Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 5)
5.
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
Answer: Option
Explanation:
Suppose, first pipe alone takes x hours to fill the tank .
Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.
 |
1 | + | 1 | = | 1 |
| x | (x - 5) | (x - 9) |
 |
x - 5 + x | = | 1 |
| x(x - 5) | (x - 9) |
(2x - 5)(x - 9) = x(x - 5)
x2 - 18x + 45 = 0
(x - 15)(x - 3) = 0
x = 15. [neglecting x = 3]
Discussion:
68 comments Page 4 of 7.
Debasish said:
9 years ago
Is there any shortcut method for solving it?
Lalitha said:
9 years ago
Why you neglect 3?
Robot said:
9 years ago
If you assume the value corresponding to the second pipe to be x,
The other two factors would be x + 5, x - 4 with respect to the 1st and the 3rd pipe.
Here,
1/ (x) + 1/ (x + 5) = 1/ (x - 4).
In this case, the answer turns out to be 10.
Help me.
The other two factors would be x + 5, x - 4 with respect to the 1st and the 3rd pipe.
Here,
1/ (x) + 1/ (x + 5) = 1/ (x - 4).
In this case, the answer turns out to be 10.
Help me.
Minion said:
9 years ago
@Sameeera.
Your solution seems to be apt to this problem. Thank you.
Your solution seems to be apt to this problem. Thank you.
Brryne said:
10 years ago
Dear Friends. The solution according to me is correct.
The reason as to why X = 3 is neglected is because there is no negative time.
For example: 3 hrs. If you substitute X with 3, then,
1/(x-5) will be 1/(3-5).
Which are 1/-2 hours (negative thirty minutes: never in real life!)
The reason as to why X = 3 is neglected is because there is no negative time.
For example: 3 hrs. If you substitute X with 3, then,
1/(x-5) will be 1/(3-5).
Which are 1/-2 hours (negative thirty minutes: never in real life!)
Santosh h said:
10 years ago
Just remember these:
1. Faster fil = lesser time.
2. If total time=x, then for 1 hr it is 1/x.
3. When you get two different answers after solving the equation just substitute the answers (like x=15) in the original equation and check whether it is feasible.
1. Faster fil = lesser time.
2. If total time=x, then for 1 hr it is 1/x.
3. When you get two different answers after solving the equation just substitute the answers (like x=15) in the original equation and check whether it is feasible.
Sagar said:
10 years ago
Looking at the question, it is quite evident that the first two pipes simultaneously fills the tank in same time.
Assuming the time taken by pipe A to fill the tank is X hours, then for every hour, it will fill 1/X of the tank.
Now, "second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe" --> So the second pipe will require lesser time than pipe A to fill tank i.e. (X-5) hours.
Second tank is also 4 hours slower than the third pipe, so third pipe will be X-5-4 = X-9 hours.
Summing up, the equation goes like:
1/X + 1/X-5 = 1/X-9.
Assuming the time taken by pipe A to fill the tank is X hours, then for every hour, it will fill 1/X of the tank.
Now, "second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe" --> So the second pipe will require lesser time than pipe A to fill tank i.e. (X-5) hours.
Second tank is also 4 hours slower than the third pipe, so third pipe will be X-5-4 = X-9 hours.
Summing up, the equation goes like:
1/X + 1/X-5 = 1/X-9.
Amol Phad said:
10 years ago
Please somebody tell me in easier way.
Vaisakh said:
10 years ago
Let first pipe alone takes x hours to fill the tank. Then if we take (x+5), it means 2nd pipe fills the tank 5 hours slower than the first pipe. So it's not possible. Just think about it.
Dipika Sinha Bhaduri said:
10 years ago
This question is a very easy one.
Let 2nd pipe's speed be x.
So 1st pipe's speed will be x+5.
And 3rd pipe's will be x-4 (as per the ques).
Now, it is given in the question that The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone.
Hence 1/x + 1/x+5 = 1/x-4.
Solving this we get x = -2 and 10. Obviously x = 5 therefore speed of 1st pipe is 10+5 = 15.
Let 2nd pipe's speed be x.
So 1st pipe's speed will be x+5.
And 3rd pipe's will be x-4 (as per the ques).
Now, it is given in the question that The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone.
Hence 1/x + 1/x+5 = 1/x-4.
Solving this we get x = -2 and 10. Obviously x = 5 therefore speed of 1st pipe is 10+5 = 15.
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