Introduction
The Alternate Working Days pattern appears when two or more workers take turns to work on consecutive days (A works Day 1, B works Day 2, A again Day 3, and so on). This pattern is important because many exam-style Time & Work problems use alternating schedules - knowing how to compute the work done in each cycle (usually a 2-day or multi-day cycle) makes these problems straightforward.
Key idea: treat one full alternate cycle as a single unit of work (sum of rates across the days in the cycle), then use cycle-counting + partial cycle handling to finish.
Pattern: Alternate Working Days
Pattern
Key concept: Compute work done per full alternation cycle, then use integer cycles + remaining partial cycle to reach total work = 1.
Steps in brief:
1. Convert each worker’s time into a daily rate (work per day).
2. Add rates for the days included in one alternation cycle (e.g., A's day + B's day = 2-day cycle).
3. Find how many full cycles are needed and then handle the leftover work day-by-day.
Step-by-Step Example
Question
A can finish a job in 10 days and B can finish the same job in 15 days. They work on alternate days starting with A (A on Day 1, B on Day 2, A on Day 3, ...). How many days will it take to finish the job?
Solution
Step 1: Identify daily rates:
A’s one-day work = 1/10. B’s one-day work = 1/15.Step 2: Compute work per full 2-day cycle (A then B):
Work in 2 days = A + B = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6.Step 3: Find how many full cycles get close to 1:
Each 2-day cycle does 1/6 of the job. Number of full cycles to approach 1: 1 ÷ (1/6) = 6 cycles would do exactly 6×(1/6)=1 job - but 6 cycles = 12 days. Check if earlier partial cycles finish: we instead look for the smallest integer cycles before exceeding 1. After 5 cycles (10 days) work done = 5×(1/6)=5/6; remaining = 1 - 5/6 = 1/6.Step 4: Handle remaining work day-by-day starting with A:
After 10 days (5 cycles) remaining = 1/6. On Day 11 A works and does 1/10 = 3/30 = 1/10 = 0.1 = 3/30; numeric comparison: 1/6 ≈ 0.1667, A’s 1/10 = 0.1, so not finished. Remaining after Day 11 = 1/6 - 1/10 = (5/30 - 3/30) = 2/30 = 1/15. Day 12 B works and does 1/15 which equals remaining - job completes at the end of Day 12.Final Answer:
The job is finished in 12 days.Quick Check:
6 full 2-day cycles would be exactly 12 days and produce 6×(1/6)=1 job. Our stepwise remainder logic matched that - answer confirmed. ✅
Quick Variations
1. Start with B instead of A - swap the order when computing the cycle and remainder.
2. Multi-person alternation (A → B → C → A → …): define cycle length equal to number of people and sum rates across that cycle.
3. Alternate with unequal shift lengths (e.g., A works 2 days, B 1 day, repeat): treat a cycle as 3 days with sum of two A-days + one B-day.
4. One worker is absent on certain cycles - subtract their contribution on those cycles and recompute.
Trick to Always Use
- Step 1 → Convert times to daily rates (1/day).
- Step 2 → Compute work per full alternation cycle (sum rates for days in the cycle).
- Step 3 → Use integer cycles to cover most of the job (cycles × cycle-work). Compute remaining fraction.
- Step 4 → Finish remaining work day-by-day in the correct turn order (respecting who starts next).
Summary
Summary
For Alternate Working Days problems:
- Always form a cycle (usually 2 days for A/B alternation) and compute the cycle work.
- Use integer cycles to reduce the job, then handle the leftover day(s) in turn order.
- If the remaining work after full cycles is less than the next worker’s full day output, the job finishes partway through that day - compute fractional day if required.
- Quick verification: recompute total contributions (cycles + leftover) and confirm they sum to 1 (the full job).
Hint: Compute cycle work (A+B), use integer cycles, then finish leftover day-by-day.
Common Mistakes: Using time ratios instead of daily work rates; forgetting turn order for remainder.