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.
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.
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.
Ronnie said:
8 years ago
@Lavanya.
Answer is 3 mins.
Answer is 3 mins.
Lavanya said:
8 years ago
Two pipes M and N can fill a cistern in 24 min and 32 min, repectively. If both the pipes are opened together, then after hoe many minutes N should be closed so that the tank is full in 18 minutes?
Please answer it.
Please answer it.
(1)
Devendar said:
8 years ago
Guys, please read the question properly their it is given as.
2nd pipe 5hrs father then 1st pipe.
2nd pipe 4 hrs slower then 3rd pipe.
Now consider 2nd pipe as x.
So,
1st pipe 5hrs slower then 2nd pipe =x-5.
3rd pipe 4hrs faster then 2nd pipe =x+4.
Now solve the equation.
1/(x-5) + 1/(x)= 1/(x+4).
x-5+x/(x (x-5))=1/(x+4).
2x-5 = x (x-5)/(x+4).
(x+4)(2x-5)=x (x-5).
2x^2 + 8x -5x - 20 = x^2 -5x.
x^2 + 8x -20 = 0.
x^2 - 2x + 10x - 20 =0.
x (x-2) + 10 (x-2) =0.
(x-2)(x+10) =0.
x =2 or.
x = -10.
This is correct.
2nd pipe 5hrs father then 1st pipe.
2nd pipe 4 hrs slower then 3rd pipe.
Now consider 2nd pipe as x.
So,
1st pipe 5hrs slower then 2nd pipe =x-5.
3rd pipe 4hrs faster then 2nd pipe =x+4.
Now solve the equation.
1/(x-5) + 1/(x)= 1/(x+4).
x-5+x/(x (x-5))=1/(x+4).
2x-5 = x (x-5)/(x+4).
(x+4)(2x-5)=x (x-5).
2x^2 + 8x -5x - 20 = x^2 -5x.
x^2 + 8x -20 = 0.
x^2 - 2x + 10x - 20 =0.
x (x-2) + 10 (x-2) =0.
(x-2)(x+10) =0.
x =2 or.
x = -10.
This is correct.
(1)
Basavaraj said:
8 years ago
Why 1/x-9 Comes to negative?
(1)
Reshma said:
8 years ago
Nice explanation @Tanu.
Bhagyasree said:
8 years ago
Suppose A takes x hours then B takes x-5 hours.
Ex: It means that A completed in 10 hours where B Completed only in 5 hours
Now C take y hours where B Completes in y+4 hours(slower),
So x-5 =y+4..............y=x-9,
Here capability of A+B = C,
1/x+1/x-5 = 1/y,
Now, You can easily solve this eq 1/x+1/x-5 = 1/x-9.
Ex: It means that A completed in 10 hours where B Completed only in 5 hours
Now C take y hours where B Completes in y+4 hours(slower),
So x-5 =y+4..............y=x-9,
Here capability of A+B = C,
1/x+1/x-5 = 1/y,
Now, You can easily solve this eq 1/x+1/x-5 = 1/x-9.
Prawesh Pradhan said:
9 years ago
@Virajgouda, if take X as 3 the answer will be negative. And time can't be negative. X - 5 and x - 9 would be -2 and -6.
Virajgouda said:
9 years ago
Why do you neglect 3? What is the reason?
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