Aptitude - Pipes and Cistern - Discussion

A in one hour = 1/12 part,
B in one hour = 1/15 part,
C in one hour = 1/20 part,

Now, A is opened continuously and B and C one hour alternately. So, 1st hour A+B, then for the next hour A+C.
A+B in one hour= 1/12 + 1/15= 3/20 parts,
A+C in one hour= 1/12 +1/20= 2/15 parts,
Now, in the first 2 hours ( ie; A+B for first hour and A+C for second hour)= 3/20 + 2/15 parts= 17/60 parts.

For the tank to be filled the part must become 60/60.
We know, 17 * 3 = 51 (which is the multiple less than 60),
So, 2 * 3 hours = 6 hours.
In 6 hours, part filled is 51/60. Now, simplifying it we get 17/20.
17/20 part is filled, so the rest part to be filled is 1-17/20= 3/20.
A+B in first hour.
A+C in second hour.
A+B in third hour.
A+C in fourth hour.
A+B in fifth hour.
A+C in sixth hour.
Now, it is the turn for A+B.
A+B can fill the rest in 1 hour, since part filled by A+B in one hour is 3/20.
So, the total time taken is 6+1 hours= 7 hours.

Hopes all got it.

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.
It's 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.

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.

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.

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.

Thanks @Vino.

Thanks @Ravi.

@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.

@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.

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