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 3 of 5.
Hemanth said:
9 years ago
Hi, I am not getting this, please give me a proper explanation.
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.
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.
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.
Deepa said:
8 years ago
1/A+1/B+1/C=1/6,
1/A+1/B=1/7.
1/6-1/7=1/42,
42/3=14.
1/A+1/B=1/7.
1/6-1/7=1/42,
42/3=14.
Vickneswari said:
7 years ago
Why did you divide 42 by 3?
And how?
And how?
A.D. said:
7 years ago
Please tell me, how to solve this question using LCM method?
Deepak MVL said:
7 years ago
first understand the question properly.
A, B, C all opened can fill in 6 hours.
ie 1/A+1/B+1/C=1/6________(x)
ie 1/6th of tank is filled if all the three operate for an hour.
So if all 3 operate for 2 hours 2/6th of the tank is filled.
So remaining is 2/3rd.(1-2/6).
After 2 hours, C is closed.
to fill the 2/3rd by A & B it takes 7 hours.
[to find the time by c alone we need to find 1/A+1/B to substitute in (x)..]
So A and B in 1 hour can complete 2/(3*7).
ie 1/A+1/B=2/21. sub in x.
1/C=1/6-2/21.
We get 1/C =1/14.
So, C takes 14 hours to fill the tank.
A, B, C all opened can fill in 6 hours.
ie 1/A+1/B+1/C=1/6________(x)
ie 1/6th of tank is filled if all the three operate for an hour.
So if all 3 operate for 2 hours 2/6th of the tank is filled.
So remaining is 2/3rd.(1-2/6).
After 2 hours, C is closed.
to fill the 2/3rd by A & B it takes 7 hours.
[to find the time by c alone we need to find 1/A+1/B to substitute in (x)..]
So A and B in 1 hour can complete 2/(3*7).
ie 1/A+1/B=2/21. sub in x.
1/C=1/6-2/21.
We get 1/C =1/14.
So, C takes 14 hours to fill the tank.
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)
Sachin said:
6 years ago
Not getting this, please explain it.
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