SBI Clerk 2016- Practice Aptitude Questions (Pipes& Cisterns)

SBI Clerk 2016- Practice Aptitude Questions (Pipes& Cisterns) Set-11:
Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

1).Two pipes can fill a cistern in 10 hours and 12 hours respectively. If both the pipes are opened together, then how much time will they take to fill the cistern ?
a)  60/11
b)  12/11
c)  65/12
d)  45/11

2).Pipe 1 and 2 can fill a tank alone in 6 hours and 9 hours respectively. Pipe 3 can empty the full tank in 18 hours. If all the three pipes are opened simultaneously, then how much time is required to fill the tank completely ?
a)  2 hours
b)  3.6 hours
c)  4.5 hours
d)  Cannot be filled

3).  Pipe A and B can fill a tank in 12 and 15 hours respectively. An outlet pipe C, can empty it in 6 hours. Initially pipes A and B are opened together, and after 5 hours pipe C is also opened. Find the time required to empty the tank ?
a)  45 hours
b)  50 hours
c)  60 hours
d)  None of these

4).Three pipes 1, 2 and 3 together take 24 hours to fill a tank. The three pipes are opened for 6 hours after which pipe 3 is closed. Pipes 1 and 2 take another 30 hours to fill the tank. In how much time would pipe 3, alone fill the tank ?
a)  40 hours
b)  30 hours
c)  68 hours
d)  None of these

5).Two pipes x and y can fill a tank in 24 and 30 minutes respectively. Both the pipes are opened for 6 minutes, after which pipe x is turned off. How much more time will pipe y take to fill the tank ?
a)  16.5 minutes
b)  22.5 minutes
c)  24 minutes
d)  Cannot be determined

6).  A tank is usually filled in 18 hours. But because of a leak in its bottom, it takes another 6 hours to fill. How much time will be taken by the leak to empty the full tank ?
a)  30 hours
b)  72 hours
c)  42 hours
d)  55 hours

7).A pipe can fill a tank in 4 hours. But due to a leak in the tank, it is filled in 5 hours. If the tank is full. How much time will the leak to empty the tank completely ?
a)  10 hours
b)  15 hours
c)  20 hours
d)  12 hours

8).Pipes 1 and 2 can fill a tank in 18 and 24 hours respectively. Both pipes work simultaneously for sometime after which pipe 1 is turned off. It takes 12 hours in all to fill the tank completely. Find the time for which pipe 1 remained turned on.
a)  9 hours
b)  10 hours
c)  11 hours
d)  12 hours

9).The volume of water flowing through a pipe is directly proportional to square of its radius. A tank has four inlet pipes with diameters as 2 cm, 4 cm, 6 cm and 8 cm. if the smallest pipe, alone, can fill a tank in 30 hours, then how much time would all the four pipes, when working together would take ?
a)  1 hours
b)  4 hours
c)  6 hours
d)  None of these

10).Two pipes A and B can fill a water tank in 20 and 24 minutes respectively and a third pipe C can empty at the rate of 3 gallons per minute. If A, B and C opened together can fill the tank in 15 minutes, the capacity (in gallons) of the tank is :
a)  180
b)  150
c)  120
d)  60

Answers:
1).a) 2).c) 3).a) 4).d) 5).a) 6).b) 7).c) 8).a) 9).a) 10).c)

Solution:

1).Work done (part of cistern filled) by first pipe in 1 hour = 1/10
Work done (part of cistern filled) by 2nd pipe in 1 hour = 1/10
Work done in 1 hour, when both the pipes are opened together = (1/10) + (1/12) = 11/60
Hence both the pipes can together fill the cistern in (60/11) or 5 5/11 hours
Answer : a)

2).Work done in 1 hour by all three pipes = (1/6) + (1/9) â€“ (1/18) = 4/18
Time required to fill the tank completely = 18/4 = 4.5 hours
Answer c)

3).Part of tank filled by A& B in 5 hours = [(1/12) + (1/15)] 5 = (9 Ã— 5) / 60 = 3/4
Work done in 1 hour when all three pipes are opened = [(1/12) + (1/15) â€“ (1/6)] = (9-10) / 60 = -1/60
Since the result (or net effect) is negative, hence tank would be emptied
So, 1/60 is emptied in 1 hour
3/4 would be emptied in [1/ (1/60)] Ã— (3/4) = 45 hours
Answer : a)

4).Tank filled by all three pipes in 6 hours = (1/24) 6 = 1/4
Remaining part = 1 – (1/4) = 3/4 is filled by pipes 1 and 2 in 30 hours
so, the entire tank would be filed by pipes 1 and 2 in = (3/4) Ã— 30 = 40 hours
1 hour work of all three pipes = 1/24
1 hour work of pipes 1 and 2 = 1/40
Hence, 1 hour work of pipe 3 = (1/24) â€“ (1/40) = 2/120 = 1/60
Pipe 3 alone would fill the tank in 60 hours.
Answer : d)

5).When pipe x is turned off (after 6 minutes)
Work done by x and y in 6 minutes = [(1/24) + (1/30)] 6 = 9/20
Remaining work = 1 â€“ (9/20) = 11/20 which would be done by pipe y alone.
1 work is done by pipe y (alone) in 30 minutes
11/20 work is done by pipe y (alone) in 30 Ã— (11/20) = 33/2 minutes = 16.5 minutes
Answer : a)

6).Consider the case when there is no leak . Then in one hour, work done = 1/18, and in 6 hours = 6/18 = 1/3.
This means 1/3rd of the tank is emptied because of the leakage in 18 + 6 = 24 hours
So, 1/3rd is emptied in 24 hours, full tank would be emptied in 24 Ã— 3 = 72 hours.
Answer : b)

7).  In 1 hour., water filled = 1/4 th of the tank.
1/4 th is emptied by leakage in 5 hours
Full tank would be emptied in 20 hours (or)
[using formula, t = (5 Ã— 4) / (5-4) = 20 hours]
Answer : c)

8).Let the time for which pipe 1 remained turned on be â€˜xâ€™ hours. Hence pipe 1 has worked for â€˜xâ€™ hours and pipe 2 has worked for 12 hours.
1/18 (x) + 1/24 (12) = 1
(x/18) + (1/2) = 1Ã  (x / 18) = (1/2)Ã  x = 9
Pipe 1 remained turned on for 9 hours.
Answer : a)

9).We are given that V = k (r2) where V is volume of water and â€˜râ€™ is radius of pipe and k is a constant.
The smallest pipe takes 30 hours to fill the tank alone, hence work done in 1 hour = 1/30
Radius = diameter / 2 = 2/2 = 1
1/30 = k (12), so k = 1/30
Work done in 1 hour by pipe 2 = 1/30 (4/2)2 = 4/30
Work done in 1 hour by pipe 3 = 1/30 (6/2)2 = 9/30
Work done in 1 hour by pipe 4 = 1/30 (8/2)2 = 16/30
In 1 hour, work done by all the four pipes = (1/30) + (4/30) + (9/30) + (16/30) = 30 / 30 = 1
Hence, the whole tank gets filled in 1 hour.
Answer : a)

10).Let the capacity of the tank be x gallons.
Quantity of water filled in the tank in 1 minute when all the pipes A, B and C are opened simultaneously
= (x/20) + (x/24) â€“ 3
According to the question
(x/20) + (x/24) â€“ 3 = x/15Ã  (x/20) + (x/24) â€“ (x/15) = 3
(6x + 5x â€“ 8x) / 120 = 3
3x = 3 Ã— 120Ã  x = (3 Ã— 120) / 3 = 120 gallons.
Answer : c)

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