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.
Karthikeyan said:
1 decade ago
How got this equation 1/x+1/x-5=1/x-9 ?
(1)
Pranesh kumar said:
1 decade ago
Why you neglect x=3 for what reasons?
(1)
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.
Pradhyumna said:
8 years ago
Can't we pick the 'B' to be as x and do the same things since When I did I got four first just by generally calculating it?
Deepi said:
1 decade ago
@ venky
See the question they have given as second pipe 5 hours faster than the first and 4 hours slower than third.
Lets consider first pipe takes X hrs for ex. 10hrs
Second pipe takes 5 hours faster than first it means 5hrs (X-5)
Third pipe is faster than second by 4hrs. That is 1hour(10-9)=(X-9).
See the question they have given as second pipe 5 hours faster than the first and 4 hours slower than third.
Lets consider first pipe takes X hrs for ex. 10hrs
Second pipe takes 5 hours faster than first it means 5hrs (X-5)
Third pipe is faster than second by 4hrs. That is 1hour(10-9)=(X-9).
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).
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).
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