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:
[A].
10
[B].
12
[C].
14
[D].
16
Answer: Option B
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 }
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
Ramya said:
(Mon, Jul 23, 2012 10:27:26 PM)
(A+B+C) - (A+B) can give you the answer.
(A+B+C) =1/6 and (A+B+C) in 2hrs=2/6 and remaining part 1-2/6=2/3.
So (A+B) in 7 hrs is 2/ (3*7) =2/21.
1/6-2/21=1/14 so answer is 14.
Sharvari said:
(Sat, Aug 25, 2012 12:12:40 AM)
How to calculate (A+B)in 7 hours ??
Nfl said:
(Fri, Jan 11, 2013 08:56:54 PM)
@Sharvari.
A+B+C = 2/6= 1/3 ..remaining part of cistern filled by A+B by 7 hours.
Remaining part 2/3 in 7 hours.. for exact 1 hour by A+B= 2/3*1/7=1/21.
(A + B)'s 7 hour's work =