Aptitude - Pipes and Cistern - Discussion

Thanks @Roshni.

It is coming as 6 hours.

Guys,
If 2 hr = 17/60 part.
Then ?hr =1 unit.
Cross multiplication we get 7 hr.

@All.

Just take a common multiple of these no. (12,15,20) ---> 60(and consider it as tanks total capacity).

Now, divide this capacity with each of these to get the rate of work of each tap
tap A ---> 60/12 --> 5 units /hr.
tap B ---> 60/15 --> 4 units /hr.
tap C ---> 60/20 ---> 3 units /hr.

Hence we can say --
A+B can fill 9 units/hr and A+C can fill 8 units /hr.

Since we are given that B and c are to be opened alternatively and all is opened for all the time.
we can do
(A+B)+(A+C)+(A+B)+........till we get their sum as 60(which is our tanks capacity)
9+ 8 + 9 + 8 +9 + 8 + 9 = > 60units.
i.e 7 hrs.

@All.

I believe its taken 6hrs straight after 2hrs because if all the three taps are kept open all the time, then it will take 5hrs to completely fill the tank. But, since in the question its given that B & C are opened alternatively every hour, that's why I guess it would take one hour more for the tank to be filled up.

Thanks @Ravi.

Thanks @Vino.

Consider the total capacity = X Litre
Amount filled in 1h by tank "A" = (X/12) Litre
Amount filled in 1h by tank "B" = (X/15) Litre
Amount filled in 1h by tank "C" = (X/20) Litre.

Now the problem says that the tank "A" is kept open and tanks "B" and "C" are opened alternatively for 1h until the tank gets full.

For the first 1 hour, the amount that both "A" and "B" can fill collectively = (X/12)+(X/15)=(27X/180)=(3X/20).

For the next 1hour the amount that both "A" and "C" can fill collectively = (X/12)+(X/20)= (32X/240)=(2X/15).

Total amount of water after 2 hours = {(3X/20)+(2X/15)} = 17X/60.
Now, this amount (17X/60), corresponding to 2hrs, repeats as a unit.
Since the max.amount (Capacity) = X Litre
(17X/60) * n = X
n = 3.5.
17X/60 Litres of 2hrs repeats for 3.5 times which means 3.5*2 = 7 hrs.

As First A and B was started then after B is turned off and C is started so, After 6 hours ((A+C)+(A+B)+(A+C)) 17/20th part of the tank is filled when c is turned off now.
As given A+B fill the 3/20'th part in 1 hour so total 6+1 hours is the time taken to fill the Cistern.

Taking the lcm(12,15,20) = 60 liters (capacity of the tank).
In 1 hr : A can fill 5 liters(60/12), B = 4 liters C = 3 liters.
Its given that A is open all time and B & C alternately.

So, in 1st hr : A+B = 9 liters.
2nd hr : A+C = 8 liters.
3rd hr : A+B = 9 liters.
4th hr : A+C = 8 liters.
5th hr : A+B = 9 liters.
6th hr : A+C = 8 liters.
7th hr : A+B = 9 liters.
Total = 60 liters (Capacity of the tank).
So Ans = C. 7 hours.