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).

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).

I AM TRYING LIKE THIS but not getting.

First pipe x to fill tank then second one is 3x.
now both 1/x+1/3x=36
4/3x=36
1/x=9???
What's the problem?

@Karthik how it is x square=12 and x =144???

Let the volume b unit meter cube
so rate of individual filling = rate of combine fillin
so
1/X +1/(X/3) = 1/36
so x=144

First pipe fills 3 times faster than second one so x=3x
If together the two pipes can fill the tank in 36 minutes so x*3x=36
x=36/3x
x=12/x
x=144

Slower pipe fills in x min ,faster fills x/3.
Let for complete the tank use symbol C.

For slower pipe:
in x min it fills C
in 1 min it fills C/x

Similarly in 1min faster pipe fills 3C/x
Both fills in 36min
so,in 36min both fills C
in 1 min both fills C/36

Now, C/x+3C/x=C/36
or 1/x+3/x=1/36

Hope you got it!

Slower pipe x,faster 3x....
1/x+1/3x=1/36
=>x=48

then 1/36-1/48
we got 1/144

@Laxman:.

Your way of approaching to the problem is good, simple and understandable.

We may also solve in this manner i.e.

x + x/3 =36
=>
4x/3 = 36
or =>

x=27