Time and Work Problems Set – 1: NABARD Grade B | IBPS PO | IBPS RRB| IBPS Clerk| SBI

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 Time and Work Quiz is important for exams such as IBPS PO, IBPS Clerk, IBPS RRB, SBI PO,RBI, NICL, SBI SO, LIC, IBPS Exams, SBI Exams and other competitive exams.

Time and Work Problems Set – 1

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1.A group of 28 men complete a job in 24 days. Another group of men, in which 14 men work with double their efficiency and remaining men work at their normal efficiency, complete the same job in 14 days. How many men are there in the group?
A. 28
B. 20
C. 34
D. 53
E. None of these

2.Some students can complete the home work in 80 days. If there were 10 students less in  the class, then it could be finished in 20 days more, Calculate the initial numbers of students in the class-

  1. 50
  2. 40
  3. 100
  4. 80
  5. None of these

3.Twin brother, Priyank and Matin, working separately can complete an assignment in 6 hr and 10 hr respectively. If they work in shift of 1 hr alternately, how much work will be remaining after 4 hr ?
A.  11/15
B.  7/15
C.  8/15
D.  2/3
E. None of these

4.  Tanmay and Jack work on a task together to complete it in some days. If Tanmay increases his efficiency by 20%, then he and Jack together would be able to complete the same task in 2 days less. Jack alone can finish the task in 30 days. In how many days can Tanmay complete the task?
A. 25
B. 27
C. 30
D. 35
E. None of these

5.To finish a work Ram will take three times as long as Sam and Tim together and Tin will take twice as long as Ram and Sam together. If the three person Ram, Sam and Tim together finish the work in 32 days, then how long will Sam take to finish the work alone?

A. 22
B. 118/7
C. 384/5
D. 211/3
E. None of these

6.Ranjeet is able to finish a work i n 65 days while Robert takes 8 days to finish the same work. If Ranjeet worked alone for 4 days and Robert joined him from 5th day, then the total number of days taken to complete the work?
A. 5  2/7
B. 1  1/7
C. 3  2/7
D. 5  1/7
E. None of these

7.Farukh worked for 2 hr and then Rasheed joined him. Farukh then worked for four more  hr and then left the work. After Farukh left, Rasheed took another 20 hr to complete the remaining work. If both of them work together they can complete the work in 18 hr. How many hr will Farukh and Rasheed take respectively to complete the work if they work separately?
A. 20 hr, 35 hr
B. 24 hr, 36 hr
C. 36 hr, 24 hr
D. 54 hr, 27 hr
E. 57 hr, 24 hr

8.5 men or 15 women could do a work in 80 days. How long it take to finish the work if 15 men and 12 women do the work on alternate days, starting from 15 men?
A. 561 / 15 days
B. 631 / 15 days
C. 100 / 11 days
D. 12 Days
E. 11 Days

9.A firm undertook a work to complete in 40 days and employed 20 men. 8 days before scheduled time one-third work was still left. How many more men are required to complete the work on time?
A. 16
B. 15
C. 12
D. 20
E. 25

10.12 men and 16 women together can complete a work in 26 days while 10 women and 18 children together can complete the work in 20 days. Also 8 men and 16 children can complete the work in 25 days. 1 man, 1 woman and 1 children start the  work and from 2nd day onwords every day 1 new man, 1 new woman and 1 children join the work. On which day the work shall be completed?
A. 20th  day
B. 21st  day
C. 22nd day
D. None of these
E. CND


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Correct Answers

  1. Let the total Number of men in the other group be Y.

1E1D1= M­2E2D2

28*1*24= {(y-14)*1+14*2}*14

Y= 34Men

Option (C)

  1. Let the initial number of students=N

1D1= M­2D2

N*80 = 100*(N-10)

8N =10N-100

10N-8N=100

N= 50

Option (A)

  1. Part of the work completed in 4hours= 1/6+1/10+1/6+1/10=8/15

Remaining work= 1-8/15=7/15

Option (B)

  1. Suppose Tanmay can finish the task in ‘x’ days. So, tanmay and Jack can finish 1/x and 1/30 of task in a day respectively.

Let the number of days taken by them to finish the task together is

‘N’ days.

N/x+N/30=1

Or,N=1/[1/x+1/30]……..Equation (1)

And [(N-2)1.2/x]+[(N-2/30)]=1

Or, (N-2)=1/[(1.2/x)+(1/30)]…..Equation (2)

Equation (1) – Equation (2)

N-(N/2)=1/[1/x+1/30]-1/[(1.2/x) + (1/30)]

2= [(30x/x+30) – (30x/(x+36)]

Or,x2-24x+1080=0

Discriminant= 576-4(1080)<0;

Thus, no real roots, Such condition is not possible

  1. Let 1/R=x, 1/S=y and 1/T=z

x = (y+z)/3, 3x=y+z, z=(x+y)/2, 2z=x+y

x+y+z=1/32

3z=1/32, z=1/96

4x=1/32, x=1/128

1/128+y+1/96=1/32

Y=5/384=1/S

Sam will take 384/5 days to finish the work.

Option (C)

  1. Work done by Ranjeet in one day=1/6

Work done by Robert in one day=1/8

Work done by Ranjeet in 4days= (1/6)*4=2/3

Remaining work after fourth day= 1-2/3=1/3

Work done by Ranjeet and Robert together in one day=1/6+1/8=7/24

If they work,together time taken to complete 1/3 part of the work=

(1/3) ÷ (7/24)

=(1/3)*(24/7)

=8/7days

Total time taken to finish the work is = 4+8/7

=36/7=5(1/7)days

Option(D)

  1. Number of hours Faizal worked= 2+4=6hours

Rasheed worked for (4+20)=24hours

Let Faizal can complete the work alone in F hours and Rasheed can complete the work alone in R hours Per hour work of Faizal=1/F

And that of Rasheed=1/R

Therefore,

1/F+1/R=1/18

Or,18/F+18/R=1……..Equation (1)

As per question, Faizal worked for 6 hours and Rasheed for 24 hours

to complete the work

  • 6*(1/F)+ 24*(R)=1…….Equation (2)

Operate (1) – 3x(2)

18/F+18/R-18/F-72/R=1-3

  • -54/R= -2 => R = 27

On substituting the value of R in eq.(1) we get F=54

Option (D)

  1. Given that 5men= 15 womenè1man= 3women

5men can do the work in 80 days so 15 men do the work in 80*5/15=

80/3days

Work done by 15 men in a day=3/80

3women=1man => 12women= 4men

5 men can do the work in 80days so 4men can do the work in 80*5/4=

100days

i.e. work done by 12women(or 4men) in a day= 1/100

if 15 men and 12 women work on alternate days,

work done in 2 days=3/80+1/100=19/400

work done in 40 days= 19/400*20=380/400

In 41 days work completed =395/400

In 42 days work completed= 399/400

Rest of the work would be completed in 1/15days by 15men

So time taken to completed the work=631/15days

Option (B)

  1. Let total number of men after 32 days be X

Work completed in 32 days=2/3

Rate of work of one man= 2/(3*32*20)

Work left=1/3

  • [2/(3*32*20)]*8*x=1/3

X=40. So, 20 extra men are required.

Option (D)

  1. LCM of 26,20 and 25=1300

Let the total work be 1300 unit

Let ‘m’ be the units of work done by 1 man in 1day

‘w’ be the units of work done by 1 woman in 1 day

And ‘c’ be the units of work done by 1 child in 1 day

26(12m+16w)=1300…….(1)

20(10w+18c)=1300……..(2)

25(8m+16c)=1300……….(3)

On solving equation (1), (2) and (3);

We get, m=1.5units, w=2units and c=2.5units

Work done on day 1=1.5+2+2.5=6*1units

Work done on day 2= (6+6*1)=6*2units

Work done on day 3 = 6*3units

Let the work is completed on nth day

6*1+6*2+…….+6n≥1300

6n(n+1)/2 ≥ 1300

3n(n+1) ≥1300

The smallest integer value of n satisfying the above equation will be the required number of days

For n=20

3n(n+1)=1260

For n=21

3n(n+1)=1386≥1300

Therefore, the work shall be finished on 21st day.

Option (B)

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