Time & Work • Topic 3 of 5

Alternate Working

Alternate-day problems are where the LCM-unit method really shines. Workers take turns — A on day 1, B on day 2, A on day 3, and so on — so you should group the work into "cycles". One full cycle here is two days, in which A + B together do a fixed number of units. Count how many complete cycles fit before the work is nearly done, then handle the leftover day by day. The classic trap is assuming the job ends exactly on a cycle boundary; usually it finishes part-way through the next cycle, and whoever’s turn it is matters because the faster worker should ideally start. Always note who works first, compute the per-cycle output, find how many whole cycles are needed, then finish off the remainder. The same idea covers "A works for 2 days, then B for 1 day" patterns — just make the cycle as long as one full rotation.

✅ Solved examples

1. A (10 days) and B (15 days) work on alternate days, A starting. When is the job done?
Total = LCM(10,15) = 30; A = 3, B = 2. Per 2-day cycle = 5 units. After 5 cycles (10 days) = 25 units; 5 left. Day 11 (A) does 3 → 28; day 12 (B) does 2 → 30. Finished in 12 days.
2. A (12 days) and B (24 days) work alternately, A first. Time to finish?
Total = 24; A = 2, B = 1. Cycle (2 days) = 3 units. 8 cycles = 16 days = 24 units. Finished exactly at 16 days.
3. A (8 days) and B (12 days), alternate days, B starting. Total time?
Total = 24; A = 3, B = 2. Cycle = 5 units but B goes first. After 4 cycles (8 days) = 20 units; 4 left. Day 9 (B) does 2 → 22; day 10 (A) does 3, only 2 needed ⇒ 2/3 day. Total = 9⅔ days.
4. A (20 days) and B (30 days) work alternately, A starting. Fraction done in first 6 days?
Total = 60; A = 3, B = 2. Per cycle = 5; 3 cycles = 6 days = 15 units ⇒ 15/60 = 1/4 of the work.

✏️ Practice — try these, take hints as needed

1. A (6 days) and B (12 days) work alternately, A first. Days to finish?
Total = 12; A = 2, B = 1; cycle = 3 units.
4 cycles = 8 days = 12 units.
Ends exactly on a cycle boundary.
8 days
2. A (9 days) and B (18 days) alternate, A starting. Total time?
Total = 18; A = 2, B = 1; cycle = 3.
6 cycles = 12 days = 18 units.
Exact finish.
12 days
3. A (10 days) and B (20 days) alternate, B starting. Time to finish?
Total = 20; A = 2, B = 1; cycle = 3 units.
6 cycles = 12 days = 18 units; 2 left.
Day 13 is B (1 unit → 19), day 14 A needs 1 of 2 ⇒ ½ day.
13½ days
4. A (16 days) and B (16 days) alternate. When finished?
Equal efficiency, so order doesn’t matter.
Total = 16; each 1/day; cycle = 2 units.
8 cycles = 16 days.
16 days
5. A (4 days) and B (6 days) alternate, A first. Fraction done after 4 days?
Total = 12; A = 3, B = 2; cycle = 5.
2 cycles = 4 days = 10 units.
10/12.
5/6

📝 Topic test — 8 questions

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