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

Total Questions: 60

31. A’ alone can complete some work in 20 days. ‘B’ is twice as efficient as ‘A’ and thrice as efficient as ‘C’. If ‘B’ and ‘C’ start working together, then after how many days ‘A’ should join them so that the work gets completed in exactly 6 days?

Correct Answer: (d) 2 days  
Solution:

Time taken by ‘B’ to complete the work alone = 20 ÷ 2 = 10 days

Time taken by ‘C’ to complete the work alone = 10 × 3 = 30 days

Let the total work be 60 units. {LCM (10, 20 and 30)}

So, efficiency of ‘A’ = (60/20) = 3 units/day

Efficiency of ‘B’ = (60/10) = 6 units/day

Efficiency of ‘C’ = (60/30) = 2 units/day

Let the number of days for which ‘A’ works be ‘x’.

ATQ:
3x + 6 × 6 + 2 × 6 = 60
3x = 60 - 36 - 12
3x = 12
x = 4

So, ‘A’ needs to work for 4 days, which means he must join ‘B’ and ‘C’ 2 days after they started the work.

32. A’ and ‘B’ alone can complete a work in 42 days and 28 days, respectively. ‘A’ and ‘B’ started working on it together and left after 7 days and then ‘C’ completed the remaining work in 14 days. Find the time taken by ‘C’ alone to complete the entire work.

Correct Answer: (a) 24 days  
Solution:

Let the total work = L.C.M of 28 and 42 = 84 units
Then, efficiency of ‘A’ alone = 84 ÷ 42 = 2 units/day
Efficiency of ‘B’ alone = 84 ÷ 28 = 3 units/day
So, combined efficiency of ‘A’ and ‘B’ = 2 + 3 = 5 units/day

So, work complete by ‘A’ and ‘B’ together in 7 days = 5 × 7 = 35 units
Remaining work = 84 - 35 = 49 units

So, efficiency of ‘C’ alone = 49 ÷ 14 = 3.5 units/day
So, time taken by ‘C’ alone to complete the entire work = 84 ÷ 3.5 = 24 days

33. A and B work on alternate days; A working on the 1st day, B on the 2nd, then A again on the 3rd followed by B on the 4th and so on. In this way, they can finish the work in 25 days. The work done by A varies every day and on any particular day d, the work done by A is d units. B works at a constant rate. The ratio of the work done by A on the 1st day to that done by B on the 2nd day is 1 : 4. Find out how much time B alone will require to finish the work?

Correct Answer: (a) 54.25 days  
Solution:

Work done by A on 1st day = 1 unit
Work done on 3rd day = 3 units, on 5th day = 5 units and so on

Since ratio of work done by A on 1st day to B on 2nd day is 1 : 4, therefore work done by B on 2nd day (and subsequent days) = 4 × 1 = 4 units

Now A works on 1st, 3rd, 5th, 7th.....25th days, i.e. on all the odd numbered days till 25.
So total work done by A = (1 + 3 + 5 + 7............+ 25) units = 169 units

Now B works on 2nd, 4th, 6th............24th days, i.e. all the even numbered days till 25.
So number of days worked by B = 12
Work done by B in one day = 4 units
So the total work done by B = 12 × 4 = 48 units

So total work done = work done by A + work done by B = 169 + 48 = 217 units

Now time taken by B alone to finish the work
= 217/4 = 54.25 days

34. Siddharth, Nikita, Vinay and Neha individually takes different number of hours to complete a task. If Siddharth, Nikita and Vinay works together then they take 6 hours 40 minutes to complete the task and if Nikita, Vinay and Neha works together then they take 8 hours to complete the task. If Siddharth and Neha working together complete the task in 13 hours 20 minutes, then how much time will all the four persons take to complete the task working together?

Correct Answer: (c) 5 5/7 hours
Solution:

Let the number of hours taken by Siddharth to complete the task = S
Work done by Siddharth in 1 hour = (1/S)

Let the number of hours taken by Nikita to complete the task = N
Work done by Nikita in 1 hour = (1/N)

Let the number of hours taken by Vinay to complete the task = V
Work done by Vinay in 1 hour = (1/V)

Let the number of hours taken by Neha to complete the task = T
Work done by Neha in 1 hour = (1/T)

Siddharth, Nikita and Vinay take 6 hours 40 minutes to complete the task
= 6 (40/60) hrs = 20/3 hrs

Work done by Siddharth, Nikita and Vinay in 1 hour = 3/20
(1/S) + (1/N) + (1/V) = (3/20) ....(1)

Nikita, Vinay and Neha take 8 hours to complete the task
Work done by Nikita, Vinay and Neha in 1 hour = 1/8

(1/N) + (1/V) + (1/T) = (1/8) ....(2)

Subtracting (2) from (1) we get
(1/S) - (1/T) = (3/20) - (1/8)
(1/S) - (1/T) = (1/40) ....(3)

Also, Siddharth and Neha take 13 hours 20 minutes to complete the task
= 13 (20/60) = (40/3) hours

Work done by Siddharth and Neha in 1 hour = (3/40)
(1/S) + (1/T) = (3/40) ....(4)

Adding (3) and (4) we get
2 × (1/S) = (4/40)
(1/S) = 1/20

Substituting in equation 4 we get
(1/T) = (3/40) - (1/20) = (1/40)

Total work done by all four persons together
= (1/S) + (1/N) + (1/V) + (1/T)
= (3/20) + (1/40) = (7/40)

Total number of hours required = (40/7) = 5 (5/7) hours

35. A group of 12 persons containing men and women can do a piece of work in 13 days, other group of 10 persons containing men and women can do the same piece of work in 16 days. Number of men in first group is 1 more than the number of men in second group. If there is another group of 2 men and 6 women which can do the same piece of work in 26 days. What is the ratio of efficiency of 1 man to that of 1 woman?

Correct Answer: (e) 5 : 1
Solution:

Let number of men in first group is ‘x + 1’ and number of men in second group is ‘x’.

Number of women in first group = 12 - (x + 1) = (11 - x)
Number of women in second group = (10 - x)

Let 1 man does ‘M’ unit and 1 women does ‘W’ unit of work in 1 day.

Total work = 26 × (2M + 6W) = 13 × [(x + 1) × M + (11 - x) × W]
= 16 × [x × M + (10 - x) × W]

Now, 26 × (2M + 6W) = 13 × [(x + 1) × M + (11 - x) × W]
52M + 156W = 13xM + 13M + 143W - 13xW
39M + 13W = 13x (M - W)

3M + W = x(M - W) ---- (1)

also,
26 × (2M + 6W) = 16 × [x × M + (10 - x) × W]
52M + 156W = 16xM + 160W - 16xW

52M - 4W = 16x (M - W) ---- (2)

From (1) and (2)-
=> 52M - 4W = 16 × (3M + W)
=> 52M - 4W = 48M + 16W
=> 4M = 20W
=> M : W = 5 : 1

36. Anil and Sunil are two painters who can paint a square wall in 6 hrs and 8 hrs respectively. They started working and after working for 2 hrs Sunil left the work and new person Ajit joins the work with Anil. Both Anil and Ajit worked for next hour and now Anil stopped working and remaining work is completed by Ajit alone in 2 hrs. Find the time taken by Ajit if he is employed to paint a circular ring in the same wall in such a way that outer diameter of the ring is equal to the side of the wall and width of the ring is 5/6 of the radius.

Correct Answer: (a) 55/6 hrs  
Solution:

Let outer radius and width of the ring is 6x and 5x respectively.
Inner radius of the ring = 6x - 5x = x

Area of the ring = π × [(6x)² - (x)²]
= (22/7) × 35x² = 110x²

Side of the wall = diameter of the ring = 2 × 6x = 12x
Area of the wall = (12x)² = 144x²

Anil and Sunil can paint the wall in 6 h and 8 h respectively.

Area of wall painted by Anil in 1 h = 144x²/6 = 24x²
Area of wall painted by Sunil in 1 h = 144x²/8 = 18x²

Let area of wall painted by Ajit in 1 h = Ax²

According to the question-
=> 2 × (24 + 18) + 1 × (24 + A) + 2 × A = 144
=> A = 12

Area of wall painted by Ajit in 1 h = 12x²

Now time taken by Ajit to paint the circular ring
= 110/12 = (55/6) h

37. Mr. Joe has three dogs namely Max, Charlie and Buddy. Max is double efficient than Charlie, Buddy alone takes 15 min more to fill a hole in the earth than the time taken by Max alone to fill the same hole. Charlie started filling the hole and after 4 min Max also joined and together they fill the hole after working for 2 more min. Find the time taken by Charlie and Buddy together to fill a hole of volume half than the volume of previous hole with double of their efficiency.

Correct Answer: (d) 5/3 min  
Solution:

Since Max is double powerful than Charlie. So, time taken by Max is half the time taken by Charlie to fill the hole.
Let time taken by Max and Charlie is ‘x’ and ‘2x’.
Time taken by Buddy = ‘x + 15’

According to question-
⇒ 4(1/2x) + 2(1/2x) + (1/x) = 1
⇒ (4/2x) + (2/2x) + (4/2x) = 1
⇒ 10/2x = 1
⇒ x = 5

Hence time taken by Max, Charlie and Buddy is ‘5’ min, ‘10’ min and ‘20’ min respectively.

Let Charlie and Buddy together take ‘t’ min to fill the hole with double their efficiencies.
⇒ (2t/10) + (2t/20) = 1/2
⇒ t = 5/3 min

38. Ram and Shyam can complete a work alone in 60 and 40 days respectively. They started doing the work together and after four days Shyam was replaced by Ajay and 8 days after that Shyam replaced Ram. Find the total number of days in which work gets complete if Ajay can complete the work alone in double the no. of days in which Ram and Shyam together complete the work

Correct Answer: (e) 23 7/11 days
Solution:

Work completed by Ram alone = 60 days
Work completed by Shyam alone = 40 days
Work completed by Ajay alone = 2/(1/60 + 1/40)
= 2×24 = 48 days

Now,
For 4 days work done by Ram and Shyam
= 4×(1/60 + 1/40) = 4/24 = 1/6

For 8 days work done by Ram and Ajay = 8×(1/60 + 1/48) = 3/10

Remaining work = 1 - (1/6 + 3/10) = 8/15

This is completed by Shyam and Ajay in ‘N’ days (let)

So,
N×(1/40 + 1/48) = 8/15
N×(11/240) = 8/15
N = 128/11 days = 11 7/11 days

Total no. of days in which work gets completed
= 4 + 8 + 128/11 = 23 7/11 days

39. 3 men and 4 women are employed to clear a forest cover of certain no. of trees. A woman alone is 25% less efficient than a man. They all started working and after 5 days of work 2 men and 1 woman more is employed such that after working for 12 more days, 1/10th of remaining forest is left to be cleared consisting of 50 trees. Find the number of trees cut down by 3 men and 2 women in a day.

Correct Answer: (a) 15  
Solution:

Let a man alone clears the forest in ‘n’ days
a man alone clears a forest = 1/n, in one day
Then a woman in 1 day alone clears = 0.75/n

In 5 days, part of forest cleared = 5×(3/n + 4×0.75/n)
= 30/n

Remaining forest = 1 - 30/n

In next 12 days, forest cleared = 12×(5/n + 5×0.75/n)
= 105/n

Remaining forest left to be cleared
= 1 - 30/n - 105/n = 1/10

9/10 = 135/n
1/n = 1/150

So, n = 150 days

A woman alone does work in = n/0.75 = 200 days

And, Let the number of trees in forest = M trees
M/10 = 50
M = 500 trees

So, work done by 3 men and 2 women in a day
= 3/150 + 2/200 = 3/100

Trees cut = 3×500/100 = 15 trees in a day

40. Lalit alone can complete a work in 48 days and Deepak alone can complete the same work in 64 days. In how many days they together will do the 1/8th part of the same work?

Correct Answer: (e) 24/7
Solution:

Lalit’s one day work = 1/48
Deepak’s one day work = 1/64

Their one day work together = 1/48 + 1/64
= 7/192

Total time taken to complete the whole work together
= 192/7 days

Total time taken to complete 1/8 of the whole work together
= 1/8 × 192/7 = 24/7 days