Time and Work

Total work = 20 × 30 = 600 units. After x days, 15 workers work for (35-x) days: 20x + 15(35-x) = 600. Solving: 5x = 75, x = 15. 5 workers should leave after 15 days.

This is an inverse proportion problem. More men means fewer days. Total work = 12 × 24 = 288 man-days. With 8 men: 288/8 = 36 days. Alternatively, using inverse proportion: (12/8) × 24 = (3/2) × 24 = 36 days.

A's work rate is 1/15 per day, B's work rate is 1/20 per day. Working alternately starting with A, in 2 days they complete 1/15 + 1/20 = 4/60 + 3/60 = 7/60 of the work. After 16 days (8 cycles of 2 days), they complete 8 × 7/60 = 56/60, leaving 4/60. On day 17, A works and completes 1/15 = 4/60, finishing the job. Total time = 17 days.

Let total work = LCM(60, 40) = 120 units. A + B's rate = 120/60 = 2 units/day. A + B + C's rate = 120/40 = 3 units/day. Therefore, C's rate = 3 - 2 = 1 unit/day. Time for C alone = 120/1 = 120 days. Option E is correct. Options A-D are common errors from incorrect rate calculations or LCM selection.

6 women complete work in 3 days, so 1 woman's 1-day work = 1/(6×3) = 1/18. Thus 3 women's 1-day work = 3/18 = 1/6. The 3 women + 18 children team completes the work in 2 days, so their 1-day work = 1/2. Therefore 18 children's 1-day work = 1/2 - 1/6 = 1/3. So 9 children's 1-day work = 1/6, meaning 9 children take 6 days. This is a classic work-rate problem using reciprocals.

A's efficiency is twice B's, so if B does 1 unit per day, A does 2 units per day. Together they do 3 units daily, completing the job in 24 days, meaning total work = 72 units. B alone would take 72/1 = 72 days. Options A, B, D, E are incorrect as they don't match the calculation.

Total work = 5 persons × 3 days × 8 hours = 120 person-hours. One person working 4 hours/day needs 120/4 = 30 days. This is an inverse proportion problem.

Let n persons complete work in 30 days. With n+5 persons, work completes in 30-10 = 20 days. Total work = n × 30 = (n+5) × 20. Solving: 30n = 20n + 100, so 10n = 100, giving n = 10 persons.

A does 1/2 work in 1 day, so A's rate = 1/2. B does full work in 1 day, so B's rate = 1. B does half work as C in 1 day, meaning C does twice what B does, so C's rate = 2. Efficiency ratio A:B:C = 1/2 : 1 : 2 = 1:2:4.

15 men in 20 days means 15M × 20 = 1 work, so 1 man-day = 1/300 work. Similarly, 24W × 20 = 1 work, so 1 woman-day = 1/480 work. Combined rate of 10 men and 8 women = 10/300 + 8/480 = 1/30 + 1/60 = 3/60 = 1/20. Time = 1/(1/20) = 20 days.

Let initial number of workers be x. The total work = 25x man-days. With (x-10) workers, work completed in 35 days, so (x-10) × 35 = 25x. Solving: 35x - 350 = 25x, giving 10x = 350, so x = 35. This uses the inverse proportion between workers and days - fewer workers means more days needed for the same work.

From the given rates: 1 man = 44 days, so 2 women = 44 days means 1 woman = 88 days, and 3 boys = 44 days means 1 boy = 132 days. Combined rate: 1/44 + 1/88 + 1/132 = 3/132 + 1.5/132 + 1/132 = 5.5/132 = 1/24. So they complete the work in 24 days.

Total work = 12 men × 12 days = 144 man-days. In the first 6 days with 12 men, work completed = 12 × 6 = 72 man-days. Remaining work = 144 - 72 = 72 man-days. After 6 men leave, only 6 men remain. Time needed for remaining work = 72/6 = 12 days. The question asks for EXTRA days beyond the original 12, so 12 more days are needed.

Let original workers = x. Total work = x × 22 worker-days. With 3 workers absent, workers = x - 3, and they complete in 24 days: (x - 3) × 24 = x × 22. Solving: 24x - 72 = 22x, so 2x = 72, giving x = 36 workers. This means 36 workers would take 22 days, and 33 workers would take 24 days - both completing the same total work of 792 worker-days.

Total work = 24 men × 35 days = 840 man-days. To complete in 21 days, men needed = 840 ÷ 21 = 40 men. This is an inverse proportion problem.