Aptitude - Pipes and Cistern - Discussion

I have solved it my way. First take the lcm of 12, 15 and 20 which is 60. So we assume the total capacity of tank to b 60 units.

That means A fills 5, B fills 4 and C fills 3 in 1 hr.
For the first hr A and B are opened so tank filled is 9 (5+4).
For the second hr A and C are opened so tank filled in 2nd hr is 8 (5+3).
So total tank filled in 2 hrs is 9+8=17. So in 6hrs 51 unit of tank is filled.

Now only 9 units of tank is to be filled which will b filled in next hr which is 7th hr and in this hr B will b opened. So in 7th hr 5 (By A) +4 (By B) =9unit will be filled which is sufficient to fill the remaining tank.

Hene total tank will be filled in 7 hrs. Hope you all got this.

Is this correct?

A is open all the time Part filled x/12.

B is open for 1 hr = 1/15.
C is open for 1 hr = 1/20.

x/12+ 1/15+ 1/20 = 1.

(5x+4+3)/60 = 1.

x = 60-7)/5 = 53/5 ~~ 5 hrs.

How come you consider 6 hrs duration for the part fillings?

For the first hour, part filled from the tank (1\12)+(1\15) = 3\20.

Second (1\12)+(3\20) = 2\15.

Third (1\12)+(1\15) = 3\20.

Fourth (1\12)+(1\15) = 2\15.

Fifth (1\12)+(1\15) = 3\20.

Lets see how much the tank full in 5 hours.

3*(3\20)+ 2*(2\15) = 43\60 So the tank is not full yet.

For the sixth hour ,part filled from the tank (1\12)+(3\20) = 2\15.

Let's see how much the tank full in 6 hour.

3*(3\20)+3*(1\15) = 17\20 so the tank is not full yet.

For the seventh hour, part filled from the tank(1\12)+(2\15) = 3\20.
Let's see how much the tank full in 7 hour.

4*(3\20)+3*(2\15) = 1 so the tank is completely full after 7 hour.

First two hours 17/60 part tank filled.

We are taking every two hour calculation so after 6 hour 17/20.

If you consider next two hours means it will become 17/15 means tank is overflowing.

So only after 6 hour we calculating remaining part that is (1-17/20) = 3/20.

Now its A and B turn to fill the tank. It fill 3/20 part in one hour. So totally (6+1) = 7 hours.

Its very simple.

Part filled in 1st and 2nd hour:

1/12+1/15+1/12+1/20 = 17/60.

Part filled : Time taken
17/60 : 2 hr
1 : ?

By solving this we will get 6.8 that is nearby 7.

Please carefully read the question A is open for all times, B and C are opened alternatively for one hour, that means if B is opened for 1st hour that time C is closed(A and B filling the tank) For 2nd hour B is closed C is opened(A and C is Filling the tank)3rd hour B opened C closed and so on...

1st hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

2nd hour:

1 hour work of A and C is 1/12+1/20 = 2/15.

3rd hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

4th hour:

1 hour work of A and C is 1/12+1/20 = 2/15.

5th hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

6th hour:

1 hour work of A and C is 1/12+1/20 = 2/15.

7th hour:

1 hour work of A and B is 1/12+1/15 = 3/20.

(Until you have to calculate 60/60).

If we add the whole we will get 1 is the answer that is capacity of tank.

3/20+2/15+3/20+2/15+3/20+2/15+3/20+2/15+3/20 = (9+8+9+8+9+8+9)/60 = 1.

Hence, time taken to fill the tank is 7 hours.

(A + B)'s 1 hour's work= 1/12+1/15 = 3/20.

(A + C)'s 1 hour's work =1/12+1/20 = 2/15.

Therefore (3/20 + 2/15)= 17/60 part of work is completed in first 2 hrs.

17/60 part of work completed in ------- 2hrs.

1 part (full work) completed in ------- x hrs.

Now cross multiply => x= 2(60/17) = 7hrs.

It written part filled in six hours then why is it multiplied by 3 and not six?

Take LCM a, b, c = 180, tank capacity 180 liters.

a - 15 liter per hrs.
b - 12.
c - 9.

1hrs a+b = 27 liter.
1hrs a+c = 24 liter.

In choice more than 6 hrs.

= 27*4 = 108.
= 24*3 = 72.

Total 180 liters in 7 hrs.