Aptitude - Pipes and Cistern - Discussion

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.

@Senthil.

Lets start afresh,

If you think in terms of speed/rate of filling the tank of 'l' Litres capacity, then the speed of slower pipe will be x litre/min and the speed of faster pipe will be 3x litre/min.

But, if you think in terms of time, then the slower pipe will take x min(or 3x min) and the faster one will take x/3 min(or x min) to fill the tank.

Now, if you consider the second case i.e.,consider in terms of time, then you will only need to find out how much portion of the tank's capacity each pipe can fill in 36 min.

Since, in 1 min the slower pipe can fill 'l'/x litres of the tank's capacity(In x min 'l' litre), and

In 1 min the faster one can fill 'l'/(x/3) litres of the tank's capacity(In x/3 min 'l' litre).

Therefore, in 36 min slower pipe will fill 36*('l'/x) liters, And
In 36 min faster pipe will fill 36*('l'/(x/3)) liters.

Since, both the pipes are contributing in filling the tank of capacity 'l' Litres.

Therefore,

36*('l'/x) + 36*('l'/(x/3)) = 'l'.

('l'/x) + ('l'/(x/3)) = 'l'/36.

(1/x) + (1/(x/3)) = 1/36.

(1/x) + (3/x) = 1/36.

Hope, it will be clear now.

Whenever we say ram works twice as fast as shyam in a given timeframe. it means it takes two shyams to complete that work in that given timeframe because single shyam can only do half of ram's capacity. kindly always keep this common sense in mind while attempting these kind of ques independent of their form.

So solution: one pipe works 3 times faster than other..let name of this pipe A
let name of other pipe (slower)B

B is equivalent to shyam here..so if both pipes can fill one tank in 36 mins. both means A+B and as per our common sense A is equivalent to 3 B's. therefore
(B+B+B)+B = 36
B = 36/4? NOT POSSIBLE
as if 4 people can do a work in 1 min then 1 person will do it in 4*1 mins. coz more people less time and less people more time.

What we understand by this is timse is inversely proportional to work done. therefore if four B's doing it in 36 mins then single B will do it in 4*36 = 144.

To all those who have not understood yet.

Try this.

It is given that- One pipe can fill a tank three times as fast as another pipe.

Not let 1st pipe is A (faster) and 2nd pipe is B (slower).

Now let pipe A fills the tank in x/3 minute.

Therefore in 1 minute pipe A will fill 1/ (x/3) part of the tank.

Similarly let pipe B fills the tank in x minute (because pipe B is slower therefore it will take more time).

Therefore in 1 minute pipe B will fill 1/x part of the tank.

Therefore in 1 minute both the pipe (A and B) fill 1/36 part of the tank.

Now it is given that- Together the two pipes can fill the tank in 36 minutes.

Therefore A+B=36 mins.

Substitute the value of A and B according to 1 min we get.

=> 1/ (x/3) + 1/x =1/36.

=> 3/x +1/x= 1/36.

=> 4/x =1/36.

=> 4*36=x.

=>x=144 (ans).

Above answer is right 144 mins.

How when the tank was empty by one slow pipe (B) & 3 times faster pipe then slow its called pipe (A).

Know imagine the tank contain 4 liters of water and time taken to empty is 36 min.

As per above condition the B pipe can take 1 liter of water is 36 min mean the A pipe will take 3 liters at the time (3 times more than B. So 3x1=3).

If the pipe A as been removed the work as gone for B. But the B is 3 times slower.

Now A pipe as take 36 min for 3 liter means B will take 3 times more time 108 min (36x3=108).

Already the pipe B was have taken 36 min (for 1 liter) and adding the 108 min for balance 3 liter after adding the answer was 144 min.

(or).

For 1 liter it take 36 min means for 4 liter its 144 min as per above calculation.

Let A be faster pipe & B be slower pipe.

We know that [b] Capacity = Work/Time [/b].

Assume the capacity of A as 3 unit/min (i.e. A pipe can fill three times more capacity than B as given in the question).

Then the capacity of B is 1 unit/min.

Total capacity (A+B) = 3+1= 4 unit/min.

If the pipes together (A+B) fill the tank in 36 min, then net work = capacity*time.

i.e Net work = 4*36 = 144 units.

Now the time taken by slower pipe B = Network/Capacity of B.

Time (B) = 144/1 = 144 min.

Similarly for A pipe = 144/3 = 48 min.

Ans: 144 min.

I have another solution,

Suppose slower pipe is taking X min, then in that time faster pipe will fill 3 tanks.

Now both start filling tanks,then when slower pipe will finishing filling till that time faster also done with filling 3 tanks.

Total number of tanks filled is Slower= 1, and faster = 3, sum total is 4 tanks, NOw time taken for filling 1 tank is 36, Therefore time taken for filling 4 tanks is 36*4 = 144.

Therefore it indicates same time taken by slower pipe also to complete fill the tanks.

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.

Let take x is the time taken by a faster pipe and 3x be the time taken by the slowest pipe.
hence 3x+x=36,
x=9.

which means in total 36 min.
9 min taken by fastest pipe and remaining 27 min taken by the slowest pipe.
So instead of using the fastest pipe, we use the slowest pipe then it takes 3 times more time than the slowest pipe.

Hence, 9*3=27.
So total time taken will be 27+27=54.

Please correct me, if I am wrong.

Let V be the volume of the tank, the pipes are filling.

If x and x/3 mins are the time taken by pipe 1 and pipe 2, then V/x and V*3/x are the volumes filled by each of the pipes per minute.

So,
V/x + V*3/x = V/36.
Here, V/36 is the volume that is filled in the tank per minute by both pipes together. V gets cancelled and that is how we get the expression,
1/x + 3/x = 1/36.
While proceeding to solve this, we obtain 144 as the answer.