# Quantitative Aptitude Questions (Time and work) for SBI Clerk 2018 Day- 50

Dear Readers, SBI is conducting Online preliminary Examination for the recruitment of Clerical Cadre. preliminary Examination of SBI Clerk was scheduled from June 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). A and B alone can do a piece of work in x and (x+10) days respectively and they work together can finish the work in 9 3/8 days. Then find the efficiency ratio of A and B?

a) 2:7

b) 5:3

c) 2:3

d) 1:3

e) 4:5

2). A, B and C all together can do a piece of work in 5 5/11 days. A takes 2 days less than the no of days taken by B and C together then find the no of days taken by A alone?

a) 12

b) 15

c) 10

d) 16

e) 20

3). A project manager hired 25 boys to complete a project in 10 days. 25 boys started working, after 8 days project manager notices only four-fifth of the work was completed, then how many extra boys required completing the remaining work on time?

a) 2

b) 4

c) 1

d) 0

e) 6

Direction (4-5): Answer these questions based on the information given below.

A and C together can do a piece of work in 24 days. B and C together can do the same work in 20 days. C can complete the same work in 60 days. After A has worked for 10 days, B for 10 days, time taken by C to complete the remaining job is x days.

4). Mani, Kalai and Shiva take (x-5) days, (x-10) days and (x+5) days respectively to complete a job. The three work in a rotation to complete the job with only 1 person working on a day. Who should start the job so that the job is completed in the least possible time?

a) 18 days

b) 18 2/3 days

c) 19 2/3 days

d) 20 days

e) None of these

5). Shivani and Vijay alone can do a piece of work in (x+5) and (x+25) days respectively. They started working alternatively starts with Vijay, how many days required completing the total work?

a) 37 3/5

b) 36 2/5

c) 35 1/5

d) 37 2/5

e) 38 2/5

6). A, B and C together can do a piece of work in 5 days less than A and B alone takes 15 days and C alone takes 30 days. Then find the no of days taken to complete the work by A alone?

a) 10

b) 15

c) 20

d) 12

e) 5

7). A, B and C alone complete a piece of work in 24, 12 and 36 days respectively. If two persons work on each day (BC, AC and AB) starts from BC and continue until the work was completed, then find the no of days taken to complete the whole work

a) 9 days

b) 9 ¾ days

c) 8 2/3 days

d) 9 1/3 days

e) 10 days

8). P can complete a piece of work in thrice the time taken to complete Q and R together. If P and Q together can complete the work in 9 3/5 days and R alone complete in 16 days, then find the no of days taken to complete the work by P alone?

a) 8

b) 12

c) 16

d) 24

e) 15

9). A, B and C gets a total wage of Rs.3600 if they work together. A and B gets a total wage of Rs. 2800 if they work together and A alone gets Rs.1600. Find the efficiency ratio of A, B and C

a) 1:2:3

b) 2:4:3

c) 4:3:2

d) 2:5:4

e) 1:5:3

10). P, Q and R alone can complete a piece of work in 24, 36 and 18 days respectively. P started working, after 2 days R joins with them and Q joins with them after another 2 days, then find the no of days P, Q and R worked together

a) 5 2/9 days

b) 6 2/9 days

c) 4 3/5 days

d) 6 4/9 days

e) None of these

A = x days   B = x+10 days

A+B = 1/x+1/(x+10) = 1/(9 3/8)

1/x+ 1/(x+10) = 8/75

(x+x+10)/(x2+10x) = 8/75

(2x+10)*75 = 8(x2+10x)

75x+375 = 4x2+40x

4x2-35x-375 =0

Simplify the above equation, we get x= 60/4 , -25/4 = 15 , -25/4 (Eliminate the –ve value)

X= 15 days

A = 15 and B = 25 days

Efficiency ratio of A and B = 1/15: 1/25 = 5:3

Let us take B and C together x days and A takes x-2

Given,

1/x + 1/(x-2) = 11/60

(x+x-2)/(x2-2x) = 11/60

(2x-2)/(x2-2x) = 11/60

11x2-142x+120 =0

Simplify the above equation, we get x= 132/11, 10/11

= 12, 10/11(Eliminate 10/11)

A takes to complete the work in (12-2) = 10 days

Given,

(25*8)/(4/5) = (25+x)*2/(1/5)

50 = 50+x

X=0

Therefore no extra boys required to complete the work

Direction (4-5):

A alone complete a work = 1/24 – 1/60 = (5-2)/120 = 3/120 = 1/40

B alone completes the work = 1/20 – 1/60 = 4/120 = 1/30

LCM of 40, 30 and 60 = 120

Total work = 120 units

A = 3 units per day

B = 4 units per day

C = 2 units per day

A’s 10 days work = 10* 3 = 30 units

B’s 10 days work = 10*4= 40 units

Remaining = 120 –(30+40) = 50 units

Remaining units completed by C alone = 50/2 = 25 days = x

From the above statement we get x= 25 days

Mani’s one day work = 1/(25-5) = 1/20

Kalai’s one day work = 1/(25-10) = 1/15

Shiva’s one day work = 1/(25+5) = 1/30

LCM of 20, 15 and 30 = 60

Total work = 60 units

Mani’s per day output = 3 units

Kalai’s per day output = 4 units

Shiva’s per day output = 2 units

Kalai is most efficient so she is working in first day and Mani is second most efficient so he is working in second day and Shiva is least efficient so he is working in third day.

1 cycle (3 days) = (4+3+2) = 9 units

6th cycle (18 days) = 9*6 = 54 units

Remaining work = 60-54= 6 units

19th day Kalai completes 4 units

20th days Mani completes (6-4=2)/3

Total no of days = 19+2/3 = 19 2/3 days

From the above statement we get x= 25 days

Shivani’s per day work = 1/(25+5) = 1/30

Vijay’s per day work = 1/(25+25) = 1/50

LCM of 30 and 50 = 150

Total work = 150 units

Shivani’s per day work = 5 units

Vijay’s per day work = 3 units

1 cycle (2 days) = 3 + 5 = 8 units

18th cycle (36 days) = 8*18 = 144 units

Remaining = 150 – 144 = 6 units

37th day = 3 units completed

38th day = 3/5 units completed

Required no of days = 37+ 3/5 = 37 3/5 days

Let us take A, B and C together can complete a piece of work in x days

A = x+5

Given,

1/(x+5) + 1/15 +1/30 = 1/x

1/(x+5) – 1/x = -1/15-1/30

-5/(x2+5x) = -1/10

50 = x2+5x

=>x2+5x-50 =0

Simplify the above equation we get x= 5, -10 (Eliminate –ve value)

No of days taken by A to complete the work in 10 days

LCM of 24, 12 and 36 = 72

A = 3 units

B = 6 units

C = 2 units

B+C = 8 units

A+C = 5 units

A+B = 9 units

1 cycle (3 days) = 8+5+9 = 22 units

3rd cycle (9 days) = 22*3 = 66 units

Remaining = 72-66 = 6 units

Remaining completed by B and C = 6/8 = ¾

Required no of days = 9 ¾ days

P+Q+R together completes in = (P+Q) + R = 5/48 + 1/16 = 8/48 = 1/6

P, Q and R together work in 6 days

Q and R alone takes x days and P takes 3x

Given,

1/x+1/3x= 1/6

4/3x = 1/6 => x= 8 days

No of days taken by P alone to complete the work in 24 days

Total share of A, B and C = 3600

Share of C = 3600 – 2800 = 800

Share of B = 3600 – 800- 1600 = 1200

Efficiency ratio of A, B and C = 1600: 1200: 800 = 16:12:8 = 4:3:2

LCM of P, Q and R = 24, 36 and 18 = 72 units

P = 3 units

Q = 2 units

R = 4 units

Let us take total no of days x

Given,

3x + 2(x-2) + 4(x-4) = 72

3x+2x-4+4x-16 =72

9x = 72+20= 92

X= 92/9 = 10 2/9 days

P, Q and R works together = x – 4 = 10 2/9 – 4 = 6 2/9 days

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