Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 11)
11.
One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
Answer: Option
Explanation:
Let the slower pipe alone fill the tank in x minutes.
| Then, faster pipe will fill it in | x | minutes. |
| 3 |
 |
1 | + | 3 | = | 1 |
| x | x | 36 |
 |
4 | = | 1 |
| x | 36 |
x = 144 min.
Discussion:
67 comments Page 2 of 7.
Shan said:
10 years ago
I am getting a different answer,
Two pipe fills the tank in 36 min.
Let's take slower one = A.
Faster one as B.
So, B = A*3.
If B+A = 36.
Then B = 27 and A = 9.
If B alone has to fill the tank then it will fill in.
27min (of B) + 9/3 of A = 27+3 = 30 min.
And if go like this and A alone has to fill the tank then it would be:
30*3 = 90 min.
Why I this is not right someone Please explain?
Two pipe fills the tank in 36 min.
Let's take slower one = A.
Faster one as B.
So, B = A*3.
If B+A = 36.
Then B = 27 and A = 9.
If B alone has to fill the tank then it will fill in.
27min (of B) + 9/3 of A = 27+3 = 30 min.
And if go like this and A alone has to fill the tank then it would be:
30*3 = 90 min.
Why I this is not right someone Please explain?
Indrajit said:
1 decade ago
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!
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!
Bhushan said:
1 decade ago
Above ans is right. Thing this way.....we want to find the time that the slower pipe alone can fill the tank not the time from total 36mins hence we take 1/A+1/b=1/36 so that we are finding the time from 100% and not from 36 mins. We are basically generalizing by taking the numerator as 1.
Aditi said:
1 decade ago
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.
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.
Rishabh said:
7 years ago
So as per ques.(simple trick).
Pipe A (slower)complete its work to 1unit/min.
Then Pipe B (faster)complete its work to 3unit/min.
Then total unit in 1 min=4units/min.
Then total units in total time=4*36=144units.
NOW B(Slower pipe) time=(1min/1unit)x144units = 144units.
Pipe A (slower)complete its work to 1unit/min.
Then Pipe B (faster)complete its work to 3unit/min.
Then total unit in 1 min=4units/min.
Then total units in total time=4*36=144units.
NOW B(Slower pipe) time=(1min/1unit)x144units = 144units.
(9)
Alan S said:
8 months ago
@All.
Here's an easy way:
Let the time for slower pipe be = x
Then the faster pipe will be = 3x
The ratio between both pipes is = 1 : 3
When we add 1 and 3 (1+3), we get 4.
Now simply, just multiply 4 by 36.
The answer is 144 min.
Hence, Option C is correct.
Here's an easy way:
Let the time for slower pipe be = x
Then the faster pipe will be = 3x
The ratio between both pipes is = 1 : 3
When we add 1 and 3 (1+3), we get 4.
Now simply, just multiply 4 by 36.
The answer is 144 min.
Hence, Option C is correct.
(3)
Pavan said:
8 years ago
Actually A=3B because A pipe fill d tank 3 times fast as B pipe.
So A=3B but in time.
Then B = A/3
Let, A= x-----(1) assume i.e B=x/3---------(2)
Formula 1/A+1/B=1/36 substitute both in
1/x+1/x/3 = 1/36,
1/x+3/x = 1/36,
4/x = 1/36.
=>x=4*36 =>144.
So A=3B but in time.
Then B = A/3
Let, A= x-----(1) assume i.e B=x/3---------(2)
Formula 1/A+1/B=1/36 substitute both in
1/x+1/x/3 = 1/36,
1/x+3/x = 1/36,
4/x = 1/36.
=>x=4*36 =>144.
Raj said:
1 decade ago
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.
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.
Sai Shodhan Rao said:
5 years ago
@All.
slower pipe = 3x . 1 part of work = 1/3x.
faster pipe = x. 1 part of work = 1/x.
total time = 36 min 1 part of time = 1/36.
Then, it should be;
total work:
1/x + 1/3x = 1/36.
4/3x = 1/36.
x = 48 mins.
And the slower pipe 3x = 3(48) = 144 mins.
slower pipe = 3x . 1 part of work = 1/3x.
faster pipe = x. 1 part of work = 1/x.
total time = 36 min 1 part of time = 1/36.
Then, it should be;
total work:
1/x + 1/3x = 1/36.
4/3x = 1/36.
x = 48 mins.
And the slower pipe 3x = 3(48) = 144 mins.
(8)
Madhu reddy said:
1 decade ago
Let's A = Fast pipe.
B = Slow pipe.
A takes x time. And B takes 3x time.
So, in 1 min. 1/x + 1/3x = 1/36.
By solving. 4x/3x^2 = 1/36.
Finally x = 4*36/3.
x = 48.
Time taken by slow pipe = 3x = 3*48 = 144 min.
B = Slow pipe.
A takes x time. And B takes 3x time.
So, in 1 min. 1/x + 1/3x = 1/36.
By solving. 4x/3x^2 = 1/36.
Finally x = 4*36/3.
x = 48.
Time taken by slow pipe = 3x = 3*48 = 144 min.
(1)
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