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

Total Questions: 60

1. ‘B’ can finish a project in ‘x’ hours, ‘A’ needs ‘0.7x’ hours to do half of the same project. ‘B’ and ‘C’ together can finish the same project in one third time taken by ‘A’ to finish the project alone. ‘A’ takes 90 hours more to finish the project alone as compared to time taken by all three together to finish the work. Find the time taken by ‘C’ to finish 80% of the project alone.

Correct Answer: (d) 60 hours  
Solution:

Time taken by ‘B’ to finish the project = ‘x’ hours
Time taken by ‘A’ to finish the project = ‘1.4x’ hours

‘B’ and ‘C’ can finish the project together in (1.4x/3) hours.

So, time taken by ‘C’ alone to finish the project =
1/[(3/1.4x) - (1/x)] = (7x/8) hours

Time taken by all three to finish the project together = (1.4x - 90) hours.

So, (1/x) + (1/1.4x) + (8/7x) = 1/(1.4x - 90)

{(1.4 + 1 + 1.6)/1.4x} = 1/(1.4x - 90)

5.6x - 360 = 1.4x
4.2x = 360
x = (600/7)

So, Time taken by ‘C’ alone to finish 80% of the project
= (7x/8) × (600/7) × 0.8 = 60 hours

2. Number of days taken by ‘P’, ‘Q’ and ‘R’ to do a certain work alone is (x + 3) days, (x - 3) days and (x + 1) days, respectively and number of days taken by ‘P’ and ‘Q’ together to do the work is (1/4) days more than the number of days taken by ‘Q’ and ‘R’ together to do the work. Find the number of days taken to complete the work if all three of them decided to do the work alternatively till the work gets completed starting with ‘P’, then ‘Q’ and then ‘R’.

Correct Answer: (c) (2x - 9.5) days  
Solution:

Let the total amount of work be {(x + 3)(x - 3)(x + 1)} units

Efficiency of ‘P’ = {(x + 3)(x - 3)(x + 1)}/(x + 3)
= {(x - 3)(x + 1)} units/day

Efficiency of ‘Q’ = {(x + 3)(x - 3)(x + 1)}/(x - 3)
= {(x + 3)(x + 1)} units/day

Efficiency of ‘R’ = {(x + 3)(x - 3)(x + 1)}/(x + 1)
= {(x - 3)(x + 3)} units/day

ATQ:

[{(x + 3)(x - 3)(x + 1)}/{(x - 3)(x + 1) + (x + 1)(x + 3))}]
= (1/4) + [{(x + 3)(x - 3)(x + 1)}/{(x + 3)(x + 1) + (x + 3)(x - 3)}]

{(x² - 9)/2x} = (1/4) + {(x - 3)(x + 1)/(2x - 2)}

2x² - 22x + 36 = 0
2x² - 18x - 4x + 36 = 0

2x(x - 9) - 4(x - 9) = 0
(2x - 4)(x - 9) = 0

x = 2 or x = 9

As number of days cannot be a negative integer, so value of ‘x’ cannot be 2.

So, x = 9

Number of days taken by ‘P’ to do the work alone = (9 + 3) = 12 days

Number of days taken by ‘Q’ to do the work alone = (9 - 3) = 6 days

And, number of days taken by ‘R’ to do the work alone = (9 + 1) = 10 days

So, let total work = 60 units (lcm of 12, 6 and 10)

Efficiency of ‘P’ = (60/12) = 5 unit/day

Efficiency of ‘Q’ = (60/6) = 10 unit/day
Efficiency of ‘R’ = (60/10) = 6 unit/day

So, work done by ‘P’, ‘Q’ and ‘R’ in consecutive 3 days, separately = (5 + 10 + 6) = 21 units

Similarly, work done by ‘P’, ‘Q’ and ‘R’ in consecutive 6 days, separately = (21 × 2) = 42 units

Work done ‘P’ and ‘Q’ in next 2 days, separately = (5 + 10) = 15 units

Total work done in 8 consecutive days = 42 + 15 = 57 units

Time taken by ‘R’ to do the rest of the work = (60 - 57)/6 = (3/6) = 0.5 days

So, total number of days taken to complete the whole work = (8 + 0.5) = 8.5 days

Since, 2x - 9.5 = (2 × 9 - 9.5) = 8.5

Hence, option c.

3. Directions (3-5): Answer the questions based on the information given below.

Ramesh and Suresh while working alone can complete a piece of work ‘P’ in ‘5x + 3’ days and 20 days, respectively. Both of them working alternately, starting with Ramesh, can complete the same work ‘P’ in (4x + (24/7)) days. Mahesh and Mukesh working together can complete a piece of work ‘Q’ in ‘x + y’ days, while Mukesh and Manish working together can complete the same work ‘Q’ in ‘y + 6’ days while Manish alone can complete the whole work ‘Q’ in ‘y + 11’ days. Mahesh worked on work ‘Q’ for ‘x - 2’ days, Mukesh worked on work ‘Q’ for ‘y + 4’ days and Manish completed the remaining work ‘Q’ in {(8y + 13)/6} days.

Ques: Time taken by Suresh alone to complete 75% of the work ‘P’ is:

Correct Answer: (a) 15 days  
Solution:

Let total amount of work ‘P’ = 20(5x + 3) units {LCM of 20 and (5x + 3)}

Efficiency of Ramesh = 20(5x + 3)/(5x + 3) = 20 units per day

Efficiency of Suresh = 20(5x + 3)/20 = (5x + 3) units per day

Since, Ramesh and Suresh together alternately worked for (4x + (24/7)) days

So, Ramesh worked for (2x + 2) days and Suresh worked for (2x + (10/7)) days

So, 20(2x + 2) + (5x + 3)(2x + (10/7)) = 20(5x + 3)

70x² - 328x - 110 = 0

70x² - 350x + 22x - 110 = 0

(70x + 22)(x - 5) = 0

x = 5 or x = -(22/70) {not possible}

So, x = 5

For Mahesh, Mukesh and Manish:

Time taken by Mahesh and Mukesh together to complete the work = x + y = (y + 5) days

Let total amount of work ‘Q’ = (y + 5)(y + 6) units

Efficiency of Mahesh and Mukesh together
= {(y + 5)(y + 6)}/(y + 5) = (y + 6) units per day

Similarly, efficiency of Mukesh and Manish together
= {(y + 5)(y + 6)}/(y + 6) = (y + 5) units per day

ATQ;
Amount of work ‘Q’ completed by Mahesh and Mukesh together in 3(x - 2 = 5 - 2) days = 3(y + 6) units

Amount of work ‘Q’ completed by Mukesh and Manish together in ‘y + 1’ (y + 4 - 3) days = (y + 1)(y + 5) units

Amount of work ‘Q’ left = (y + 5)(y + 6) - 3(y + 6) - (y + 1)(y + 5) = (2y + 7) units

So, Time taken by Manish to complete (2y + 7) units of work
= {(8y + 13)/6} - (y + 1) = (2y + 7)/6 days

So, efficiency of Manish = {(2y + 7)}/{((2y + 7)/6)} = 6 units per day

So, {(y + 5)(y + 6)}/{(y + 11)} = 6

y² + 5y - 36 = 0

y² + 9y - 4y - 36 = 0

y(y + 9) - 4(y + 9) = 0

(y + 9)(y - 4) = 0

y = 4 or y = -9 (not possible)

Efficiency of Mukesh = y + 5 - 6 = 4 + 5 - 6 = 3 units per day

Efficiency of Mahesh = y + 6 - 3 = 4 + 3 = 7 units per day

For Suresh:

Total work = 20 × (5x + 3) = 20 × (5 × 5 + 3) = 560 units

Efficiency = 5x + 3 = 5 × 5 + 3 = 28 units per day

Desired time = (0.75 × 560)/28 = 15 days

Hence, option a.

4. Time taken by Mukesh alone to complete the whole work ‘Q’ is

Correct Answer: (e) None of these
Solution:

Let total amount of work ‘P’ = 20(5x + 3) units {LCM of 20 and (5x + 3)}

Efficiency of Ramesh = 20(5x + 3)/(5x + 3) = 20 units per day

Efficiency of Suresh = 20(5x + 3)/20 = (5x + 3) units per day

Since, Ramesh and Suresh together alternately worked for {4x + (24/7)} days

So, Ramesh worked for (2x + 2) days and Suresh worked for {2x + (10/7)} days

So, 20(2x + 2) + (5x + 3){2x + (10/7)} = 20(5x + 3)

70x² - 328x - 110 = 0

70x² - 350x + 22x - 110 = 0

70x(x - 5) + 22(x - 5) = 0

(70x + 22)(x - 5) = 0

x = 5 or x = -(22/70) {not possible}

So, x = 5

For Mahesh, Mukesh and Manish:

Time taken by Mahesh and Mukesh together to complete the work = x + y = (y + 5) days

Let total amount of work ‘Q’ = (y + 5)(y + 6) units

Efficiency of Mahesh and Mukesh together = {(y + 5)(y + 6)}/(y + 5) = (y + 6) units per day

Similarly, efficiency of Mukesh and Manish together = {(y + 5)(y + 6)}/(y + 6) = (y + 5) units per day

ATQ:

Amount of work ‘Q’ completed by Mahesh and Mukesh together in 3(x - 2 = 5 - 2) days = 3(y + 6) units

Amount of work ‘Q’ completed by Mukesh and Manish together in ‘y + 1’ (y + 4 - 3) days = (y + 1)(y + 5) units

Amount of work ‘Q’ left = (y + 5)(y + 6) - 3(y + 6) - (y + 1)(y + 5) = (2y + 7) units

So, Time taken by Manish to complete (2y + 7) units of work = [(8y + 13)/6] - (y + 1) = (2y + 7)/6 days

So, efficiency of Manish = [(2y + 7)]/[(2y + 7)/6] = 6 units per day

So, [(y + 5)(y + 6)]/(y + 11) = 6

y² + 5y - 36 = 0

y² + 9y - 4y - 36 = 0

y(y + 9) - 4(y + 9) = 0

(y + 9)(y - 4) = 0

y = 4 or y = -9 {not possible}

Efficiency of Mukesh = y + 5 - 6 = 4 + 5 - 6 = 3 units per day

Efficiency of Mahesh = y + 6 - 3 = 4 + 3 = 7 units per day

Desired time = 90/3 = 30 days

Hence, option e.

5. Time taken by Mahesh, Mukesh and Manish together to complete the work ‘Q’ given the condition that Mahesh works 20% more efficiently, Mukesh works 40% more efficiently and Manish works 60% less efficiently, is:

Correct Answer: (d) 6 days  
Solution:

Let total amount of work ‘P’ = 20(5x + 3) units {LCM of 20 and (5x + 3)}

Efficiency of Ramesh = 20(5x + 3)/(5x + 3) = 20 units per day

Efficiency of Suresh = 20(5x + 3)/20 = (5x + 3) units per day

Since, Ramesh and Suresh together alternately worked for {4x + (24/7)} days

So, Ramesh worked for (2x + 2) days and Suresh worked for {2x + (10/7)} days

So, 20(2x + 2) + (5x + 3){2x + (10/7)} = 20(5x + 3)

70x² - 328x - 110 = 0

70x² - 350x + 22x - 110 = 0

70x(x - 5) + 22(x - 5) = 0

(70x + 22)(x - 5) = 0

x = 5 or x = -(22/70) (not possible)

So, x = 5

For Mahesh, Mukesh and Manish:

Time taken by Mahesh and Mukesh together to complete the work = x + y = (y + 5) days

Let total amount of work ‘Q’ = (y + 5)(y + 6) units

Efficiency of Mahesh and Mukesh together
= {(y + 5)(y + 6)}/(y + 5) = (y + 6) units per day

Similarly, efficiency of Mukesh and Manish together
= {(y + 5)(y + 6)}/(y + 6) = (y + 5) units per day

ATQ;

Amount of work ‘Q’ completed by Mahesh and Mukesh together in 3(x - 2 = 5 - 2) days = 3(y + 6) units

Amount of work ‘Q’ completed by Mukesh and Manish together in ‘y + 1’ (y + 4 - 3) days = (y + 1)(y + 5) units

Amount of work ‘Q’ left = (y + 5)(y + 6) - 3(y + 6) - (y + 1)(y + 5) = (2y + 7) units

So, Time taken by Manish to complete (2y + 7) units of work
= {(8y + 13)/6} - (y + 1) = (2y + 7)/6 days

So, efficiency of Manish = {(2y + 7)}/{((2y + 7)/6)} = 6 units per day

So, {(y + 5)(y + 6)}/{(y + 11)} = 6

y² + 5y - 36 = 0

y² + 9y - 4y - 36 = 0

y(y + 9) - 4(y + 9) = 0

(y + 9)(y - 4) = 0
y = 4 or y = -9 (not possible)
Efficiency of Mukesh = y + 5 - 6 = 4 + 5 - 6 = 3 units per day
Efficiency of Mahesh = y + 6 - 3 = 4 + 3 = 7 units per day
Desired time = 90/(1.2 × 7 + 1.4 × 3 + 0.4 × 6) = 6 days
Hence, option d.

6. A, B and C alone can do a certain piece of work in 35 days, 28 days and 20 days respectively. All of them started work together but after ‘x’ days B and C left the job. Now, the remaining work is done by A with 25% more efficiency. Find the value of ‘x’, if the whole work is completed in ‘3x + 2’ days.

Correct Answer: (b) 5 days  
Solution:

Let the total work be 140 units (LCM of 35, 28 and 20)
Amount of work done by A alone in one day = 140/35 = 4 units
Amount of work done by B alone in one day = 140/28 = 5 units
Amount of work done by C alone in one day = 140/20 = 7 units

Amount of work done by A alone in one day with increased efficiency = 4 × 1.25 = 5 units

So according to question,
(4 + 5 + 7) × x + 5 × (3x + 2 - x) = 140
16x + 10x + 10 = 140
26x = 130, x = 5

So, the value of ‘x’ is 5 days.
Hence, option b.

7. Directions (7-8): Answer the questions based on the information given below.

Ajay and Vijay together can complete 75% of the work in 36 days while Vijay and Rakesh together can complete 60% of the work in 24 days. Vijay is 25% less efficient than Tushar. Ajay and Rakesh together can complete 70% of the work in 24 days.

Ques. Rakesh is how much percent less/more efficient than Tushar?

Correct Answer: (c) 50%
Solution:

Time taken by Ajay and Vijay together to complete the work = 36/0.75 = 48 days
Time taken by Vijay and Rakesh together to complete the work = 24/0.60 = 40 days
Time taken by Ajay and Rakesh together to complete the work = 24/0.70 = 240/7 days

Let the total work = 240 units (LCM of 40, 48 and 240)

Amount of work done by Ajay and Vijay together in one day = 240/48 = 5 units
Amount of work done by Ajay and Rakesh together in one day = 240/(240/7) = 7 units
Amount of work done by Vijay and Rakesh together in one day = 240/40 = 6 units

Amount of work done by Ajay, Vijay and Rakesh together in one day = (5 + 7 + 6)/2 = 9 units

Amount of work done by Ajay alone in one day = 9 - 6 = 3 units
Amount of work done by Vijay alone in one day = 9 - 7 = 2 units
Amount of work done by Rakesh alone in one day = 9 - 5 = 4 units

Amount of work done by Tushar alone in one day = 2/0.75 = 8/3 units

Desired percentage = (4 - 8/3)/(8/3) × 100
= 4/8 × 100 = 50%

Hence, option c.

8. In how many days Ajay, Vijay, Rakesh and Tushar together can complete 70% of the work?

Correct Answer: (b) 14.4 days  
Solution:

Time taken by Ajay and Vijay together to complete the work = 36/0.75 = 48 days
Time taken by Vijay and Rakesh together to complete the work = 24/0.60 = 40 days
Time taken by Ajay and Rakesh together to complete the work = 24/0.70 = 240/7 days

Let the total work = 240 units (LCM of 40, 48 and 240)

Amount of work done by Ajay and Vijay together in one day = 240/48 = 5 units

Amount of work done by Ajay and Rakesh together in one day = 240/(240/7) = 7 units
Amount of work done by Vijay and Rakesh together in one day = 240/40 = 6 units
Amount of work done by Ajay, Vijay and Rakesh together in one day = (5 + 7 + 6)/2 = 9 units

Amount of work done by Ajay alone in one day = 9 - 6 = 3 units
Amount of work done by Vijay alone in one day = 9 - 7 = 2 units
Amount of work done by Rakesh alone in one day = 9 - 5 = 4 units

Amount of work done by Tushar alone in one day = 2/0.75 = 8/3 units

Amount of work done by Ajay, Vijay, Rakesh and Tushar together in one day = 3 + 2 + 4 + 8/3
= 35/3 units

So, the time taken by Ajay, Vijay, Rakesh and Tushar together to complete 70% of the work
= 0.70 × 240/(35/3) = 14.4 days

Hence, option b.

9. A can complete 25% of the work in 7.5 days, B can complete 50% of the work in 10 days and C can complete 20% of the work in 3 days. All together with the help of D can complete the whole work in 4 days. If total wages obtained by A, B, C and D together is Rs. 2250, then find the share of D.

Correct Answer: (a) Rs. 900  
Solution:

Time taken by A to complete the whole work = 7.5 × 4 = 30 days
Time taken by B to complete the whole work = 10 × 2 = 20 days
Time taken by C to complete the whole work = 3 × 5 = 15 days

Let total amount of work = 60 units (LCM of 30, 20, 15)

Amount of work done by A in one day = 60/30 = 2 units
Amount of work done by B in one day = 60/20 = 3 units
Amount of work done by C in one day = 60/15 = 4 units

Amount of work done by A, B, C and D in one day = 60/4 = 15 units
Amount of work done by D in one day = 15 - (2 + 3 + 4) = 6 units

Wages of D = (6/15) × 2250 = Rs. 900

Hence, option a.

10. Bhanu, Dhani and Girish alone can complete a work in (2B - 0.5A) days, 0.5C days and (0.75A + B) days respectively. Bhanu started the work alone and left after 4 days and after that Dhani and Girish joined the work together. After 10 days from starting the work, Dhani was replaced with Bhanu and after 8 days more Girish was replaced with Dhanu such that the remaining work is completed in 8 days.

Note: (12B + A = 7A + 4B) and (4C + B = 11B – C)


Which of the following is true according to the given information?

Correct Answer: (e) Both (b) and (c
Solution:

Given, (12B + A = 7A + 4B)
12B - 4B = 7A - A
8B = 6A
A/B = 4/3

Given, (4C + B = 11B - C)
4C + C = 11B - B
5C = 10B
B/C = 1/2

Let A = 4x, B = 3x, C = 3x × 2 = 6x

Time taken by Bhanu alone to complete the work
= (2B - 0.5A) = [2 × 3x - 0.5 × 4x] = 4x days

Time taken by Dhani alone to complete the work
= 0.5C = 0.5 × 6x = 3x days

Time taken by Girish alone to complete the work
= (0.75A + B) = (0.75 × 4x + 3x) = 6x days

Let the total amount of work = 12x units

Bhanu’s efficiency = 12x/4x = 3 units per day

Dhani’s efficiency = 12x/3x = 4 units per day
Girish’s efficiency = 12x/6x = 2 units per day

According to the question,
3 × 4 + (4 + 2) × 6 + (3 + 2) × 8 + (4 + 3) × 8 = 12x
12 + 36 + 40 + 56 = 12x
12x = 144
x = 12

Time taken by Bhanu alone to complete the work = 48 days
Time taken by Dhani alone to complete the work = 36 days
Time taken by Girish alone to complete the work = 72 days

(a) Time taken by Bhanu alone to complete the work is 60 days.
Time taken by Bhanu alone to complete the work = 48 days
This is not true.

(b) (0.75A + C) = 108
0.75 × 48 + 72 = 108
This is true.

(c) Time taken by Dhani and Girish together to complete the work is 24 years.
Time taken by Dhani and Girish together to complete the work = (12 × 12)/6 = 24 days
This is true.

Both (b) and (c) are true.