BANK & INSURANCE (TIME AND WORK & PIPE AND CISTERN) PART 2

Total Questions: 60

21. A’ alone can complete 60% of a work in 18 days while ‘B’ takes 15 days more than ‘A’ to complete it. If ‘B’ and ‘C’ together can complete the work in 20 days, then find the time taken by ‘C’ alone to complete the same work

Correct Answer: (a) 36 days  
Solution:

Time taken by ‘A’ alone to complete the entire work =
18 ÷ 0.6 = 30 days

Time taken by ‘B’ alone to complete the entire work =
30 + 15 = 45 days

Let the total work = L.C.M of 30, 45 and 20 = 180 units
Then, efficiency of ‘B’ = (180/45) = 4 units/day

Combined efficiency of ‘B’ and ‘C’ = (180/20) = 9 units/day

So, efficiency of ‘C’ alone = 9 - 4 = 5 units/day

So, time taken by ‘C’ alone to complete the entire work = (180/5) = 36 days

Hence, option a.

22. A’ and ‘B’ together can complete 50% of a work in 45 days while ‘B’ and ‘C’ together can complete 80% of the same work in 56 days. If ‘A’, ‘B’ and ‘C’ worked on it for 30 days, 60 days and 60 days, respectively and completed the whole work, then find time taken by ‘A’ alone to complete the whole work

Correct Answer: (a) 210 days  
Solution:

Time taken by ‘A’ and ‘B’ together to complete the whole work = 45/0.50 = 90 days

Time taken by ‘B’ and ‘C’ together to complete the whole work = 56/0.80 = 70 days

Let total work = 630 units (LCM of 90 and 70)
Combined efficiency of ‘A’ and ‘B’ = 630/90 = 7 units per day
Combined efficiency of ‘B’ and ‘C’ together = 630/70
= 9 units per day

Amount of work completed by ‘A’ and ‘B’ together in 30 days = 30 × 7 = 210 units
Amount of work completed by ‘B’ and ‘C’ together in 30 days = 30 × 9 = 270 days

Remaining work = 630 - 210 - 270 = 150
So, 150 units of work is completed by ‘C’ in 30 days
So, efficiency of ‘C’ = 150/30 = 5 units per day

Efficiency of ‘B’ = 9 - 5 = 4 units per day
Efficiency of ‘A’ = 7 - 4 = 3 units per day

Required time = 630/3 = 210 days

23. A’ alone can complete a work in 56 days and ‘B’ is 75% more efficient than ‘A’. If ‘A’ and ‘B’ work together for 16 days and left, then ‘C’ alone can complete the remaining work in 8 days. Find the time taken by ‘C’ alone to complete the entire work.

Correct Answer: (c) (112/3) days  
Solution:

Let the efficiency of ‘A’ = ‘4x’ units/day
Then, total work = 4x × 56 = 224x units

Efficiency of ‘B’ = 4x × 1.75 = 7x units/day

Work done by ‘A’ and ‘B’ together in 16 days = (4x + 7) × 16 = 176x units

Remaining work = 224x - 176x = 48x units

So, efficiency of ‘C’ alone = 48x ÷ 8 = ‘6x’ units/day
And so, time taken by ‘C’ alone to complete the entire work = (224x/6x) = (112/3) days

24. A’ alone takes 4 more days than time taken by ‘A’ and ‘B’ together to do the work. If ‘B’ alone takes 24 days to do the work, then find the time taken by ‘A’ to complete the work alone.

Correct Answer: (d) 12 days 
Solution:

Let the time taken by ‘A’ to do the work alone be ‘x’ days.
ATQ,
(1/x) + (1/24) = {1/(x - 4)}

(1/24) = {1/(x - 4)} - (1/x)

(1/24) = {(x - (x - 4))}/{(x - 4) × x}

x² - 4x = 96

x² - 4x - 96 = 0

x² - 12x + 8x - 96 = 0

x(x - 12) + 8(x - 12) = 0

(x - 12) + (x + 8) = 0

So, x = 12 or x = -8

So, ‘A’ takes 12 days to finish the work alone

25. 12 women working 6 hours a day take 12 days to complete a task whereas 18 children working 8 hours a day take 12 days to complete the same task. How much time will be taken by 7 women and 10 children working together to complete the same task.

Correct Answer: (a) 72 hours  
Solution:

Let efficiency of a woman and a child be ‘W’ units per hour and ‘C’ units per hour,respectively

According to the question,
12W × 6 × 12 = 18C × 8 × 12

864W = 1728C

1W = 2C

5W = 10C

Therefore, efficiency of (7 women + 10 children) = (7W + 5W) = 12W units per hour

Total work = 12W × 6 × 12 = 864W units

The time taken by 7 women and 10 children working together to complete the same task = (864W/12W) = 72 hours

26. 16 men working 8 hours a day take 10 days to cut a tree whereas 32 children working 10 hours a day take 16 days to cut the same tree. Find the time taken by 13 men and 28 children to cut the same tree

Correct Answer: (c) 64 hours
Solution:

Let the efficiency of a man and a child be ‘M’ units per hour and ‘C’ units per hour, respectively

16M × 8 × 10 = 32C × 10 × 16

1280M = 5120C

1M = 4C

1C = (M/4)

Efficiency of 28 children = 28C = (28M/4) = ‘7M’ units per hour

efficiency of 13 men and 28 children = 13M + 28C
= (13M + 7M) = ‘20M’ units/hour

Let the time taken by 13 men and 28 children to cut the same tree be ‘t’ hours

So, 16M × 8 × 10 = (13M + 28C) × t

16M × 8 × 10 = (13M + (28M/4)) × t

16M × 8 × 10 = (13M + 7M) × t

16M × 8 × 10 = 20M × t

t = {(16M × 8 × 10)/20M} = 64

27. B’ alone can complete a work in 15 days while ‘A’ and ‘B’ together can complete the same work in 8 days. ‘A’ alone started the work and left after 12 days and the rest of the work was completed by ‘B’ alone. If both of them are paid Rs. 7,200 for completing the work in this manner, then find the wages received by ‘A’.

Correct Answer: (a) Rs. 5,040  
Solution:

Let the total work = L.C.M of 15 and 8 = 120 units

Then, efficiency of ‘B’ alone = 120 ÷ 15 = 8 units/day

Combined efficiency of ‘A’ and ‘B’ = 120 ÷ 8 = 15 units/day

So, efficiency of ‘A’ alone = 15 - 8 = 7 units/day

Work done by ‘A’ alone in 12 days = 7 × 12 = 84 units

So, payment received by ‘A’ = (84/120) × 7200
= Rs. 5,040

28. A work can be completed by ‘Q’ alone in 20 days and with the help of ‘P’, it can be completed in (60/7) days. Find the total time taken to complete the whole work if they decide to work on alternate days, starting with ‘P’

Correct Answer: (d) 17 days  
Solution:

Total work = 60 units (LCM of 20 and 60)

Efficiency of ‘Q’ = (60/20) = 3 units/day

Combined efficiency of ‘P’ and ‘Q’ = {60/(60/7)} = 7 unit/day

So, efficiency of ‘P’ = (7 - 3) = 4 unit/day

Work done by ‘P’ and ‘Q’ together in 2 days (starting with ‘P’) = (4 + 3) = 7 units

Work done by ‘P’ and ‘Q’ in 16 days i.e. (2 × 8) days = 7 × 8 = 56 unit

Remaining work = 60 - 56 = 4 unit

Work done by ‘P’ on 17th day = 4 unit

Required time = (16 + 1) = 17 days

29. A’ can complete 40% of work in 4 days whereas ‘B’ can complete 20% of the same work in 3 days. Find the time taken to complete the work when both are working together.

Correct Answer: (b) 6 days  
Solution:

Time taken by ‘A’ to complete the whole work = (4/0.4) = 10 days

Time taken by ‘B’ to complete the whole work = (3/0.2) = 15 days

Let total work = 30 units (LCM of 10 and 15)

A’s efficiency = (30/10) = 3 units per day

B’s efficiency = (30/15) = 2 units per day

(A + B)’s 1 day work = 3 + 2 = 5 units

The time taken to complete the work when both are working together = (30/5) = 6 days

30. A’ can do some work in 30 days. ‘B’ can do 75% of the same work in 25 days. They both start working together and after 5 days, ‘B’ leaves. Find the time taken by ‘A’ alone to finish the rest of the work.

Correct Answer: (c) 20.5 days
Solution:

Let the total work be 240x units.

Efficiency of ‘A’ = 240x ÷ 30 = 8x units/day

Efficiency of ‘B’ = (240x × 0.75) ÷ 25 = 7.2x units/day

Total work done by ‘A’ and ‘B’ together in a day = (7.2x + 8x) × 5 = 76x units

Amount of work left = 240x - 76x = 164x units

Time taken by ‘A’ to complete the remaining work = (164x/8x) = 20.5 days