SBI CLERK PRELIMS – APTITUDE – 17

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Section : Quantitative Aptitude

Topic : Time and work

Time: 20 Minutes

Q.1)  Popeye started a work and left it after 9 days. The remaining work is completed by pope. So, the work gets completed in 25 days. Pope can alone finish the work in 30 days. Find the number of days in which Popeye can alone finish the work?

a) 14 days

b) 16 days

c) 18 days

d) 20 days

e) None of these

Q.2) Damon and Oliver work together in a farm, 25% of the work is completed in two days, if they work together. Due to some illness Oliver started at  1/3 rdof his efficiency and Damon starts working at twice his efficiency so that the work is completed in time, then what is the difference between the number of days in which both of them working together originally completed the work and number of days in which Damon alone would have completed the work if he works at his original efficiency?

a) 14days

b) 16days

c) 12days

d) 20days

e) None of these

Q.3) A certain number of people get together to contribute in the construction of a charity hospital. But for every month four people step out of this plan. Due to this the task is completed in  1/2  more years instead of one year. Then how many people were originally involved in this group?

a) 143

b) 102

c) 120

d) 133

e) None of these

Q.4) Pipe A can fill the tank in 30 hours when it works at 60% of its efficiency. Pipe B is one third as efficient as pipe A. How long will it take if both the pipes operate simultaneously at their 100% efficiency?

a) 14 hours

b) 12 hours

c) 11 hours

d) 10 hours

e) None of these

Q.5) 10 men can do 66.66% work in 10 days. 5 women can do 37.5% of same work in 9 days. Then in how many days can 80% of same work be done by 5 men and 2 women working together?

a) 14 Days

b) 15 days

c) 16 days

d) 18 days

e) None of these

Q.6)The amount of work to be done in a company is increased by 90%. By what percentage is it necessary to increase the number of workers to complete the new work in the same time as before if the new workers are 50% more efficient?

a) 75%

b) 60%

c) 50%

d) Cannot be determined

e) None of these

Q.7) 10 men can complete a work in 48 days. 16 women can complete the same piece of work in 60 days. 8 men and 20 women work together for 20 days, if only the women were to complete the remaining work in 4 days, then how many women would be required?

a) 40 women

b) 60 women

c) 80 women

d) 50 women

e) none of these

Q.8) Two farmers want to install tube wells of equal size in their farms. The first one employs 6 men and 4 women and gets the work completed in 14 days. The second one employs 8 men and 7 women and gets the work completed in 10 days to complete the same work. If a third farmer employs 1 man and 1 woman, how long would it take them to build the tube well for him?

a) 700/9 days

b) 500/9 days

c) 800/9 days

d) 75 days

e) None of these

Q.9) P, Q and R together can complete a piece of work in 8 days. Q and R started working and P joined them after 6 days and it took them another 6 days to complete the work. Find the number of days which p alone can complete the work?

a) 14 days

b) 12 days

c) 15 days

d) 18 days

e) None of these

Q.10) A chair manufacturer signs a contract to build 20000 chairs in 30 days and employs 40 men for the purpose. After 10 days, he finds that only 2000 chairs have been made. He employs some extra man to finish the work in 30 days. If the manufacturer had hired the extra men from the beginning , how many days would it have taken to complete the work?

a) 200/9 days

b) 100/3 days

c) 125/6 days

d) 34 days

e) None of these.

D.11-15) Study the following passage and answer the questions accordingly.

Five members of a family live in Mumbai namely A, B, C, D and E. A and B together can do a piece of work in 80 days. B and C together can do a piece of work in 60 days. C and D together can do a piece of work in 40 days. D and E together can do a piece of work in 20 days. A alone can do a piece of work in 120 days.

Q.11) If A, B and C together can do a piece of work in ‘x’ days then how much work could be done in the same days when E do the same work?

a) 2/5

b) 1/3

c) 2/3

d) 4/3

e) None of these

Q.12) B, C and D can complete a piece of work in ‘x’ days. If all of them work together and after three days B left and the remaining work be completed by C and D with help of E. In how many days can C, D and E do the remaining work?

a) 11 3/5 days

b) 12 3/5 days

c) 13 3/5 days

d) 14 3/5 days

e) None of these

Q.13) A, C and D can do a piece of work in x, y and z days, respectively. They work alternately in a way that first day A , second day C and third day D, fourth day A and so on. How many days will be needed to complete the work in this way?

a) 90 days

b) 80 days

c) 70 days

d) 60 days

e) None of these

Q.14) A, B and C can do a piece of work in ‘x’ days, ‘y’ days and ‘z’ days respectively. As they were ill, they could do 90% , 75% and 80% of their efficiency, respectively. How many days will they take to do the work together?

a) 43 16/33 days

b) 33 1/33 days

c) 48 16/33 days

d) 12 12/33 days

e) None of these

Q.15) C can do 1/4  of a work in 80 days, D can do 40% of the same work in 80 days and E can do 1/3 of a work in 800/3 days. Who will complete the work first?

a) C

b) D

c) E

d) Both C and D

e) None of these

D.16-20) Study the following information carefully and answer the question given below.

Given table shows percentage of number of students from 5 different schools attended 5 different tournaments.

Q.16) Number of students from C who attended chess, hockey and cricket is equal to the number of students who attended football, cricket and tennis from school E. Total number of students in school E is 160. Find the number of students who attended tennis from school C.

a) 50

b) 31

c) 48

d) 45

e) 43

Q.17) Number of students who attended hockey tournament from school A is the sum of the number of students who attended football tournament from school D and the number of students who attended cricket tournament from the same school. Then find the number of students from school D is what per cent of number of students from school A?

a) 40 20/37%

b) 41%

c) 40 5/9%

d) 20 20/37%

e) 44%

Q.18) Number of students from school B who attended hockey tournament is approximately what percentage less than the number of students from school C who attended tennis tournament. If the number of students who attended football from school B is 52 and the number of students who attended chess from school C is 15 more than the number of students who attended chess from school B.

a) 24%

b) 23%

c) 19%

d) 15%

e) 17%

Q.19) If the difference between the number of students who attended cricket from school E and number of students who attended football from same school is 25, and the total number of students from school B is 40% more than the total number of students from school E. In school E percentage of number of students who attended tennis is equal to percentage of number of students who attended cricket. Find the number of students who attended football from school B

a) 150

b) 125

c) 180

d) 140

e) 175

Q.20) Find the number of students who attended cricket from school C, if the number of students who attended tennis from school C is 80?

a) 46

b) 38

c) 15

d) 48

e) 20

1. e

Let Popeye can alone complete the work in x days

Work done by Popeye in 9 days = 9/x of the complete work

Work left =1- 9/x=(x-9)/x

1/30 of the work is done by pope in 1 day

(x-9)/x of the work is done by Popeye in 1 day

30 =16× (x-9)/x

30x = 16x-144

14x = 144

X  = 10 2/7

So, Popeye can alone complete the work in 10 2/7 days.

1. c

Let work done by Damon = x%

Work done by Oliver = y%

Total number of days required to complete the work =2× 100/25=8

X+y = 25/2

also, 2x+y/3=25/2

Solving them,

Y = 7.5 , x = 5

So, 100% of the work will be completed by Damon alone in 20 days.

Required difference = 20-8 =12 days.

1. b

Let the total number of people = x

12x = (x+(x-4)+(x-8)+(x-12)+(x-16)+(x-20)+….18 times)

12x = 18x – 4 (1+2+3+…17)

6x = 4×17× 18/2

X = 102

1. e

Pipe A can fill the tank when its efficiency is 100% =30× 60/100=18 hours

Pipe B can fill the tank in 54 hours.

When both the pipes operate simultaneously.

Time = (54×18)/(54+18)=13.5 hours.

1. c

10 men can do  2/3 of work in 10 days

10 men can do complete work in =10× 3/2=15 days

10M×15 = 150 men can do work in one day

5 Women can do  3/8 of work in 9 days

5 women can do complete work in =9× 8/3=24 days

5W×24 = 120 women can do work in one day

150M = 120 W

M = 4/5×W

5M =5× 4/5 W=4W

(4+2) W × days = 120 W

Days = 20 So, 80% of the work can be done in 20× 4/5=16 days

1. b

Let initially 100 units of work were done by 100 workers in 1 day, so efficiency of each worker was 1%

Now the work becomes 190 units.

So, they need to complete 190 units of work in 1 day.

So, the 100 workers do the 100 units of work

The remaining x new workers need to this 90 units of work in 1 day whose efficiency is 1.5 %each

So, number of new workers needed = 90/1.50=60 newworkersneeded.

So, percent by which workers need to be increased = 60/100×100=60%

1. b

10×M ×48 = W×16×60

M = 2W

So, 1 man work = 2 women work.

So, total work done by 8+10 = 18 men in 20 days = 18×20 = 360

Total required work = 10×48 =480

Work left =120

30 men can complete this in 4 days

So, women required = 60 [since 1 man work = 2 women work]

1. a

84 men days + 56 women days = 80 men days + 70 women days.

4 men days = 14 women days

1 man day = 3.5 women days

Total work = 80 men days + 70 women days = 350 women days.

1 man day + 1 women day = 4.5 women days

Number of days = 3500/45=700/9 days

1. b

One day’s work

1/p+1/Q+1/R=1/8

Also,

6 (1/P+1/Q+1/R)+ 6 (1/Q+1/R)= 1

6/8+6 (1/Q+1/R)= 1

1/Q+1/R=1/24

1/P=1/8-1/24

1/p=1/12

P= 12 days.

1. a

After 10 days, 400 men hours, only 10% of the work has been completed.

To complete the work in time, rest 90 % work must be completed in 20 days.

Man days required = 3600.

Hence he requires 180 men to work for 20 days

He must hire 140 extra men

If he hired 180 men from the beginning, number of days required = 4000/180=200/9 days.

1. Alone work of Each Person

(A+B)’s one day work=1/80;

(B+C)’s one day work=1/60;

(C+D)’s one day work=1/40;

(D+E)’s one day work=1/20

A’s one day work=1/120

Now, According to question,

B’s one day work=1/80-1/120=1/240

C’s one day work=1/60-1/240=1/80

D’s one day work=1/40-1/80=1/80

E’s one day s work=1/20-1/80=3/80

According to question,

(A+B+C)’s one day work=1/120+1/240+1/80=(2+1+3)/240=6/240=1/40

Required Answer, E alone works to finish,

E=(1/40)/(3/80)=1/40×(80/3)=2/3  of the work

Ans=c)2/3

1. (B+C+D)’s one days work=1/240+1/80+1/80=7/240

According to quesiton,

(B+C+D)’s three days work=7/240×3=7/80

Then, remaining work, =1-7/80=73/80

(73/80)/(1/80+1/80+3/80)=73/80×80/5=73/5=14 3/5  days

Ans=d) 14 3/5  days

1. A’s one day work=1/120;

C’s one day work=1/80;

D’s one day work=1/80

According to question,

work done in first 3 days,

=1/120+1/80+1/80=(2+3+3)/240=8/240=1/30

Time taken to complete 1/30  part of work=30 days

Required Answer, (Time taken to complete the whole work)

=3×30=90 days

Ans=a) 90 days

1. According to question,

A’s one day work=90% of 1/120=90/100×1/120=3/400

B’s one day work=75% of 1/240=75/100×1/240=1/320

C’s one day work=80% of 1/80=80/100×1/80=1/100

(A+B+C)’s one day’s work=3/400+1/320+1/100=(12+5+16)/1600=33/1600

Hence, time taken by them to complete the work=1600/33=48 16/33

Ans=c) 48 16/33  days.

1. Time taken to complete the work by C

=80×4=320 days

Time taken to complete the work by

D=80×100/40=200 days

Time taken to complete the work by E

=800/3×3=800 days

1. Number of students attended chess, hockey and cricket from school C =14%+41%+12%=67%

Number of students attended football, cricket and tennis from school E= 20%+25%+20%=65%

Total students in school E= 160

67% of School C = 65% of school E

67% of school C=65/100×160=104

Total students in school C =104/67×100=155.22≈155

Students attended tennis from School C=20% of 155

=20/100 ×155=31

1. Number of students attended hockey tournament from school A=15% of school A

Number of students attended football and

Cricket tournament from school D  = 37% of school D

15% of school A= 37% of school D

Required percentage =(school D total )/(school A total)×100

=(15/37)A /A×100=15/37×100=40 20/37%

1. Number of students attended football from B =52

Number of students attended hockey from B=52/25×21=44

Number of students attended chess tournament from school C=15+52/25×11=38Number of students attended tennis from school C=38/14×20=54

Required percentage =(54-44)/54×100=19% (approx.)

1. Difference between the number of students who attended football and number of students who attended cricket from school E = (25%-20%) = 5% of school E

Total students from school E=25/5×100=500

Total students from school B = 500×1.4=700

Number of students attended football from school B=25/100×700=175

1. Number of students attended tennis from school C=80

Total number of students from school C=80/20×100=400

Number of students attended cricket from school C=12/100×400=48

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