Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 8)
8.
Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
Answer: Option
Explanation:
Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
 |
1 | + | 1 | = | 1 |
| x | (x + 6) | 4 |
 |
x + 6 + x | = | 1 |
| x(x + 6) | 4 |
x2 - 2x - 24 = 0
(x -6)(x + 4) = 0
x = 6. [neglecting the negative value of x]
Discussion:
18 comments Page 1 of 2.
Vishwa said:
9 years ago
@Neha.
Let A = X , B= X+6.
Total Work= LCM of X, X+6 = X(X+6) ----->(1)
Work done by A in 1hr= X(X+6)/X = X+6.
Work done by B in 1hr= X(X+6)/X+6 =X.
Now the total work done by A and B in 1 hr: X+X+6 = 2X+6.
The total work should be completed in : eq(1)/2X+6 = x(X+6)/2X+6 = 4.
Now solving the quadratic eq x^2-2X-24 = 0.
Then, x= 6 answer, neglect -ve value.
Let A = X , B= X+6.
Total Work= LCM of X, X+6 = X(X+6) ----->(1)
Work done by A in 1hr= X(X+6)/X = X+6.
Work done by B in 1hr= X(X+6)/X+6 =X.
Now the total work done by A and B in 1 hr: X+X+6 = 2X+6.
The total work should be completed in : eq(1)/2X+6 = x(X+6)/2X+6 = 4.
Now solving the quadratic eq x^2-2X-24 = 0.
Then, x= 6 answer, neglect -ve value.
(5)
Vaishnavi keshari said:
1 month ago
Let the time of A=x.
So the time of B will be x+6.
since time A : B = x : x+6.
therefore efficiency = x+6 : x.
total work = 4*(2x+6).
Since the difference of time is 6, so,
total work *(1/x-1/6+x) = 6,
x= 6 (time of A to filled the pipe alone).
So the time of B will be x+6.
since time A : B = x : x+6.
therefore efficiency = x+6 : x.
total work = 4*(2x+6).
Since the difference of time is 6, so,
total work *(1/x-1/6+x) = 6,
x= 6 (time of A to filled the pipe alone).
RashmiBs said:
1 decade ago
A+B's 1 hours work = 1/4.
B's 1 hours work = 1/6.
First let us find out the B's Part = (1/6)-(1/4).
This is actual B's part of work = 1/12.
Now, find actual A's part = new value of B-(A+B).
= (1/12) - (1/4).
= 2/12 = 1/6 = 6 Hours.
B's 1 hours work = 1/6.
First let us find out the B's Part = (1/6)-(1/4).
This is actual B's part of work = 1/12.
Now, find actual A's part = new value of B-(A+B).
= (1/12) - (1/4).
= 2/12 = 1/6 = 6 Hours.
(8)
Leela said:
8 years ago
@Rashmibs.
Why do you subtract 1/4 from 1/6 to know b's actual part? According to your logic---- ( (a+b) - (b's part) ) will only give you a's part no?
Why do you subtract 1/4 from 1/6 to know b's actual part? According to your logic---- ( (a+b) - (b's part) ) will only give you a's part no?
(1)
Achal Kalpande said:
9 years ago
A x x+6 u/m.
B x+6 x u/m.
LCM = x(x + 6).
x(x + 6)/(x + 3) = 8.
Now put values in options in x.
When both sides equate, we get x = 6.
B x+6 x u/m.
LCM = x(x + 6).
x(x + 6)/(x + 3) = 8.
Now put values in options in x.
When both sides equate, we get x = 6.
Paandu said:
8 years ago
1/A + 1/B = 1/4
a:b = 6:12 = 1:2;
a/b = 1/2.
Here we have to find b
b = 2a;(keep in the place of b)
1/a + 1/2a = 1/4
2 + 1/2a = 1/4
a = 6.
a:b = 6:12 = 1:2;
a/b = 1/2.
Here we have to find b
b = 2a;(keep in the place of b)
1/a + 1/2a = 1/4
2 + 1/2a = 1/4
a = 6.
(1)
ABDULLAH said:
9 years ago
@Aman.
Because we have to obtain work done in one hour and we have to find the total time to complete a full work by A or B.
Because we have to obtain work done in one hour and we have to find the total time to complete a full work by A or B.
Abhijeet said:
6 years ago
If we take factor 12x 2 = 24.
10 and 2 is a factor.
And the answer will be 2.
10 and 2 is a factor.
And the answer will be 2.
(3)
Anshul said:
9 years ago
Pipe B takes 6 hours more than Pipe A to fill the same tank so x+6 for pipe B.
Rohit said:
1 decade ago
How to solve x+6+x/x(x+6)-1/4. to form equation. Please tell me the steps..
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