BANK & INSURANCE (TIME AND WORK PIPE AND CISTERN) PART 3

Total Questions: 60

51. 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 number of days taken to complete the whole work.

Correct Answer: (b) 9 3/4 days
Solution:

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 = 3/4

Required no of days = 9 3/4 days

52. The ratio of efficiency of Arun is to Chitra is 5:3. The ratio of number of days taken by Chitra to Banu is 2:3. Arun takes 8 days less than Chitra, when Arun and Chitra complete the work individually. Banu and Chitra started the work and left after 2 days. The number of days taken by Arun to finish the remaining work is?

Correct Answer: (d) 10 days
Solution:

The ratio of efficiency of Arun and Chitra = 5:3
The ratio of number of days taken by Arun and Chitra = 3:5
The ratio of number of days taken by Chitra and Banu = 2:3

Ratio of number of days taken by Arun : Chitra : Banu
= 6:10:15

According to the question,
⇒ Chitra – Arun = 8 days
⇒ 10’s – 6’s = 8
⇒ 4’s = 8
⇒ 1’s = 2

Number of days taken to finish the whole work,
⇒ Arun = 12 days, Chitra = 20 days, Banu = 30 days

Work done by Banu and Chitra in two days,
⇒ (2/20) + (2/30) = (1/10) + (1/15) = 25/(10×15)
= (1/6)

Rest of the work = 5/6

The number of days taken by Arun to finish the remaining work is,
Number of days = (5/6)×12 = 10 days

53. A piece of work has to be completed in 50 days, a number of men are employed but it is found that only half of the work is done in 30 days, then additional 20 men were joined to complete the work on time. How many men initially put to work?

Correct Answer: (c) 40
Solution:

Let the no of men be x,
A piece of work has to be completed in 50 days

According to the question,

Men  days  work
x   30   (1/2)
(x+20) 20  (1/2)

Work = men × days
⇒ 30x/(1/2) = (x+20)×20/(1/2)
⇒ 30x = (x+20)×20
⇒ 30x = 20x + 400
⇒ 10x = 400
⇒ x = 40 men

54. 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?

Correct Answer: (c) 4
Solution:

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) = 1/2

(Kiran + Kumar + Sanjay)’s one day work,
⇒ (1/6) + (1/3) = 3/6 = 1/2 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 + 1 day = 4 days

55. 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)?

Correct Answer: (a) 2400, 1600, 2000
Solution:

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.

56. A alone can do a piece of work in 20 days. B alone can do the same work in 5 days less than C. A and C together can do the same work in 12 days. B started working, after 2 days A joins with him for 2 days only and then C joins with him for last day only. If the remaining work completed by D alone in 5 days, then find the number of days taken by D alone to complete whole work?

Correct Answer: (b) 7 1/2 days
Solution:

A’s one day output = 1/20
B’s one day output = 1/25
C’s one day output = 1/30

Given,
=> 5/25 + 2/20 + 1/30 = (60+30+10)/300
=> 100/300 = 1/3

Remaining = 1 - 1/3 = 2/3

(2/3) × D’s whole work = 5

D’s whole work = 15/2 days = 7 1/2 days

57. Vinoth is 50% less efficient than Gokul and they together can complete the work in 10 days. If they both working together can complete whole work in 5 days, then find the number of days taken by Prabha and Vinoth working alternatively begins with Prabha

Correct Answer: (d) 14 2/3
Solution:

Efficiency ratio of Vinoth and Gokul = 1:2

Given,
1/x + 1/2x = 1/10

Simplify the above equation we get x = 15 days

Prabha one day work = 1/5 - 1/10 = 1/10
Vinoth one day work = 1/30

1 cycle (2 days) = (1/10 + 1/30) = 4/30

7th cycle (14 days) = 28/30

Remaining = 1 - 28/30 = 2/30

Remaining work completed by Prabha
= (2/30)/(1/10) = 2/3

Required number of days
= 14 + 2/3 = 14 2/3 days

58. Manu and Sharan together can do a piece of work in 6 2/3 days. Sharan and Moni together can do the same work in 12 days. Manu and Moni together can do the same work in 7 1/2 days. If Manu, Sharan and Moni working together, then find the share of Manu, out of total share of Rs.5500

Correct Answer: (b) Rs.3000
Solution:

Manu, Sharan and Moni together work
= 1/2 (3/20 + 1/12 + 2/15)
=> 1/2 × (11/30) = 11/60

Manu’s per day work = 1/10
Sharan’s per day work = 1/20
Moni’s per day work = 1/30

Efficiency ratio of Manu, Sharan and Moni
= 1/10 : 1/20 : 1/30 = 6:3:2

Share of Manu = 6/11 × 5500 = Rs.3000

59. A, B and C together can do a piece of work in 5 1/4 days. A and B together can do the same work in 6 2/5 days and B takes 4 days more than A. A and B started working together and every 3rd day C alone working until the work was completed. How many days are required to complete the whole work?

Correct Answer: (c) 9
Solution:

A, B and C together can do a piece of work in 5 1/3 days

(A + B + C)’s one day work = 3/16

A and B together can do the same work in 6 6/7 days

(A + B)’s one day work = 7/48

C’s one day work = (3/16) - (7/48) = 1/24

B takes 4 days more than A
=> B = A + 4 (Let us assume A = x, B = x+4)

(1/x) + (1/(x+4)) = 7/48

7x² - 68x - 192 = 0
x = 12

A’s per day work = 1/12
B’s per day work = 1/16
C’s per day work = 1/24

1 cycle (3 days) = 2(1/12 + 1/16) + 1/24
= 16/48 = 1/3

3rd cycle (9 days) = 3/3

Required number of days 9 days

60. Sharmi alone can do a piece of work in 12 days who is 50% and 200% more efficient than Sheeba and Akila respectively. Sheeba started working on first day, second day Sharmi joins with her and Akila joins with her on third day, how many days required to completing the whole work?

Correct Answer: (e) 8 1/2
Solution:

LCM of 12, 18 and 36 = 36

Sharmi’s per day work = 1/12 = 3 units
Sheeba’s per day work = 1/18 = 2 units
Akila’s per day work = 1/36 = 1 unit

1st day = 2 units
2nd day = 5 units
3rd day = 6 units

At the end of 3rd day 13 units completed out of 36 units

Required no. of days = 8 1/2