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__Practice Sets on Time& Work / Pipes& Cisterns | IBPS PO Mains Spl – Download in PDF__Dear Readers, Here we have given the important practice set questions on Time& Work / Pipes& Cisterns, aspirants those who are preparing for the examination can also download in pdf and make use of it.

**1).**Working together, Arjun and Ramesh can complete a task in 15 days. Arjun, working alone, can complete the task in 20 days, while Manoj can complete the same task by himself in 30 days. How many days will Manoj and Ramesh working together take to complete the task?

a)15 days

b)20 days

c)24 days

d)25 days

e)26 days

**2).**Ajay and Vijay are assigned the task to build a wall. Ajay can complete the task in 30 days, while Vijay can complete the same task in 60 days. When they work together, their efficiency drops to 80%. How long will they take to complete the task working together?

a)15 days

b)20 days

c)25 days

d)32 days

e)34 days

**3).**12 Men can complete a piece of work in 36 days. 18 women can complete the same piece of work in 60 days. 8 men and 20 women work together for 20 days. If only women were to complete the remaining piece of work in 4 days. How many women would be required?

a)70

b)28

c)66

d)40

e)None of these

**4).**8 men and 4 women together can complete a piece of work in 6 days. Work done by a man in one day is double the work done by a woman in one day. If 8 men and 4 women started working and after 2 days, 4 men left and 4 new women joined, in how many more days will the work be completed?

a)5 days

b)8 days

c)6 days

d)4 days

e)9 days

**5).**P can complete a piece of work in 18 days and B can do the same piece of work in 15 days. They started working together but after 3 days P left and Q alone completed the remaining work. The whole work was completed in (approximately)

a)12.5 days

b)8 days

c)14 days

d)14.5 days.

e)18 days

**6).**A is thrice as good a workman as B and therefore is able to finish a job in 45 days less than B. Working together, they can finish the job in

a)15 days

b)11.25 days

c)30 days

d)45 days

e)60 days

**7).**64 men can do a piece of work in 30 days working 12 hours day. In how many days 80 men complete the same piece of work if they work for 16 hours a day ?

a)16 days

b)18 days

c)20 days

d)24 days

e)25 days

**8).**A road of 5 km length will be constructed in 80 days. So, 420 workers were employed. But after 60 days, it was found that only 3 ½ km road was completed. Now, how many more people were needed to finish the work in the specified time?

a)480

b)b)80

c)200

d)d)100

e)e)120

**9).**A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?

a)18 days

b)24 days

c)30 days

d)32 days

e)36 days

**10).**A and B are working on an assignment. A takes 6 hours to type 32 pages on a computer, while B takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?

a)5 hours 15 minutes

b)4 hours 30 minutes

c)8 hours 15 minutes

d)7 hours 45 minutes

e)None of these

**11).**A tank has a leak which would empty it in 8 hours. A tap is turned on which admit 6 liters a minute into the tank and it now emptied in 12 hours. How many literes does the tank holds?

a)8260 ltr

b)8640 ltr

c)8560 ltr

d)8800 ltr

e)8720 ltr

**12).**Two pipes A and B can fill a cistern in 37 ½ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if pipe B is turned off after

a)5 min

b)9 min

c)10 min

d)13 min

e)15 min

**13).**Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?

a)6/11

b)5/11

c)7/11

d)8/11

e)9/11

**14).**A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Time required by the first pipe to fill the tank is

a)30 hours

b)15 hours

c)10 hours

d)6 hours

e)5 hours

**15).**Two pipes A and B can fill a tank in 15 minutes and 40 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

a)10 min 10 sec

b)25 min 20 sec

c)29 min 20 sec

d)20 min 10 sec

e)30 min 05 sec

**Answers:**

**1)b 2)c 3)a 4)a 5)a 6)b 7)b 8)e 9)a 10)c 11) 12) 13) 14) 15) 16) 17) 18) 19) 20)**

**Solutions:**

**1. B)**

Work completed by Arjun and Ramesh in 1 day = 1/15

Work completed by Arjun in 1 day = 1/20

So, work completed by Ramesh in 1 day = 1/15 – 1/20 = 4/60 – 3/60 = 1/60

So, Ramesh will take 60 days to complete the task himself.

Ramesh and Manoj in 1 day can complete 1/30 + 1/60 = 1/20 of the task.

So, working together, they can complete the task in 20 days.

**2. C)**

Work done by Ajay and Vijay at 100% efficiency in 1 day = 1/30 + 1/60 = 1/20

So, at 100% efficiency they can complete the task in 20 days.

At 80% efficiency, they will take 20/0.8 = 25 days.

**4. A)**From question, 1 man = 2 women

Hence, 8men + 4women = (16 + 4) women = 20 women

And, 4 men + 8 women = 16 women

20 women’s 2 days work = 2/6 = 1/3 part

Remaining work = 1 – 1/3 = 2/3

Since 20 women completed 1 work in 6 days

Hence 16 women will do 2/3 part of work in = (20 * 6)/16 * (2/3) = 5 days

**5. A)**Both together worked for 3 days.

In 3 days , P can do = 3 x 1/18 = 1/6

^{th}workIn 3 days Q can do = 3 * 1/15 = 1/5

^{th}workIn 3 days work finished = 1/6 + 1/5 = 11/30

Balance work = 19/30

Balance work finished by Q => Time taken by Q to finish the balance work = 19/30 x 15 = 9.5 days

The whole work was completed in 9.5 + 3 = 12.5 days.

**6. B)**A is thrice as good a workman as B =>A = 3B

The ratio of efficiencies of A and B = 3:1

The ratio of the days taken by A and B to finish work is = 1: 3

B takes 45 days to finish the work => 3 parts = 45 days

A alone takes => 45/3 * 1 = 15 days

A and B alone can finish the work is = AB/(A + B) = 15 * 45/(15 + 45) = 11.25

Here W1=W2 because they all performing same piece of work.

Here M1 = 64 D1= 30 and H1= 12; M2 =80 D2= ? H2= 16

M1D1H1 = M2D2H2

64 x 30 x 12 = 80 X D2 X 16

On solving, we get D2 = 18 days

**9. A)**

**10. C)**

Number of pages typed by A in 1 hour = 32/6 = 16/3

Number of pages typed by B in 1 hour = 40/5 = 8

Number of pages typed by both in 1 hour = 16/3 + 8 = 40/3

Therefore Time taken by both to type 110 pages = 110 * 3/40 = 8 ¼ hours = 8 hours 15 minutes

**11. B)**In I hour the part filled by the tap = 1/8 -1/12 = 1/24

Hence, the tap can fill the tank in 24 hours.

Therefore, capacity of the tank = 24 x 60 x 6 = 8640 litres.

**12. B)**Pipe A can fill the cistern in 37 ½ = 75/2 minutes. Since it was open for 30 minutes, part of the cistern filled by pipe A = 2/75 x 30 = 4/5

So the remaining 1/5 part is filled by pipe B.

Pipe B can fill the cistern in 45 minutes. So, time required to fill 1/5

^{th}part = 45/5 = 9 minutesi.e., pipe B is turned off after 9 minutes.

**13. A)**

Ratio of water filled by A,B,C per minute = 1/30 : 1/20 : 1/10 = 2 : 3 : 6

Proportion of R any time = 6/(2 + 3 + 6) = 6/11

**14. B)**Suppose the first pipe alone can fill the tank in x hours. Then, second pipe alone can fill the tank in (x – 5) hours, third pipe alone can fill the tank in (x – 5) – 4 = (x – 9) hours

Part filled by first pipe and second pipe together in 1 hr

= Part filled by third pipe in 1 hr

=> 1/x + 1/(x – 5) = 1/(x – 9)

From here, better to find the value of x from the given choices which will be easier. Or we can solve it as follows.

(x – 5)(x – 9) + x(x – 9) = x(x – 5)

=>x

^{2}– 18x + 45 => x = 15 or 3As we take x = 9, then (x – 9) is negative which is not possible Hence x = 15

**15. C)**LCM (15, 40) = 120

Suppose the capacity of the tank is 120 litre.

Then,quantity filled by pipe A in 1 min = 120/15 = 8 litre

quantity filled by pipe B in 1 min = 120/40 = 3 litre

Quantity filled by both the pipes together in 4 minutes = 4 (8 + 3) = 44 litre

Remaining quantity to be filled = 120 – 444 = 76 litre

Time required for pipe B alone to fill this remaining quantity = 76/3 minutes = 25 1/3 minutes = 25 minutes 20 sec

Therefore, total time required = 4min + (25 minutes 20 sec) = 29min 20 sec

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