# Practice Aptitude Questions (Pipes& cisterns) for Bank PO/Clerk Exams 2016

Practice Aptitude Questions (Pipes& cisterns) for Bank PO/Clerk Exams 2016 Set-5:
Dear Readers, Important Practice Aptitude Questions for Upcoming Bank PO/Clerk Exams was given here with Explanations. Aspirants those who are preparing for the examination can use this.

1).Pipe A and B can fill a tank in 10 hours and 12 hours respectively. When a third pipe C, which is an outlet pipe is also opened, then the tanks can be filled in 15 hours. Pipe C can empty the full tank in
a)  6 (1 / 4) hours
b)  8(4 / 7) hours
c)  20 hours
d)  None of these
2).A pipe can fill a cistern in 8 hours, but due to a leak in the bottom, it is filled in 10 hours. If the cistern is full and then pipe is turned off, then in how many hours would the leak empty the cistern ?
a)  20 hours
b)  24 hours
c)  40 hours
d)  45 hours
3).Pipe A and B can fill a tank in 10 hours and 8 hours respectively. After certain time, pipe A was closed. It took a total of 6 hours to fill the tank completely. For how many hours did pipe A work ?
a)  4(1 / 4) hours
b)  2(1 / 2) hours
c)  3(1 / 3) hours
d)  5(1 / 3) hours
4).There are three pipes x, y and z, attached to a tank. while pipes x and y are inlet pipes, pipe z is an outlet pipe. For the 1st hour, pipe x, alone is turned on. For the 2nd hour, pipe y, alone is turned on. For the 3rdhour, pipe z alone is turned on. Then again pipe x is opened for 4thhour, then pipe y and son on. This continuous till the tank is completely full. Find the time required to fill the tank completely if it is given that x takes 4 hours& y takes 5 hours respectively to fill the tank when each is working alone, and z takes 6 hours to empty the tank.
a)  10 hours
b)  12(1 / 4) hours
c)  9(3 / 5) hours
d)  None of these
5).Pipe A and B can fill a tank in 10 hours and 8 hours respectively. A pipe C can empty the full tank in 40 hours. All the three pipes were opened for 3 hours after which pipe C was turned off. Find the difference in the time required now and time required, had pipe C not be opened at all.
a)  1 / 3 hours
b)  15 / 2 hours
c)  2 hours
d)  3(1 / 2) hours
6).Two pipes A& B are attached to a tank. While pipe A can empty the tanks in 4 hours, pipe B can fill the tank in 2 hours. If half of the tank was already full when both the pipes were opened, then how much time is required by both the pipes to fill the tank, thereafter?
a)  2  hours
b)  3 hours
c)  4.2 hours
d)  None of these
7).Pipe A can fill a tank in 3 hours. Due to leakage at bottom of the tank, the full tank can be emptied in 12 hours when pipe A is turned off. Pipe B is an outlet pipe attached to the tank which works at twice the speed of the leakage. Find the time required to fill the tank when both the pipes are opened simultaneously.
a)  12 hours
b)  6 hours
c)  3 hours
d)  5 hours
8).A cistern has three pipes, A, B and C. The pipes A and B can fill it in 4 and 5 hours respectively and pipe C can empty it in 2 hours. If the pipes are opened in an order at 1, 2 and 3 A.M., when will the cistern be empty ?
a)  3 P.M
b)  7 P.M
c)  4 P.M
d)  5 P.M
9).Two taps A and B can fill a tank in 25 min and 20 min respectively. But the taps are not opened properly, so the taps A and B allow (5 / 6)th and (2 / 3)rd part of water, respectively. How long will they take to fill the tank ?
a)  13 min
b)  15 min
c)  12 min
d)  16 min
10).A water tank has three taps : A, B, and C. A fills 4 buckets in 24 min, B fills 8 buckets in 1 hr and C fills 2 buckets in 20 min. If all the taps are opened together, a full tank is emptied in 2 hours. If a bucket contains 5 litres water, what is the capacity of the tank ?
a)  120 L
b)  240 L
c)  180 L
d)  60 L
1). b) 2). c) 3). b) 4). c) 5). a) 6). a) 7). a) 8). d) 9). b) 10). b)
Solution:

1).   Let the Pipe C take x hours to empty the tank, then when all 3 are opened simultaneously, work done in one hour = (1 / 10) + (1 / 12) â€“ (1 / x) = (11 / 60) â€“ (1 / x)
(11 / 60) â€“ (1 / x) = (1 / 15)
7 / 60 = 1 / x
x = 60 / 7 hours
= 8(4 / 7) hours

2).Method 1 : Using shortcut
Time = ab / (b â€“ a) = (8 Ã— 10) / (10 â€“ 8) = 80 / 2 = 40 hours
Method 2 : Let â€˜xâ€™ be number of hours, in which the leak empties the cistern, then
(1 / 8) â€“ (1 / x) = 1 / 10
(1 / x) = (1 / 8) â€“ (1 / 10) = (5 â€“ 4) / 40 = 1 / 40
x = 40 hours

3).1 hourâ€™s work of Pipe A and B = (1 / 10) + (1 / 8) = 9 / 40
(hence both the pipes would take 40 / 9 hours to fill the tank.)
Now, pipe B has worked for a total of 6 hours, and, let pipe A work a total of â€˜xâ€™ hours (which is less than 6)
(1 / 10)x  + (1 / 8)6 = 1
x / 10 = 1 â€“ (3 / 4)
x / 10 = 1 / 4
x = 10 / 4 = 2(1 / 2) hours

4).Work done or tank filled in 1 hour = 1 / 4 , tank filled in 2nd hour = 1 / 5 and tank emptied in 3rd hour = 1 / 6
Hence in 3 hours, work done = (1 / 4) + (1 / 5) â€“ (1 / 6)  = (15 + 12 â€“ 10) / 60 = 17 / 60
Now (17 / 60) + (17 / 60) + (17 / 60) = 51 / 60 which is filled in 3 + 3 + 3 = 9 hours
Work left = 1 â€“ (51 / 60) = 9 / 60 = 3 / 20
Next would be pipe xâ€™s turn, (1 / 4) is filled in 1 hour
3 / 20 is filled in [1 / (1 / 4) ] Ã— 3 / 20 = 3 / 5 hour
Time required to fill the tank completely = 9 + (3 / 5) = 9(3 / 5)hrs

5).In 3 hours pipe C has emptied (1 / 40) Ã— 3 = 3 / 40th, which would be filled by Pipe A and B together in 1 / 3 hour
[ 9 / 40 th part is filled by A& B in 1 hour, so(3 / 40)th part would be filled in (40 / 9) Ã— (3 / 40) = 1 / 3 hour ]
Hence, extra time taken when Pipe C is opened for 3 hours is 1 / 3 hour.

6).Work done in hour = (1 / 2) â€“ (1 / 4) = 1 / 4
1 / 4 is filled in 1 hour,
1 / 2 would be filled in 2 hours. [half is already filled]

7).Pipe B works at twice the speed of leakage, hence it can alone empty the full tank in 12 / 2 = 6 hours
Work done in 1 hour = (1 / 3) â€“ (1 / 12) â€“ (1 / 6) = (1 / 3) â€“ (1 / 4) = 1 / 12
It will take 12 hours to fill the tank

8).Part of the tank filled in 2 hours = (1 / 4) + (1 / 4) + ( 1 / 5) = 7 / 10
Part of the tank emptied in an hour when all three pipes are opened = (1 / 4) + (1 / 5) – ( 1 / 2) = 1 / 20
7 / 10 the part would be emptied in (7 / 10) Ã— 20 after 3 A.M = 14 hours = 5 P.M

9).Actual time taken by tap A = (6 / 5) Ã— 25 = 30 minutes
Actual time taken by tap B = (3 / 2) Ã— 20 = 30 minutes
Part of the tank filled in 1 minute = (1 / 30) + (1 / 30) = 2 / 30 = 1 / 15
Hence, the tank will be filled in 15 minutes. Or work done in 1 min by Tap A
= (5 / 6) Ã— (1 / 25) = 1 / 30 and by Tap B
= (2 / 3) Ã— (1 / 20) = 1 / 30

10).In 2 hours water emptied by tap A, B, and C is as follows
Tap A fills 1 bucket in (24 / 4) = 6 min and 20 buckets in 120 min or 2 hours.
Tap B empties 8 Ã— 2 = 16 buckets in 2 hours
Tap C empties 1 bucket in 20 / 2 = 10 min and 12 buckets in 120 min.
Total buckets emptied in 2 hours = 20 + 16 + 12 = 48
Hence, total capacity = 48 Ã— 5 = 240L