Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 15)
15.
Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
Answer: Option
Explanation:
| Part filled in 2 hours = | 2 | = | 1 |
| 6 | 3 |
| Remaining part = |  |
1 - | 1 |  |
= | 2 | . |
| 3 | 3 |
 (A + B)'s 7 hour's work = |
2 |
| 3 |
| (A + B)'s 1 hour's work = | 2 |
| 21 |
C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }
| = | |
1 | - | 2 | |
= | 1 |
| 6 | 21 | 14 |
C alone can fill the tank in 14 hours.
Discussion:
44 comments Page 2 of 5.
Bansul yadav said:
6 years ago
A+B+C=6hr-------6ltr TC------1ltr/he fill tank.
A+B+C can run 2hr only - 2 LTR tank fill both.
Remaining part 6 - 2 = 4.
Now C is closed after 2he now remaining part fill A+B in 7hr.
So A+B = fill 4ltr ----7hr , so 1hr---4/7 effi, C efficiency is 3/7, So C can fill the tank Alone 6/3/7 = 14.
A+B+C can run 2hr only - 2 LTR tank fill both.
Remaining part 6 - 2 = 4.
Now C is closed after 2he now remaining part fill A+B in 7hr.
So A+B = fill 4ltr ----7hr , so 1hr---4/7 effi, C efficiency is 3/7, So C can fill the tank Alone 6/3/7 = 14.
(1)
Md. Hussain Ahmed said:
6 years ago
Let C alone can fill the tank in x hours.
2 hours work of A,B & C + 7 hours work of A & B = 1,
According to the condition,
2/6 + 7(1/6 - 1/x) = 1,
2/6 + 7 (x - 6/6x) = 1,
2/6 + 7x - 42/6x = 1,
2x + 7x - 42/6x = 1,
9x - 42 = 6x,
9x - 6x = 42,
3x = 42,
x = 42/3,
x = 14.
2 hours work of A,B & C + 7 hours work of A & B = 1,
According to the condition,
2/6 + 7(1/6 - 1/x) = 1,
2/6 + 7 (x - 6/6x) = 1,
2/6 + 7x - 42/6x = 1,
2x + 7x - 42/6x = 1,
9x - 42 = 6x,
9x - 6x = 42,
3x = 42,
x = 42/3,
x = 14.
Sugirtha said:
9 years ago
A pipe takes 9 mins to fill the tank. B takes 11.24 mins to fill the tank. C is a draining pipe. Three pipes are opened simultaneously after some time C is closed, A and B fills the remaining tank in 3.75 mins. How much Time C takes to empty the tank? Anyone please solve this.
Nurul said:
7 years ago
Part filled in 2 hours = 2/6 = 1/3. Remaining part = 2/3.
(A + B)'s 7 hours' work = 2/3.
So, (A+B)'s 1 hr work = 2/21,
Therefore C's 1 hour's work
= [(A+B+C)'s 1 hour]-[(A+B)'s 1 hr work].
=(1/6 - 2/21) = 1/14.
C alone can fill the tank in 14 hours.
(A + B)'s 7 hours' work = 2/3.
So, (A+B)'s 1 hr work = 2/21,
Therefore C's 1 hour's work
= [(A+B+C)'s 1 hour]-[(A+B)'s 1 hr work].
=(1/6 - 2/21) = 1/14.
C alone can fill the tank in 14 hours.
(1)
Rechu said:
9 years ago
Assume total work = 6.
So, efficiency of a + b + c = 1.
we know w = e * t.
w = 1 * 2 {given}
= 2.
Balnce = 4 work.
a + b = 4 work.
w = e * t.
4 = e * 7.
e = 4/7.
Then c = 1 - 4/7 = 3/7.
c alone = 6/3/7 = 14 hrs.
So, efficiency of a + b + c = 1.
we know w = e * t.
w = 1 * 2 {given}
= 2.
Balnce = 4 work.
a + b = 4 work.
w = e * t.
4 = e * 7.
e = 4/7.
Then c = 1 - 4/7 = 3/7.
c alone = 6/3/7 = 14 hrs.
Shiny said:
9 years ago
A takes 9 mins to fill. B takes 11.24 mins to fill. C is draining pipe, three pipes are opened after sometime C alone closed, A and B takes 3.75 mins to fill the tank. How much C alone take to empty the filled tank? Can you please solve this.
Varidhi Deshpande said:
5 years ago
This can be easily solved.
1/A + 1/B + 1/C = 1/6.
1/A + 1/B = 1/6 - 1/C.
Therefore, a+b+c worked for 2hrs + a+b worked for 7hrs ( a+b=1/6 - 1/c) = Total work i.e., 1
i.e., 2(1/6) + 7(1/6 - 1/C ) = 1.
2/ 6 - 7/6 -1 = 1/C.
c = 14.
1/A + 1/B + 1/C = 1/6.
1/A + 1/B = 1/6 - 1/C.
Therefore, a+b+c worked for 2hrs + a+b worked for 7hrs ( a+b=1/6 - 1/c) = Total work i.e., 1
i.e., 2(1/6) + 7(1/6 - 1/C ) = 1.
2/ 6 - 7/6 -1 = 1/C.
c = 14.
(3)
Jayasurya said:
6 years ago
Assume total work = 6.
So, efficiency of a + b + c = 1.
We know w = e * t.
w = 1 * 2 {given}
= 2.
Balance = 4 work.
a + b = 4 work.
w = e * t.
4 = e * 7.
e = 4/7.
Then c = 1 - 4/7 = 3/7.
c alone = 6/3/7 = 14 hrs.
So, efficiency of a + b + c = 1.
We know w = e * t.
w = 1 * 2 {given}
= 2.
Balance = 4 work.
a + b = 4 work.
w = e * t.
4 = e * 7.
e = 4/7.
Then c = 1 - 4/7 = 3/7.
c alone = 6/3/7 = 14 hrs.
Surekha Merla said:
5 years ago
@All.
Here, I've taken tank capacity as 42 litres.
A + B + C = 6 hours.
Remaining part by A + B in 7 hours.
So, LCM of 6 and 7 is 42.
If we take LCM's, cancellations will be easy in further steps.
Here, I've taken tank capacity as 42 litres.
A + B + C = 6 hours.
Remaining part by A + B in 7 hours.
So, LCM of 6 and 7 is 42.
If we take LCM's, cancellations will be easy in further steps.
Dinesh said:
1 decade ago
A&B&C can fill a tank together in 6 hrs
but they worked together for 2 hrs only
so u hv to calculate the time finished in 2 hrs so
A&B&C=1/6
For two hrs =2/6=1/3
but they worked together for 2 hrs only
so u hv to calculate the time finished in 2 hrs so
A&B&C=1/6
For two hrs =2/6=1/3
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