Aptitude - Pipes and Cistern - Discussion

So simple,

Time=work/capacity,

So total capacity=1/x+3/x.

And total work=1.

So total time taken by both of them=36=1/(1/x+3/x).

Another simpler approach:

(slower) pipe A takes 'x' min to fill complete tank alone.
Then pipe A in 1 min fills (1/x) portion of tank alone.

(faster) pipe B takes (x/3) min to fill the complete tank alone.
then pipe B in 1 min fills (3/x) portion of tank alone.

When both pipes together fills the tank then in 1 min they fill (1/x + 3/x) portion of tank.

This means complete tank can be filled in (x + x/3) minutes.
(4x/3) = 36.... so x = 144 min . Answer.

The best approach to these kinda questions are the logic of replacing the variables with the numbers but be careful in doing so because one should be aware of the numbers which are being replaced to the variables should be satisfying the condition of the variables.

So guys the easiest approach to the above problem is to:
P1 is three times faster than P2 right?

So I will write down a statement as follows,
P1=3P2.

So am gonna replace these with numbers as below,
P1=3. If p1=3 the above statement becomes 3=3P2.

By the cancellation we get the value of P2 as 1,
P2=1.

If a tank is to be filled by the two pipes together in 36 minutes then the values P1=3 becomes 3L/M P2 = 1L/M.

So my next statement is,
P1+P2 = 3+1.
P1+P2 = 4L/M.

If both the pipes are working together then the tank is being filled at a rate of 4L/M.

So the time taken by the to pipes to fill the tank is 36M.

Then we need to multiply the combined rate of pipes with the time to get the capacity of the tank.

4L/M*36M=144L which the capacity of the tank.

Now coming to the last part of the question.

"slower pipe alone will be able to fill the tank in:"

I already mentioned the slower pipe as P2 and the rate of P2=1L/M so the time taken by P2 to fill the tank of 144L capacity is (tank capacity/P2)=144L/1L/M=144M.

And finally sorry for the big explanation and guys please pardon me if there are any grammatical errors.
In brief,

P1 = faster pipe P2 = slower pipe.
P1=3P2.

If P1 = 3 =>3 = 3P2 =>P2 = 1.
Time = 36M.

P1+P2 = 1+3 =>P1+P2 = 4.
As the units of the pipes are liters/minute L/M
Find tank capacity,

(P1+P2)*Time=4*36=>Tank capacity = 144L.

144L capacity tank will be filled by slower pipe at its slower rate right?
rate of slower pipe P2?

P2 = 1L/M.

Time=Tank capacity/rate of pipe P2 => Time = 144/1 = 144.

Which is the required answer.

If slower pipe is x. Than faster 3x. Take both 36 min. means x = 9 let 9+3 = 12 and 12 square = 144.

If we take faster one as x,
Then d slower one will become 3x,

Thus the work done by them is equal to 1/36.

Therefore,

1/x + 1/(3x) = 1/36.
(3x + x)/(3x * x) = 1/36.
4/3x = 1/36.
x = 48.

Thus the slower one will become 144, thats the solution for taking x and 3x as the time.

First pipe x.
Second pipe 3 times faster then it is 3x.

Together will fill it in 36 min.
1/x+1/3x=1/36.

Then we will get x=48 minutes and 3x= 144 minutes faster pipe is 48 minutes Which takes less time and slower pipe is 144 minutes which takes more time.

If we take x+x/3 instead of 1/x+3/x we cannot get answer because the method of solving will be more complicated.