Lesson 52 of 100 | Quantitative Aptitude / Work and Time / कार्य और समय
Work, rate and efficiency
Learning outcome
Convert completed work into a rate, calculate unfinished work, and use efficiency comparisons to find individual completion times.
Concepts and assumptions
Treat one complete job as 1 unit of work. A day means the same fixed number of working hours for every worker. Assume constant productivity, no interruptions, and work that can be measured proportionately. Workers being compared perform the same job.
If a worker finishes the job in T days, equal daily contributions must total 1:
T × daily rate = 1, so daily rate = 1/T job per day.
If the rate is r, working for t days completes rt of the job. Conversely, the time required for W units is W/r: we divide the required amount by the amount produced each day. Check that completed work does not exceed 1; work stops when the job is finished.
Efficiency means work produced per unit time. For the same job, a higher rate means a shorter completion time. Since rate × completion time = 1 for either worker, the time ratio is the reverse of the rate ratio. A percentage increase in efficiency is not an equal percentage decrease in time.
All results below are exact.
Worked examples
Example 1 — Reading a daily rate. A worker completes a job in 16 days. What fraction is completed in 6 days?
Daily rate = 1/16. Work in 6 days = 6 × 1/16 = 3/8. As a percentage, 100 × 3/8 = 37.5%. The remaining fraction is 1 − 3/8 = 5/8.
Example 2 — Recovering the whole from a part. A worker completes 7/15 of a job in 14 days. Find the full-job time and additional time needed.
Daily rate = (7/15) ÷ 14 = 1/30. Full-job time = 1 ÷ (1/30) = 30 days. Remaining work = 1 − 7/15 = 8/15. Additional time = (8/15) ÷ (1/30) = 16 days. Check: 14 + 16 = 30.
Example 3 — Efficiency with a time difference. A is 25% more efficient than B and finishes the same job 6 days sooner. Find both times.
Rate ratio A:B = 125:100 = 5:4. Therefore time ratio A:B = 4:5. Write their times as 4k and 5k. The difference gives 5k − 4k = 6, so k = 6. A takes 24 days; B takes 30 days. Check: (1/24)/(1/30) = 5/4. A’s time reduction relative to B is 6/30 = 20%, not 25%.
Common mistakes
Using completion time as the rate; dividing time by work instead of work by time; reversing the wrong ratio; treating “60% as efficient” as “60% more efficient”; overlooking unequal working hours.
Practice questions
- A worker finishes a job in 20 days. What percentages are completed and remaining after 7 days?
- A worker completes 5/12 of a job in 10 days. Find the full-job time and additional time required.
- A is 60% as efficient as B. B needs 18 days for the job. How long does A need?
- A is 40% more efficient than B and takes 8 fewer days. Find both completion times.
Worked solutions
- Rate = 1/20. Completed work = 7/20 = 35%. Remaining work = 1 − 7/20 = 13/20 = 65%.
- Rate = (5/12)/10 = 1/24. Full time = 24 days. Remaining work = 7/12, requiring (7/12)/(1/24) = 14 more days.
- A’s rate = (60/100) × (1/18) = 1/30. Therefore A needs 30 days; lower efficiency correctly gives a longer time.
- Rate ratio = 140:100 = 7:5, so time ratio = 5:7. The difference of 2 parts equals 8 days; one part is 4 days. A needs 20 days, B 28 days.