, 3V/7.

Work by leak = V/2-3V/7 = V/14.

V = 14 h.

Pipe A could fill an empty cistern in 18 hours while Pipe B can drain a filled cistern in 30 hours. When the cistern is empty, Pipe A is turned on for an hour and then turned off. Now Pipe B is allowed to drain out water from the cistern for an hour and then turned off. The pipes were alternately left open for an hour each time till the cistern was full. How much time did it take for the cistern to be full?

Can anyone help me to solve this?

Guys, it's simple.

First thing;
A pump can fill water in 2 hours.
Due to the leak, it takes 2 1/3 hours.

They are asking in how many hours it can empty the tank through leak only.

So,
1. Tap ---> 2 hours.
2. Tap+leak ---> 2 1/3 hours (i.e) 7/3 hours.
3. Leak ---> ?

Leak = ( tap + leak ) - tap
= 1/7/3 - 1/2.
= 3/7 - 1/2,
= 6-7/14,
= -1/14.
Total leak = -14 - indicates leak or outlet etc.

If the total area of pump=1 part
The pumop take 2 hrs to fill 1 part
The pumop take1 hour to fill 1/2 portion
Due to lickage
The pumop take 7/3 hrs to fill 1 part
The pumop take1 hour to fill 3/7 portion
Now the difference of area = (1/2-3/7)=1/14
This 1/14 part of water drains in 1 hour
Total area=1 part of water drains in (1x14/1)hours= 14 hours
So the leak can drain all the water of the tank in 14 hours.

First, let's find the rate of filling up with and without leakage.

Assume that the volume of the tank is 2 liters (for easy calculation).
Then the rate of filling without leak = 1 liter/hour.
Rate of filling with leakage = 2 liters/ (7/3) hours = 6/7 liters per hour.
Now we get the leakage rate ie = 1 - 6/7 = 1/7 liters/hour.

So, time taken = total volume/rate of leakage = 2/ (1/7) = 14 hours.

Let consider 12-litre tank and pump fill in 2 hours(120 mins) w/o leakage.
with leakage, it takes 2 hours 20 mins,
means its take 20 mins extra,
so pump fill 2 litres extra in leakage condition,
total 14 litres fill by the pump but in tank 12 litres left,
means in 2hrs 20mins leakage leaks only 2 litres,
the total is 12 litres,
2 litres x6=12 litres.
7/3 hours(2 hrs 20 mins)x6 = 14 hours.