Time and Work

This is a direct proportion problem: more persons × more hours × more days = more machines. Initial rate: 2 persons × 2 hours × 2 days = 2 machines, so rate = 2/(2×2×2) = 1/4 machine per person-hour-day. New scenario: 6 × 6 × 6 × (1/4) = 216/4 = 54 machines. Alternatively, scaling each factor by 3 gives 3×3×3 = 27 times the output, so 2 × 27 = 54.

Let 1 boy's work = b, 1 girl's work = g per day. From statements: 6b + 8g = 1/10 and 26b + 48g = 1/2. Solving gives b = 1/200, g = 1/160. For half work by 15 boys + 20 girls: (15/200 + 20/160) * days = 1/2. Days = 2.1 ≈ 2 days.

A completes 1/12 of job per day. B is 60% more efficient, so B does 1.6 × (1/12) = 2/15 per day. C is 60% less efficient, so C does 0.4 × (1/12) = 1/30 per day. Together they do 1/12 + 2/15 + 1/30 = 5/60 + 8/60 + 2/60 = 15/60 = 1/4 per day. Time = 1 ÷ 1/4 = 4 days.

A's 1-day work = 1/6, B's = 1/9, C's = 1/12. In 2 days, A+B+C complete 2(1/6 + 1/9 + 1/12) = 2(6/36 + 4/36 + 3/36) = 2(13/36) = 13/18 of work. Remaining 5/18 is done by fourth person. Fourth person's share = (5/18) × 1350 = Rs 375. This is based on work contribution proportion.

Annu's 1-day work = 1/6, Bhanu's 1-day work = 1/8. Together they do 1/6 + 1/8 = 7/24 per day. In 3 days with Charu, they complete the work, so combined 1-day work = 1/3. Charu's 1-day work = 1/3 - 7/24 = 1/24. Charu's 3-day work = 3/24 = 1/8. Charu's share = (1/8) * 600 = Rs. 75. Option D is correct. Option A (80) would be 1/7.5, B (85) is 1/7.06, C (70) is 1/8.57 - none match the calculation.

In time and work problems, work = rate × time. Total work = LCM(18, 9, 3) = 18 units. Ajay's rate = 1 unit/day, Divya's rate = 2 units/day, Sanjana's rate = 6 units/day. First 3 days: Ajay + Divya = 3 units/day. Work done = 3 × 3 = 9 units. Remaining = 18 - 9 = 9 units. With Sanjana: combined rate = 1 + 2 + 6 = 9 units/day. Time = 9/9 = 1 day. Total = 3 + 1 = 4 days. Option B is correct.

Each person completes 1/91 of job per day. Day 1: 1 person works = 1/91 completed. Day 2: 2 people work = 2/91 completed. Total after 2 days = 3/91. Day n: n people work. Total work = sum of (1+2+3+...+n)/91 = n(n+1)/(2×91). We need this = 1, so n(n+1) = 182. Testing n=13: 13×14 = 182. So 13 days needed. Option C (13) is correct.

12 men × 10 days = 120 man-days of work. To complete in 8 days: 120/8 = 15 men required. This is an inverse proportion problem - fewer days means more men. Option A (14) and C (16) are close but don't satisfy the equation exactly.

Using the work formula: M1D1 = M2D2, where M is number of workers and D is days. Initially: 76 × 33 = M2 × 44. Solving: M2 = (76 × 33) / 44 = 76 × (33/44) = 76 × (3/4) = 57 ladies actually worked. Therefore, ladies who did not report = 76 - 57 = 19. The inverse relationship between workers and time is key here.

Let efficiencies be A, B, C. Given A+C = 2B and A+B = 3C. From A = 2B - C, substituting: 2B - C + B = 3C, so 3B = 4C, B = 4C/3. Then A = 2(4C/3) - C = 5C/3. Combined efficiency A+B+C = 5C/3 + 4C/3 + C = 4C. Since they finish in 12 days, 4C × 12 = 1 work, so C = 1/48 work per day. C alone needs 48 days.

If A takes 70 days and B is 40% more efficient, then B's efficiency is 140% of A's efficiency (or 1.4 times A). Time is inversely proportional to efficiency. So B's time = A's time / 1.4 = 70 / 1.4 = 50 days. Option C is correct.

A man's 1-day work is 1/20, a woman's is 1/30, and a boy's is 1/60. In 2 days, 2 men complete 2*(2/20) = 1/5 of the work, and 8 women complete 2*(8/30) = 8/15. Let x boys assist; they complete 2*(x/60) = x/30 of the work. Total: 1/5 + 8/15 + x/30 = (22+x)/30 = 1, so x = 8. Eight boys are needed.

Using work = people × days, total work W = (x-5)(x). Also, 75% of W = (x+5)(x-11), so 0.75x(x-5) = (x+5)(x-11). Simplifying: 3x² - 15x = 4x² - 11x + 5x - 55, which gives x² - x - 55 = 0. Solving: x = 10 (valid) or x = -9 (invalid). Total work W = 10(5) = 50 person-days. For (x+10) = 20 people: days = 50/20 = 2.5 days.

Work rates: A = 1/6 per day, B = 1/10 per day, C = 1/15 per day. Combined daily rate = 1/6 + 1/10 + 1/15 = (5+3+2)/30 = 10/30 = 1/3. Work completes in 3 days. Wages are distributed in ratio of work rates: A:B:C = 1/6:1/10:1/15 = (5:3:2)/30 = 5:3:2. Sum = 10 parts. A's 2-day wage = (5/10) × (2/3) × 300 = 100. B's 2-day wage = (3/10) × (2/3) × 300 = 60. C's 2-day wage = (2/10) × (2/3) × 300 = 40. Sum = 100 + 60 + 40 = 200. Option A would be just A's 2-day wage. Option C would be half the total earned.

Z's 30% work takes 0.30 × 25 = 7.5 days. Y's 20% work takes 0.20 × 20 = 4 days. X does remaining 50%, taking 0.50 × 15 = 7.5 days. Total = 7.5 + 4 + 7.5 = 19 days. Option D is correct.