# Quantitative Aptitude Questions (Time and Work) for SBI Clerk / IDBI Executive 2018 Day-53

Dear Readers, SBI is conducting Online preliminary Examination for the recruitment of Clerical Cadre. preliminary Examination of SBI Clerk was scheduled from June/July 2018. To enrich your preparation here we have providing new series of Time and Work- Quantitative Aptitude Questions. Candidates those who are appearing in SBI Clerk Prelims Exam can practice these Quantitative Aptitude average questions daily and make your preparation effective.

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1) 2 men or 5 women or 7 boys can finish a work in 469 days, and then the number of days taken by 7 men, 5 women and 2 boys to finish the work is?

1. 134
2. 106
3. 100
4. 98
5. None of these

2) 6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 15 men finish the same work-?

1. 2
2. 4
3. 7
4. 9
5. None of these

3) 12 men can complete a work in 9 days. 3 days after they started the work, 2 men left and 8 more men joined. How many days will it take to complete the remaining work?

1. 2 days
2. 3 days
3. 4 days
4. 5 days
5. None of these

4) 25 women earned 143500 by working 35 days. How many men must work for 12 days to receive 70848 provided the daily wages of a man is twice that of a woman?

1. 18
2. 20
3. 15
4. 22
5. None of these

5) Kiran can do a piece of work in 9 days and Kumar can do the same work in 18 days. They started the work. After 3 days Sanjay joined them, who can complete the same work in 3 days. What is the total number of days in which they had completed the work?

1. 12
2. 8
3. 4
4. 6
5. None of these

6) Arun can do a piece of work in 10 days, Bala in 15 days. They work together for 5 days, the rest of the work is finished by Chitra in two more days. If they get Rs. 6000 as wages for the whole work, what are the daily wages of Arun, Bala and Chitra respectively (in Rs)?

1. 2400, 1600, 2000
2. 2000, 1800, 2200
3. 1800, 2600, 1600
4. 1500, 2500, 2000
5. 2800, 1400, 1800

7) (x-2) person can do a work in x days and (x+7) person can do 75% of the same work in (x-10) days. Then in how many days can (x+10) person finish the work?

1. 27 days
2. 12 days
3. 25 days
4. 18 days
5. None of these

8) 20 Men can complete a work in 12 days and 24 Boys can complete same work in 20 days. 16 men and 8 boys started working and worked for 12 days. How many more days are needed to complete the work?

1. 4 days
2. 7 days
3. 9 days
4. 2 days
5. No more days needed.

9) A and B each working alone can do a work in 12 and 36 days respectively. They started the work together but A left after sometime and B finished the remaining work in 4 days. After how many days from the start did A leave?

1. 10 days
2. 8 days
3. 12 days
4. 3 days
5. None of these

10) A man ‘A’ can do a piece of work in 10 days, man ‘B’ can do the same piece of work in 12 days while man ‘C’ can do it in 15 days. They started work together but after 2 days ‘A’ left the work and the remaining work was completed by ‘B’ and ‘C’ together. In how many days remaining work will be completed?

1. 5 days
2. 7 days
3. 2 1/6 days
4. 3 1/3 days
5. None of these

2 men = 7 boys => 1man= 7/2 boys

5 women= 7boys=> 1 women=7/5boys

7men + 5women + 2boys= (7×7/2) + (5×7/5) + 2 = 67/2boys

Now, B1*D1= B2*D2

7*469=67/2*D2

D2= 98days

6c + 2m = 6days

1M=2C

3m + 2m = 6 days

5m can do a piece of work in 6 days

15 men can do it in,

= > 5*6/15= 2 days

Total work = men * days

12 men can complete a work in 9 days

Total work = 12*9 = 108

3 days work = 12*3 = 36

Remaining work = 108 – 36 = 72

Now, the total men = 12 – 2 + 8 = 18 men

Remaining work can be completed in,

= > 72/18 = 4 days

1 women’s 1 day wages=Total wages/(no of days*total women)

=>143500/(25*35)=164

1 men’s 1 day wages=2*1 women’s 1 day wages=2*164=328

Total men=70848/(12*328)=18 men

Kiran’s one day work = (1/9) days

Kumar’s one day work = (1/18) days

Sanjay’s one day work = (1/3) days

(Kiran + Kumar)’s one day work = (1/9) + (1/18) = 3/18 = 1/6 days

3 day work = (1/6)*3 = (1/2)

(1/2) of the work can be completed. Remaining is,

= > 1-(1/2) = ½

(Kiran + Kumar + Sanjay)’s one day work,

= > (1/6) + (1/3) = 3/6 = ½ days

Whole work can be completed by three of them in 2 days.

According to the question,

(1/2)*2 = 1 day

Total days = 3 days + 1day = 4 days

Arun’s 1 day work = (1/10) days

Bala’s 1 day work = (1/15) days

(Arun + Bala)’s 1 day work = (1/10 + 1/15) = 25/(10*15) = 5/6

Remaining work = 1- (5/6) = 1/6 work

(1/6)* Chitra’s whole work = 2

Chitra’s whole work = 12 days

Chitra’s 1 day work = (1/12) days

Arun : Bala : Chitra = (1/10) : (1/15) : (1/12) = 6: 4: 5

15’s = 6000

1’s = 400

The daily wages of Arun, Bala and Chitra is, Rs. 2400, Rs. 1600 and Rs. 2000 respectively.

3/4 * (x-2)x = (x+7)(x-10)

x – 6x – 280 = 0

x = 20; x = -14

So x = 20 days.

Person        days

18               20

30               ?

(18*20) = 30*y

Y = 360/30 = 12 days

20M * 12 days = 24B *20 days

Then 1M=2B

Now 16M + 8B==>16M+4M=12 days i.e. 20M=12 days.

So no more days needed to complete the work.

(A+ B)’s one day work = (1/12) + (1/36) = 1/9 days

After x days of the work, A left the job and B finished the remaining work in 4 days.

(A+ B)’s x day work = x/9 days

Remaining work = 1- (x/9) = (9-x)/9 days

[(9-x)/9]*36 = 4

X = 8 days

(A+B+C)’s one day work = (1/10) + (1/12) + (1/15)

= > (6 + 5 +4)/60 = 15/60 = ¼

(A+B+C)’s 2 day work = 2/4 = ½

Remaining work = ½

(B+C)’s one day work = (1/12) + (1/15) = 3/20

According to the question,

= > (1/2)*(20/3)

= > 10/3 days

= > 3 1/3 days

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