Lesson 55 of 100 | Quantitative Aptitude / Work and Time / कार्य और समय
Alternate-day and changing-team work
Learning outcome
Calculate alternate-day completion times, including partial days, and track remaining work as teams change.
Concepts and assumptions
Count the job as 1. Assume constant rates, equal working shifts, additive contributions and no skipped working days or changeover delays. Only the named worker or team works in each stage. A worker needing T days alone has rate 1/T.
For alternating workers, identify who starts. A two-day cycle contains one full day each, so cycle work is the sum of their daily rates. Count complete cycles without exceeding the job.
If cycles finish exactly, stop. Otherwise check the next worker against the remainder. If they can finish that day, final-day fraction = remainder ÷ their daily rate. Otherwise subtract their full-day output and continue in order.
A fractional day is part of a working shift, not necessarily of 24 hours. Distinguish the numbered finishing day from total working time. A cycle average is unreliable for an incomplete final cycle.
For changing teams, calculate each stage’s work separately.
Worked examples
Example 1 — Finishing on the first day of a cycle. A needs 12 days alone and B 20 days. They alternate, starting with A.
Cycle work = 1/12 + 1/20 = 2/15. Seven cycles complete 14/15 in 14 days. Remainder = 1/15. A works next and needs (1/15)/(1/12) = 4/5 day. Total = 14 4/5 working days, finishing during day 15.
Example 2 — Finishing on the second day. A needs 8 days alone and B 12 days. They alternate, starting with B.
Cycle work = 1/12 + 1/8 = 5/24. Four cycles complete 20/24 = 5/6 in 8 days. Remainder = 1/6. B works day 9, completing 1/12. Remaining work is now 1/6 − 1/12 = 1/12. A needs (1/12)/(1/8) = 2/3 of day 10. Total = 9 2/3 working days, not 10 full days.
Example 3 — Three team stages. A, B and C alone need 18, 24 and 36 days. A+B work for 4 days, then B+C for 3 days, then A+C finish.
Take total work as 72 units. Daily outputs are A = 4, B = 3, C = 2. First-stage work = (4 + 3) × 4 = 28. Second-stage work = (3 + 2) × 3 = 15. Remainder = 72 − 28 − 15 = 29 units. Final team produces 4 + 2 = 6 units daily, requiring 29/6 days. Total = 4 + 3 + 29/6 = 71/6 = 11 5/6 days.
Common mistakes
Ignoring who starts; treating alternation as joint work; counting a full final day; omitting earlier work; rounding early.
Practice questions
- A and B need 6 and 12 days alone respectively. They alternate, starting with A. Find completion time.
- A and B need 9 and 12 days alone respectively. They alternate, starting with A. Find completion time.
- A and B need 10 and 20 days alone respectively. They alternate, starting with B. Find completion time.
- A, B and C alone need 15, 20 and 30 days. A+B work for 2 days, then B+C for 4 days, then A finishes alone. Find total time.
Worked solutions
- Cycle work = 1/6 + 1/12 = 1/4. Four complete cycles finish exactly, taking 8 days. No extra day is needed.
- Cycle work = 7/36. Five cycles take 10 days and complete 35/36. A needs (1/36)/(1/9) = 1/4 day. Total = 10 1/4 days.
- Cycle work = 3/20. Six cycles take 12 days and complete 9/10. B’s next full day leaves 1/10 − 1/20 = 1/20. A needs (1/20)/(1/10) = 1/2 day. Total = 13 1/2 days.
- Use 60 work units: rates are 4, 3 and 2. Completed work = 2(4 + 3) + 4(3 + 2) = 34. Remainder = 26. A needs 26/4 = 6 1/2 days. Total = 2 + 4 + 6 1/2 = 12 1/2 days.