Quantitative Aptitude

Time and Work

2,195 Questions
All Pairwise rates and staged schedules 440 Workers-days-hours chain rule 439 LCM units and efficiency ratios 409 Different worker categories 311 Combined rates and an unknown worker 257 Division of wages 219 Alternate work 110 Opposing rates 10

Time and Work Questions

Multiple choice

There are two walls to be built. Area of one wall is twice that of the other. The labourers work on the larger wall for half a day. For the remaining half of the day, half of the labourers work on the larger wall, while the remaining work on the smaller wall. At the end of the day, the larger wall is completely built, while some portion of the smaller wall is left to be built, which is completed by one labourer the next day. How many labourers worked on the first day?

  1. 4

  2. 8

  3. 6

  4. 10
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

20 men could do a piece of work in 30 days working 4 hours a day, 30 men worked for 20 days for 3 hours a day. Then all were laid off, except one. How many more days would it take the last man to complete the remaining work, working 10 hours a day?

  1. 20 days

  2. 30 days

  3. 40 days

  4. 60 days

  5. 80 days
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total work = 20 * 30 * 4 = 2400 man-hours. Work done = 30 * 20 * 3 = 1800 man-hours. Remaining work = 600 man-hours. One man working 10 hours a day takes 600 / 10 = 60 days.

Multiple choice

Some men can do a piece of work in a certain number of days working for a certain number of hours a day. If the number of men, their efficiency and the working hours per day, all are increased by 25%, they complete their work in 61 days less time than estimated. What is the estimated number of days?

  1. 125

  2. 120

  3. 100

  4. 80

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let M, E, H, D be men, efficiency, hours, and days. Work = M*E*H*D. New work = (1.25M)(1.25E)(1.25H)(D-61). Since work is constant: M*E*H*D = 1.25^3 * M*E*H(D-61). D = 1.953125 * (D-61). D = 1.953125D - 119.14. 0.953125D = 119.14, D = 125.

Multiple choice

20 men can complete a work in 50 days, working for 5 hours a day. After 15 days from the start of work, 10 men quit. How many hours per day do the remaining men have to work so as to complete the work on time?

  1. 6 hours

  2. 8 hours

  3. 10 hours

  4. 12 hours

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial work is 20 men * 50 days * 5 hours = 5000 man-hours. After 15 days, 20 * 15 * 5 = 1500 man-hours are done, leaving 3500 man-hours. With 10 men remaining, they must work 3500 / (10 * 35 days) = 10 hours per day.

Multiple choice

A is two times as efficient as B and three times as efficient as C. If it takes 20 days for B and C to finish a piece of work together, how many days will all of them take to finish the same piece of work together?

  1. 9 1 11 days

  2. 9 8 9 days

  3. 10 2 11 days

  4. 12 days

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the efficiency of A be 6 units/day, which means B's efficiency is 3 units/day and C's efficiency is 2 units/day. Since B and C together (5 units/day) take 20 days to complete the work, the total work is 5 * 20 = 100 units. Working together, all three have a combined efficiency of 6 + 3 + 2 = 11 units/day, so they will complete the work in 100/11 = 9 1/11 days.

Multiple choice

If three men A, B, and C work for 8 hours everyday, they individually take 10 days, 15 days and 20 days, respectively, to complete a piece of work. Firstly, A and B together started the work and worked for 6 days at 6 hours per day. Then, A and B were replaced by C; who works for 10 hours a day. How many days does C have to work to complete the work?

  1. 4 days

  2. 5 days

  3. 6 days

  4. 8 days

  5. 9 days
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total work = 60 units (LCM of 10, 15, 20). A's rate = 6/day, B's rate = 4/day, C's rate = 3/day. A+B work 6 days at 6 hours/day (equivalent to 36 hours). Work done = (6+4) * (36/8) = 45 units. Remaining work = 15 units. C works 10 hours/day, rate = 3 * (10/8) = 3.75 units/day. Days for C = 15 / 3.75 = 4.