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

Total Questions: 60

41. Time taken by ‘A’ to do a work is half the time as taken by ‘B’ to do the same work. ‘B’ and ‘C’ together can do the same work in 18 days. ‘A’ and ‘C’ together can do it in 14 days. If ‘A’ and ‘B’ works together, then the time taken by them to finish the work would be:

Correct Answer: (c) 21 days
Solution:

Let the total work be 126 units. {LCM (14 and 18)}

Efficiency of ‘B’ and ‘C’ together = 126 ÷ 18 = 7 units/day

Efficiency of ‘A’ and ‘C’ together = 126 ÷ 14 = 9 units/day

ATQ:
Let the efficiency of ‘B’ and ‘C’ be ‘b’ units/day and ‘c’ units/day, respectively.

So, efficiency of ‘A’ = b × 2 = 2b units/day

So,
(b + c) = 7 ...(i)
(2b + c) = 9 ...(ii)

On subtracting equation (i) from equation (ii), we have;
b = 2

So, efficiency of ‘A’ = 2 × b = 4 units/day
So, required time = 126 ÷ (2 + 4) = 21 days
Hence, option c.

42. A factory has four workers namely ‘A’, ‘B’, ‘C’, and ‘D’ and they can finish a work in 12 hours, 10 hours, 15 hours and 8 hours, respectively. The workers started working in pairs, starting with ‘B’ and ‘C’. If after every hour, the less efficient worker leaves the work and the more efficient worker from the remaining two replaces him, then which of the following pairs will finish the work?

Correct Answer: (d) ‘A’ and ‘D’
Solution:

Let the total work be 120 units. {LCM (12, 10, 15 and 8)}
Efficiency of ‘A’ = 120 ÷ 12 = 10 units/hour
Efficiency of ‘B’ = 120 ÷ 10 = 12 units/hour
Efficiency of ‘C’ = 120 ÷ 15 = 8 units/hour
Efficiency of ‘D’ = 120 ÷ 8 = 15 units/hour

Work done by ‘B’ and ‘C’ together in first hour = 12 + 8 = 20 units
Work done by ‘B’ and ‘D’ together in second hour = 12 + 15 = 27 units
Work done by ‘D’ and ‘A’ together in third hour = 15 + 10 = 25 units
Work done by ‘D’ and ‘B’ together in fourth hour = 15 + 12 = 27 units
Work done by ‘D’ and ‘A’ together in fifth hour = 15 + 10 = 25 units

Total work done in 5 hour = 20 + 27 + 25 + 27 + 25 = 124 units > 120 units

Therefore, ‘D’ and ‘A’ will finish the work.
Hence, option d.

43. A man is twice as efficient as a woman. 12 women take 30 days to build a wall. If __ men work together, then it would take them __ days to build the same wall. The values given in which of the following options will fill the blanks in the same order in which it is given to make the statement true:

I. 9, 10  II. 24, 6  III. 15, 12

Correct Answer: (d) Only III
Solution:

Let the efficiency of each woman be ‘x’ units/day.
So, total work = x × 12 × 30 = 360x units
And, efficiency of each man = x × 2 = 2x units/day

Let the number of days required time be ‘d’ days

For statement I:
360x ÷ (2x × 9) = d
d = 20
So, ‘I’ is false.

For statement II:
360x ÷ (2x × 24) = d
d = 7.5
So, ‘II’ is false.

For statement III:
360x ÷ (2x × 15) = d
d = 12
So, ‘III’ is true.

Hence, option d

44. A’ and ‘B’ together can complete 40% of a work in 28 days while ‘B’ and ‘C’ together can complete 80% of the same work in 36 days. ‘A’ started working alone and replaced by ‘B’ after 18 days. After working for 45 days, ‘B’ was replaced by ‘C’ who completed the remaining work alone. Find the time taken by ‘A’ alone to complete 40% of the work.

Correct Answer: (a) 50.4 days  
Solution:

Time taken by ‘A’ and ‘B’ together to complete the whole work = 28/0.40 = 70 days

Time taken by ‘B’ and ‘C’ together to complete the whole work = 36/0.80 = 45 days

Let total amount of work = 630 units {LCM of 70 and 45}

Amount of work done by ‘A’ and ‘B’ together in one day = 630/70 = 9 units

Amount of work done by ‘B’ and ‘C’ together in one day = 630/45 = 14 units

So, amount of work done by ‘A’ and ‘B’ together in 18 days = 18 × 9 = 162 units

Amount of work done by ‘B’ and ‘C’ together in 27 days i.e. (45 – 18) days = 27 × 14 = 378 units

Remaining work (630 – 162 – 378 = 90 units) is completed by ‘C’ alone in 9 days i.e. (36 – 27) days

So, efficiency of ‘C’ = 90/9 = 10 units per day

Efficiency of ‘B’ = 14 – 10 = 4 units per day

Efficiency of ‘A’ = 9 – 4 = 5 units per day

Required time = (0.40 × 630)/5 = 50.4 days

Hence, option a.

45. A certain work can be completed by ‘P’ and ‘Q’ together in (140/9) days while ‘R’ alone can complete it in 30 days. ‘P’ started doing the work alone and after 5 days ‘Q’ joined him such that they worked together for another 12 days. After that both ‘P’ and ‘Q’ were replaced by ‘R’ who completed the rest work in (18/7) days. Find the time taken by ‘P’ alone to complete 80% of the same work.

Correct Answer: (c) 28 days  
Solution:

Let the total work = 420 units {LCM of (140/3) and 30}

Let efficiency of ‘P’ be ‘a’ units/day

Efficiency of ‘R’ = (420/30) = 14 units/day

Combined efficiency of ‘P’ and ‘Q’ = 420/(140/9) = 27 units/day

According to question:
5 × a + 12 × 27 + (18/7) × 14 = 420
5a + 324 + 36 = 420
5a = 420 – 360
5a = 60
a = 12

Required time taken = {(420 × 0.8)/12} = 28 days

Hence, option c.

46. The ratio of efficiency of Ajay and Sneha is 6:5. The ratio of number of days taken by Prabha to Sneha is 3:2. Ajay takes 3 days less than Sneha, when Ajay and Sneha complete the work individually. Prabha and Sneha started the work and left after 3 days. The number of days taken by Ajay to finish the remaining work is?

Correct Answer: (c) 10 5/6 days  
Solution:

The ratio of efficiency of Ajay and Sneha = 6 : 5

The ratio of number of days taken by Ajay and Sneha = 5 : 6

The ratio of number of days taken by Prabha and Sneha = 3 : 2

Ratio of number of days taken by Ajay : Sneha : Prabha
= 5 : 6 : 9

According to the question,
⇒ Sneha – Ajay = 8 days
⇒ 6’s – 5’s = 3
⇒ 1’s = 3

Number of days taken to finish the whole work,
⇒ Ajay = 15 days, Sneha = 18 days, Prabha = 27 days

Work done by Prabha and Sneha in one day,
⇒ (1/18) + (1/27) = 45/(18×27) = 5/54

Prabha and Sneha’s 3 day work
⇒ (5/54) × 3 = 5/18

Rest of the work = 13/18

The number of days taken by Ajay to finish the remaining work is,
Number of days = (13/18) × 15 = 65/6 = 10 5/6 days

47. Ragu and Rajesh can separately do a piece of work in 12 and 15 days respectively. They worked together If the work was finished in next 2 days, then the number of days in which Rohit alone could do the work?

Correct Answer: (b) 24 days
Solution:

Ragu and Rajesh worked together

1/12 + 1/15 = (12 + 15)/(12×15) = 3/20

Ragu and Rajesh’s 5 days work = (3/20) × 5
= (3/4)

Remaining work 1/4 done by Ragu and Rohit

Ragu and Rohit finished it in 2 days

(1/4) × (Ragu + Rohit)’s whole work = 2
(Ragu + Rohit)’s whole work = 8

Rohit’s one day work = (1/8) – (1/12) = 1/24

Rohit alone can complete the work in 24 days

48. 7 women and 5 men can do a work in 8 days. 6 women and 9 men can do a same work in 6 days. How long will 6 women and 7 men can take to do the work?

Correct Answer: (a) 7 1/14 days  
Solution:

Total work = (men (or) women) × days

Work equal, so,
(7w + 5m) × 8 = (6w + 9m) × 6

56w + 40m = 36w + 54m

20w = 14m

10w = 7m

1w = (7/10) m ⇒ 1m = 10/7 w

7w + 5m = 7w + (50/7)w = 99/7 w

6w + 7m = 6w + 10w = 16w

Women  days
(99/7)  8
16    ?

(99/7) × 8 = 16x
x = 99/14 days = 7 1/14 days

49. Direction (49-50): 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.

Ques: 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?

Correct Answer: (c) 19 2/3 days
Solution:

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 day Mani completes (6-4=2)/3

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

50. 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?

Correct Answer: (a) 37 3/5  
Solution:

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