BANK & INSURANCE (TIME AND WORK & PIPE AND CISTERN) PART 1

Total Questions: 60

1. To fill a cistern, pipes X, Y and Z take 10 minutes, 30 minutes and 15 minutes respectively. What is the time that the three pipes together will take to fill the cistern?

Correct Answer: (b) 5 min
Solution:X can do a piece of work in 10 minutes.
Y can do a piece of work in 30 minutes.
Z can do a piece of work in 15 minutes.
Minute’s work of all the three pipes together = 1/10 + 1/30 + 1/15 = 6/30 = 1/5 part
To fill the whole cistern, time taken = 1/(1/5) = 5 minutes

2. A can do a work in 21 days and B in 28 days. If they work on it together for 3 days, then the fraction of the work that is left is,

Correct Answer: (c) 3/4
Solution:Let the LCM of 21 and 28 (which is 84) be the work done.
So A does 84/21 = 4 units per day and B does 84/28 = 3 units per day.
So in 3 days they can do 3 × (4+3) = 21 units.
So the work done is 21/84 = 1/4 so the remaining work is 3/4.

3. A certain job was assigned to a group of men to do in 60 days. But 18 men did not turn up for the job and the remaining men did the job in 96 days. The original number of men in the group was?

Correct Answer: (e) 48
Solution:

Let the original number of men was m.
∴ 60m = 96(m - 18)
m = 48

4. A and B working together take 8 days and 18 days less than A alone and B alone respectively to do a work. The number of days taken by A and B together to do the work is:

Correct Answer: (c) 12
Solution:

Let (A + B) days taken = m days
So, A alone days taken = (m + 8) days
So, B alone days taken = (m + 18) days
1/(m+8) + 1/(m+18) = 1/m
1/(m+18) = 1/m - 1/(m+8)
1/(m+18) = 8/m(m+8)
8m + 144 = m×m + 8m
m = 12

5. A certain number of men can complete a piece of work in 80 days. Had there been 15 men less, it would have taken 16 days more. How many men were there initially?

Correct Answer: (d) 90 men  
Solution:Let the number of men = p
Then p men do the work in 80 days.
Now (p - 15) men do the work in (80 + 16) = 96 days
Thus, p×80 = (p - 15)×96
80p = 96p - 1440
16p = 1440
p = 90 men
Hence initially there were 90 men.

6. X can do a piece of work in 10 days and Y can do the same in 24 days. If Z also works with them then it takes only 6 days to complete the whole work. In how many days Z alone can complete the whole work?

Correct Answer: (e) 40
Solution:Amount of work done in a day if X, Y and Z work together = 1/X + 1/Y + 1/Z = 1/6
Substituting values of X, Y and Z we get,
1/10 + 1/24 + 1/Z = 1/6
1/Z = 1/6 - (1/X + 1/Y) ⇒ 3/(120)
Z alone can do the piece of work in 120/(3) = 40 days

7. A’s efficiency is half of B and C’s efficiency is thrice of A to finish a piece of work. If they all work together, they can finish the work in 5 days. B can do the work alone in:

Correct Answer: (d) 15 days  
Solution:A’s efficiency is half of B i.e. 2A = B…(i)
C’s efficiency is thrice of A i.e. 3A = C…(ii)
It is given that working together they can finish the work in 5 days.
⇒ A + B + C = 1/5
⇒ A + 2A + 3A = 1/5 (from i and ii)
⇒ 6A = 1/5
⇒ A = 1/30
⇒ 2A = 1/15
⇒ B = 1/15 (from i)
⇒ B takes 15 days to complete the work alone.

8. If 9 men or 16 boys can do a piece of work in 20 days, then 10 men and 8 boys together will take how many days to finish the same work?

Correct Answer: (c) 12 12/29 days
Solution:1 day work of 16 boys = 1/20 part
1 day work of 1 boy = 1/(16×20) = 1/320 part
Similarly, 1 day work of 1 man = 1/(9×20) = 1/180
1 day work of 8 boys = 8/320 = 1/40
1 day work of 10 men = 10/180 = 1/18
Therefore, 10 men and 8 boys together will take
= 1/40 + 1/18
⇒ 12 12/29 days

9. Aman can do a piece of work in 40 days. Naman is 25% more efficient than Aman. What is the number of days taken by Naman to do the same piece of work?

Correct Answer: (d) 32 days  
Solution:Let’s assume total work = 1 unit
Amount of work done by Aman in one day = 1/40 unit
Hence, amount of work done by Naman in one day
= 125% more than 1/40 unit = 1.25/40 = 1/32 unit
Hence time taken by Naman to complete the same piece of work = 32 days

10. If A and B together can do a piece of work in ‘m’ days and A alone can do it in ‘n’ days, then B will complete the same work in how many days?

Correct Answer: (e) mn/(n−m)
Solution:

Let us suppose B can complete the given work in ‘B’ days and total work is equal to 1 unit.
Therefore: (1/n) + (1/B) = (1/m)

⇒ (1/B) = (1/m) - (1/n)
⇒ (1/B) = (n - m)/mn
⇒ B = mn/(n - m)