Time and Work

Rohan's rate = 1/36 work per day. Combined rate (Rohan + Mohan) = 1/24 work per day. Mohan's rate = (1/24) - (1/36) = (3 - 2)/72 = 1/72 work per day. So Mohan alone takes 72 days to complete the work. Option D is correct.

Using the work formula: M1 × D1 × H1 = M2 × D2 × H2. Given: 24 men × 12 days × 8 hours = 2304 man-hours. For 36 men working 6 hours per day: 36 × 6 × D2 = 2304, so D2 = 2304 / 216 = 10.67 days. However, this doesn't match 11. Let me recalculate: 24 × 12 × 8 = 2304. 36 × 6 × D = 2304, so D = 2304/(36×6) = 2304/216 = 10.67. The answer should be approximately 11 days (rounding up since you can't have a fraction of a day in practice).

Total work = 24 men × 8 hours/day × 10 days = 1920 man-hours. Required: complete in 6 days at 10 hours/day. Men needed = 1920 ÷ (6 × 10) = 1920 ÷ 60 = 32 men. Option A (30): would give 1800 man-hours, insufficient. Option C (34): would give 2040 man-hours, more than needed. Option D (36): would give 2160 man-hours.

This is an inverse proportion problem - fewer workers means more days. The total work is constant at 36 × 25 = 900 man-days. With only 15 men, divide 900 by 15 to get 60 days. Think of it as redistributing the same total work across fewer people.

Total work = 12 men × 6 days = 72 man-days. If 10 men work alone: 72 ÷ 10 = 7.2 days. The second statement (10 men + 21 women in 8 days) confirms different work rates but is not needed to find 10 men's time. 7.2 is not listed in A-D.

Ram's rate = 1/20 work per day, Mohan's rate = 1/30 work per day. Combined rate = 1/20 + 1/30 = 5/60 = 1/12. Therefore, working together they complete the work in 12 days.

16 men in 8 days means 1 man does 1/(16×8) = 1/128 work per day. 20 women in 16 days means 1 woman does 1/(20×16) = 1/320 work per day. 12 men + 10 women for 6 days complete: 12/128×6 + 10/320×6 = 9/16 + 3/16 = 12/16 = 3/4 of work. Remaining 1/4 work by 10 women: (1/4)/(10/320) = 8 days.

B and C together do 2 days of work: B's rate = 1/20, C's rate = 1/30, combined = 1/12 per day. In 2 days they complete 2/12 = 1/6 of work. Remaining = 5/6. A's rate = 1/18, so time for 5/6 = (5/6) × 18 = 15 days.

Work = men × days. Original: 7 × 12 = 84 man-days. New work: 2 × 84 = 168 man-days (double work). Required men = 168/8 = 21 men. Additional men needed = 21 - 7 = 14. Option D is correct.

Combined rate of A+B = 1/20+1/30 = 1/12. In 5 days, they complete 5/12 work. Remaining 7/12 is done by C at 1/12 rate, taking 7 days.

45 men × 16 days = 720 man-days of total work. After 6 days, work done = 45 × 6 = 270 man-days. Remaining work = 720 - 270 = 450 man-days. Now 30 more men join, so 75 men work. Days needed = 450/75 = 6 days. Total days = 6 + 6 = 12 days. Options are 8, 3, 5, 10, 'None of these' - so 'None of these' (E) is correct.

Let total work be LCM(36,52,72)=468 units. A's rate=13, B's rate=9, C's rate=6.5 units/day. If total time is x days, A worked x-8 days, B worked x-12 days, C worked x days. 13(x-8)+9(x-12)+6.5x=468. Solving: 28.5x-232=468, x=24. C worked all 24 days.

3 men or 6 boys do work in 40 days. So 1 man = 2 boys (since 3 men = 6 boys). 6 men + 8 boys = 6 men + 4 men = 10 men. If 3 men take 40 days, 10 men take (3×40)/10 = 12 days.

P's rate: (2/3)/6 = 1/9 per day. Q's rate: (1/3)/8 = 1/24 per day. R's rate: (3/4)/12 = 1/16 per day. Combined work in 4 days: 4(1/9 + 1/24 + 1/16) = 4(1/9 + 7/48) = 4(55/144) = 55/36. Remaining work: 1 - 55/36 = -19/36 (this is negative, indicating an error in my calculation). Let me recalculate: 1/9 = 16/144, 1/24 = 6/144, 1/16 = 9/144. Combined = 31/144 per day. In 4 days: 124/144 = 31/36 done. Remaining: 5/36. Q's time for 5/36 = (5/36)/(1/24) = 5/36 × 24 = 10/3 = 3.33 days.

Work = Examiners × Days × Hours. Let W = work for original books. 4 × 8 × 5 = 160 examiner-hours. New work is twice: 2W = 320 examiner-hours. With 2 examiners for 20 days: 2 × 20 × h = 320, so 40h = 320, h = 8 hours. Option C is correct.