Typist A takes twice as much time as another typist B, or thrice as much time…

Concept: each worker's rate of work (amount completed per hour) is the reciprocal of the time that worker alone needs for that same amount of work. When several people work together, their individual rates simply add up to give the combined rate; and if a fixed amount of work is finished together in a known time, that combined rate can be read off directly.

1. Let A's time to type 200 pages be x hours. Since A takes twice as long as B, B needs x/2 hours for 200 pages; since A takes thrice as long as C, C needs x/3 hours for 200 pages.

2. Convert these times to rates for 200 pages: A's rate = 200/x pages per hour, B's rate = 200/(x/2) = 400/x pages per hour, C's rate = 200/(x/3) = 600/x pages per hour.

3. Add the three rates to get the combined rate: (200 + 400 + 600)/x = 1200/x pages per hour.

4. The problem states that all three working together type 200 pages in 2 hours, so their combined rate is 200/2 = 100 pages per hour.

5. Equate the two expressions for the combined rate: 1200/x = 100, so x = 12 hours -- this is A's time to type 200 pages.

6. C's time to type 200 pages is x/3 = 12/3 = 4 hours, so C's rate is 200/4 = 50 pages per hour.

7. Therefore, C alone takes 400/50 = 8 hours to type 400 pages.

Cross-check: with x = 12, A's rate is 200/12 which is about 16.67 pages per hour, B's rate is 400/12 which is about 33.33 pages per hour, and C's rate is 600/12 = 50 pages per hour. Their sum is 100 pages per hour, matching the given combined rate of 200 pages in 2 hours, confirming that C alone needs 8 hours for 400 pages.