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 2 of 7.
Nehal said:
8 years ago
By taking pipe b=x.
Then,
For others, it will be ...(x+5) for a.
And (x-4) for c.
Eq: 1/(x+5) +1/(x)=1/(x-4).
After solving ans comes as x=10.
Is it correct?
Then,
For others, it will be ...(x+5) for a.
And (x-4) for c.
Eq: 1/(x+5) +1/(x)=1/(x-4).
After solving ans comes as x=10.
Is it correct?
(3)
Vijay said:
1 decade ago
As 2nd one is 4 hrs slower than 3rd pipe, 3rd pipe should be 4 hrs faster than 2nd. So, 3rd pipe is 9 hrs faster than 1st one.
(2)
SHUBHADIP DAS said:
5 years ago
In how much time 2nd and 3rd pipe together fo the work? Please explain the answer.
(2)
Saimanasa said:
5 years ago
Thank you @Manasa.
(2)
Pankaj said:
7 years ago
A+B = C.
Now,
Let A pipe fill the tank in =X hr.
2nd pipe 5 hr slower than A,
B = (X-5) hr
3rd pipe 4 hr slower than B,
C= X- (5+4) hr = (X-9) hr.
LCM Of (A+B) is;
A= X
X(X-5) LCM.
B=X-5.
A's 1hr work= X(X-5)/X = X-5.
B's 1hr work= X(X-5)/(X-5) = X.
_______________________________
A+B. = 2X-5
_________________________________
A+B =C.
X(x-5)/2x-5 = x-9,
X^2-5x/2x-5 = x-9 cross multiple,
2x^2-5x-18x+45 = x^2-5x (5x cancel out),
2x^2-18x+45,x^2 = 0,
X^2-18x+45 = 0,
X^2-(15+3)x+45= 0,
X^2-15x-3x+45 = 0,
X(x-15) - 3(x-15) = 0,
(X-15) (x-3) = 0,
Ans= 15 hr.
Now,
Let A pipe fill the tank in =X hr.
2nd pipe 5 hr slower than A,
B = (X-5) hr
3rd pipe 4 hr slower than B,
C= X- (5+4) hr = (X-9) hr.
LCM Of (A+B) is;
A= X
X(X-5) LCM.
B=X-5.
A's 1hr work= X(X-5)/X = X-5.
B's 1hr work= X(X-5)/(X-5) = X.
_______________________________
A+B. = 2X-5
_________________________________
A+B =C.
X(x-5)/2x-5 = x-9,
X^2-5x/2x-5 = x-9 cross multiple,
2x^2-5x-18x+45 = x^2-5x (5x cancel out),
2x^2-18x+45,x^2 = 0,
X^2-18x+45 = 0,
X^2-(15+3)x+45= 0,
X^2-15x-3x+45 = 0,
X(x-15) - 3(x-15) = 0,
(X-15) (x-3) = 0,
Ans= 15 hr.
(1)
Basavaraj said:
9 years ago
Why 1/x-9 Comes to negative?
(1)
Devendar said:
9 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)
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)
Seenam goel said:
7 years ago
Let us suppose, The 2nd pipe can fill the tank in X hrs.
Hence, 1st pipe will take (X-5)hrs.
& , 3rd pipe will take (X+4)hrs.
Now, acc to the question.
work done by 1st and 2nd pipe together = work done by 3rd pipe alone.
Hence, 1/(x-5)+1/x =1/(x+4) ;
after solving the above part we will get a quadratic equation,
i.e. x^2+8x-20=0 ;
After solving the quadratic equation we get,
x=2 & x= -10 as the result and we can see that both of them are not adequate.
Hence, acc to me there is a problem with the data given in the question.
So, DATA IS INADEQUATE.
Hence, 1st pipe will take (X-5)hrs.
& , 3rd pipe will take (X+4)hrs.
Now, acc to the question.
work done by 1st and 2nd pipe together = work done by 3rd pipe alone.
Hence, 1/(x-5)+1/x =1/(x+4) ;
after solving the above part we will get a quadratic equation,
i.e. x^2+8x-20=0 ;
After solving the quadratic equation we get,
x=2 & x= -10 as the result and we can see that both of them are not adequate.
Hence, acc to me there is a problem with the data given in the question.
So, DATA IS INADEQUATE.
(1)
Swapnil said:
7 years ago
1/x + 1/ (x-5) = 1/(x-9).
Why 1/(x-9) equate with 1/x + 1/ (x-5)?
Why 1/(x-9) equate with 1/x + 1/ (x-5)?
(1)
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