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 3 of 7.
Chandni said:
1 decade ago
I solved it by taking 2nd pipe's speed as x.
Then 1st pipe speed would be x-5 and 3rd pipe = x+4.
If we resolve now, we get y = -10 and y = 2.
So, I was finding answer according to 2.
Then 1st pipe speed would be x-5 and 3rd pipe = x+4.
If we resolve now, we get y = -10 and y = 2.
So, I was finding answer according to 2.
(1)
Sachin said:
1 decade ago
How can a pipe fill in both 3 and 9 hours. Hows it possible?
Only one answer must be true?
which one how to get?
Only one answer must be true?
which one how to get?
(1)
Pranesh kumar said:
1 decade ago
Why you neglect x=3 for what reasons?
(1)
Karthikeyan said:
1 decade ago
How got this equation 1/x+1/x-5=1/x-9 ?
(1)
NAGARJUNA said:
7 years ago
Can anyone help me?
If suppose A and B are filling a tank started at same time.
A alone can fill it in 'a' minutes B alone can fill it in 'b' minutes.
So we are taking (1/a)+(1/b)=1 to find the time.
But when they are filling it simultaneously why are we taking same (1/x+1/x-5=1/x-9)?
I think after completion of a ,b is started it means we have to take a+b right? But not (1/a)+(1/b).
If suppose A and B are filling a tank started at same time.
A alone can fill it in 'a' minutes B alone can fill it in 'b' minutes.
So we are taking (1/a)+(1/b)=1 to find the time.
But when they are filling it simultaneously why are we taking same (1/x+1/x-5=1/x-9)?
I think after completion of a ,b is started it means we have to take a+b right? But not (1/a)+(1/b).
Prasad said:
1 decade ago
How you have taken 9 hours?
Ronnie said:
8 years ago
@Lavanya.
Answer is 3 mins.
Answer is 3 mins.
Monu dhaka said:
8 years ago
@Lavanya.
Answer is 10 minutes {(18/24)+(8/32)}=1.
18-8 = 10minutes.
Answer is 10 minutes {(18/24)+(8/32)}=1.
18-8 = 10minutes.
Sana said:
8 years ago
Suppose, second pipe alone takes x hours to fill the tank.
Then, first pipe and third pipe will take (x + 5) and (x - 4) hours respectively to fill the tank.
1/(x+5) + 1/x = 1/(x-4)
( x + x + 5 ) / ( x^2 + 5x ) = 1/(x-4)
Srossmultiply ( 2x + 5 ) * ( x - 4 ) = 1 * (x^2 + 5x )
Solve, you get
(x - 10)(x + 2) = 0
x = 10. [neglecting x = -2]
1st pipe = x + 5 = 15 hours.
Then, first pipe and third pipe will take (x + 5) and (x - 4) hours respectively to fill the tank.
1/(x+5) + 1/x = 1/(x-4)
( x + x + 5 ) / ( x^2 + 5x ) = 1/(x-4)
Srossmultiply ( 2x + 5 ) * ( x - 4 ) = 1 * (x^2 + 5x )
Solve, you get
(x - 10)(x + 2) = 0
x = 10. [neglecting x = -2]
1st pipe = x + 5 = 15 hours.
Rofiqul Hoque said:
1 decade ago
Let 1st pipe take time = x hours.
2nd pipe is 5 hours faster means its take 5 hours less than 1st pipe.
i.e., 2nd = (x-5) hours.
3rd pipe is 4 hours slower than the 2nd pipe.
i.e., 3rd +4 = (x-5).
=>3rd = (x-5) -4.
=>3rd = x-9.
Now,
1/x + 1/ (x-5) =1/ (x-9).
Dear friends now just solve it. Bye friends meet you again.
2nd pipe is 5 hours faster means its take 5 hours less than 1st pipe.
i.e., 2nd = (x-5) hours.
3rd pipe is 4 hours slower than the 2nd pipe.
i.e., 3rd +4 = (x-5).
=>3rd = (x-5) -4.
=>3rd = x-9.
Now,
1/x + 1/ (x-5) =1/ (x-9).
Dear friends now just solve it. Bye friends meet you again.
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