P and Q working alone can complete a work in 9 days and 18 days respectively. They work on alternate days one at a time. P works on the first day. In how many days the whole work will be completed?
12
Take total work as the LCM of the two times so the daily rates are whole numbers: \(\text{LCM}(9,18) = 18\) units.
P finishes in 9 days, so P's rate \(= \dfrac{18}{9} = 2\) units/day; Q finishes in 18 days, so Q's rate \(= \dfrac{18}{18} = 1\) unit/day.
They alternate starting with P, so one full 2-day cycle (P then Q) does \(2 + 1 = 3\) units.
After 5 cycles = 10 days, work done \(= 5 \times 3 = 15\) units, leaving \(18 - 15 = 3\) units.
Day 11 is P's turn: \(15 + 2 = 17\) units. Day 12 is Q's turn: \(17 + 1 = 18\) units — the work is exactly complete.
Key idea: fix a convenient total, group alternate days into cycles, then finish the remainder day by day. The whole work is completed in 12 days.
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