A can complete a piece of work in 19 days and B can complete it in 21 days. If…

Concept

When two people work on alternate days, convert each person's individual number of days into a whole-number daily output by choosing the total work as the LCM of their two day-counts; then add the two workers' daily outputs cycle by cycle, in the order they alternate, until only a fraction of the last worker's day is needed to finish what remains.

Step-by-step solution

1. Take the total work as LCM(19, 21) = 399 units, so that both A’s and B’s daily output come out as whole numbers.

2. A’s work per day = 399 ÷ 19 = 21 units; B’s work per day = 399 ÷ 21 = 19 units.

3. Since B starts, each 2-day cycle (B then A) completes 19 + 21 = 40 units.

4. Number of complete 2-day cycles = floor(399 ÷ 40) = 9, so 9 × 40 = 360 units are completed in 18 days.

5. Work remaining after 18 days = 399 − 360 = 39 units.

6. Day 19 is an odd day, so it is B’s turn: B adds 19 units, taking the total to 360 + 19 = 379 units, leaving 399 − 379 = 20 units.

7. Day 20 is A’s turn; A’s full-day output is 21 units, but only 20 units remain, so A needs 20/21 of that day to complete the remaining work.

8. Total time taken = 19 complete days + 20/21 of the 20th day = 19 20/21 days.

Cross-check

Expressed as fractions of the whole job: 9 cycles give 360/399 of the work, B’s day 19 adds 19/399 (total 379/399), leaving 20/399 still to do. Since A completes 21/399 of the job per full day, the remaining 20/399 needs (20/399) ÷ (21/399) = 20/21 of a day — confirming a total time of 19 20/21 days.