Aptitude - Pipes and Cistern - Discussion

See......LCM of 12,15,20 is 60.
Therefore Total Tank Capacity is 60.

The individual capacity of A in One hour is 60/12=5.
The individual capacity of B in One hour is 60/15=4.
The individual capacity of C in One hour is 60/20=3.

A+B=5+4=9 in one hour.
A+C=5+3=8 in one Hour.

in two hours 17 liter. 17*3=51 liter in 6 hour and turns is A+B=9 in one Hour means.

6+1=7 Hour Final Answer.

According to me, the solution is

A * full time work + B * half time work + C * half time work = full part work.
=> (1/12) * x + (1/15) * (x/2) + (1/20) * (x/2) = 1.
x = (120/17) = 7.06

@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.

Suppose it takes "x" time for the complete tank to fill,so now tap"A" runs for all the time that is A*x whereas B and C runs alternatively so B*(x/2) &C*(x/2) respectively.

Therefore Ax +(B+C)*x/2=1.
That is x/12 +(1/15+1/20)*x/2=1.

Solving x, we get 7.

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.

How to solve this by lcm method? Please explain me.

A-12,B-15,C-20 TOTAAL FACTOR IS 60,
A+B+C 1 HOUR WORK IS 5,
SO B one hour work is 15/5=3, C=20/5=4.

Here one thing mentioned A is working all time so we do not take that value so A+B =4+3=7 HOURS.

I think the Answer is 7(1/9) hours.

Thanks for the answer @Piyush.

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.