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Friday, 29 December 2017

PIPES & CISTERN PART 1 NEW PATTERN (with Detailed Solutions)


New Pattern PIPES & CISTERN QUESTIONS for IBPS PRE,IBPS MAINS,IBPS SO,SSC CGL,FCI,PSSB Exams,Punjab Kanongo exams

1. Two pipe p and q can fill a cistern in 12 and 15 minutes respectively. If both are opened together and at the end of 3 minutes the first one is closed; how much longer will cistern take to fill ? 

A) 35/4 minutes
B) 33/4 minutes
C) 30 minutes
D) 9 minutes

Solution :In such a question find the work of a person or thing to be finished in 1 unit of time Now p’s 1 minute work = 1/12and q’s 1 minute work = 1/15

They work together for 3 minutes So the work to be finished by both in 3 minutes = p’s 3 minute work + q’s 3 minute work = (1/12)*3 + (1/15)*3= (1/4) + (1/5) = 9/20

Therefore the remaining work (to be finished by ‘q’ alone) = 1 – (9/20) = 11/20

Now q’s 1/15 work takes 1 minuteSo its 11/20 work will take = 15*(11/20) = 33/4 minutes (option ‘B’)

2. A and B can fill a cistern in 75/2 min and 45 min respectively. Both the pipes are opened together. The cistern will be filled just in half an hour if the pipe B is turned off after? 
A) 9 minutes
B) 5 minutes
C) 10 minutes
D) 15 minutes

Solution :The part of cistern to be filled by A in 1 minute = 2/75Therefore in half an hour i.e. 30 minutes it will fill the part of cistern = 30*(2/75) = 60/75 = 4/5

So the empty part of the cistern still, which B will fill = 1 – 4/5 = 1/5

Now the part of the cistern to be filled by 1 minute = 1/45Hence the time taken by it to fill 1/5 of the whole = 45/5 = 9 minutes.

Obviously after 9 minutes it must be turned off. (option ‘A’)

3.A cylindrical overhead tank is filled by two pumps P1 and P2. P1 can fill the tank in 8 hr while P2 can fill the tank in 12 hr. There is another pipe P3 which can empty the tank in 8 hr. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open P3 when the tank is half filled so that the tank is exactly filled up by the time he is back. Due to technical fault P3 opens when the tank is one-third filled. If the supervisor comes back as per the plan what percent of the tank is still empty?
A) 20%
B) 15%
 C) 8%
D) 10%

Solution:To know this we need to know the time in which the supervisor had to come back.

Pump P3 was to open when the tank was filled half according to the supervisor’s plan. Therefore calculate the time to be taken by the pumps 1 and 2 working together in which half of the tank will be filled.

Now part of the tank to be filled by P1 + P2 in one hour = 1/8 + 1/12 = 5/24Therefore the time in which half of the tank was to filled by both of them working together = (24/5)/2 = 12/5 hours

Now the pump P3 was to open Obviously, part of the tank to be filled by all of them in one hour = 1/8 + 1/12 – 1/8 = 1/12Therefore time to be taken by all of them to fill the remaining half of the tank = 12/2 = 6 hours

Thus, total time in which the supervisor had to return = 12/5 + 6 = 42/5 hours

Now we just have to calculate the part of the tank filled by all the tanks in 42/5 hours

1/3 of the tank filled by the pumps P1 + P2 in = (24/5)/3 = 8/5 hours (from above)Remaining time in which all the pumps have to work together = 42/5 – 8/5 = 34/5 hours

Part of the tank to be filled by all the pumps working together in 34/5 hours = (1/12)*(34/5) = 17/30

So the part of the tank filled in total = 1/3 + 17/30 = 27/30 Therefore empty part of the tank on return of the supervisor = 1 – 27/30 = 1/10In percentage = (1/10)*100 = 10% (option ‘D’)

4. A and B can fill a cistern in 20 and 30 minutes respectively, while C can empty it in 15 minutes. If A, B, C are kept open successively for 1 minute each, how soon will the cistern is filled? 

A) 165 minutes
B) 170 minutes
C) 167 minutes
D) 175 minutes
 Solution: As all the cisterns are kept open successively for 1 minute each for 3 minutes, we should find the part of the cistern filled in 3 minutes i.e. one interval of time; so it is 1/20 + 1/30 – 1/15 = 1/60

It’s obvious that the pipe C which is emptying the cistern will not have to be opened in the last interval, otherwise the cistern will never fill; so we need to know the combined work of A and B in the last interval i.e. last 3 minutes.

So, A’s 1 minute work + B’s 1 minute work i.e. 2 minute work = 1/20 + 1/30 = 5/60 = 1/12It does mean all the pipes A, B and C work simultaneously to fill = 1 – 1/12 = 11/12 of the cistern

Time taken to fill 1/60 of the cistern by all the pipes working together = 3 minutesHence, time taken in all to fill 11/12 of the cistern = 165 minutes

So, total time to fill the full cistern = Time taken for 11/12 of the part + time taken for 1/12 of the part= 165 + 2 = 167 minutes (option ‘C’)

5. One pipe can fill a tank three times as fast as another pipe. If the two pipes together can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in? 
A) 81 minutes B) 108 minutes C) 144 minutesD) 192 minutes

Solution:Let the slower pipe alone fills the tank in x minutes

As the faster pipe is three times efficient as the slower one is; time taken by it to fill the tank = x/3 minutes

Therefore 1 minute work of the slower pipe + 1 minute work of the faster pipe = 1 minute work of both the pipes => 1/x + 3/x = 1/36solving x = 144

So the slower pipe will fill the tank in = 144 minutes (option ‘C’)

FASTER APPROACH RULE: If A is x times efficient than B, and working together they finish a work in y days; then time taken by A = [y(x+1)]/x; and time taken by B = y(x+1)

As the faster pipe i.e. A is faster 3 times than B; so x = 3 and y = 1

Here the slower pipe is B; so time taken by it = 36*(3+1) = 144 minutes (option ‘C’)

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