Time and Work

This is a direct proportion problem where work output depends on both number of workers and days worked. First, find the rate: 60 kg ÷ (5 workers × 3 days) = 4 kg per worker-day. For 8 workers working 5 days: 8 × 5 × 4 = 160 kg.

Statement I: Combined rate of Reetesh + Rahul. Statement III: Reetesh's individual rate. From I and III, Rahul's rate can be found. Statement II confirms this but is not needed - once we have individual rates, Rahul's time alone can be calculated. So I and III are sufficient. I and II together also work: from combined rate (I) and the work pattern (II), Rahul's contribution can be derived. Thus I + either II or III works.

A's work in (3/4)t time = (1/2) of B's work in t time → A's rate : B's rate = (1/2)/(3/4) = 2/3. Together in 18 days: (2/3 + 1) × B's work = 18 days → (5/3)B = 18 → B = 18 × 3/5 = 54/5 = 10.8 days for full work. This is wrong; recalculate: If B's rate = 1 work per x days, A's rate = (1/2)/(3/4x) = 2/3x. Combined: 1/x + 2/3x = 18 days → 5/3x = 18 → x = 30 days. B takes 30 days alone.

A's 1-day work = 1/6, B's 1-day work = 1/12. Together in 3 days: 3(1/6 + 1/12) = 3(3/12) = 9/12 = 3/4 work done. Remaining 1/4 work by B alone: (1/4)/(1/12) = 3 days. Total = 3 + 3 = 6 days.

Let Q's efficiency = 1 unit/day. Then P's efficiency = 3 units/day. Combined efficiency = 4 units/day. Together they complete work in 16 days, so total work = 4 * 16 = 64 units. Q alone would take 64/1 = 64 days. Option C is correct.

18 men = 36 women, so 1 man = 2 women. 12 men + 16 women = 12(2) + 16 = 40 women equivalents. 36 women take 40 days, so 40 women take (36×40)/40 = 36 days. Option D is correct.

First, calculate each person's daily work capacity: Person 1 = 1/8, Person 2 = 1/12, Person 3 = 1/16. In 3 days, they complete 3×(1/8 + 1/12 + 1/16) = 3×(13/48) = 39/48 of the work. The fourth person contributed 1 - 39/48 = 9/48 = 3/16 of the total work. Therefore, the fourth person gets (3/16)×1200 = Rs. 225.

Total work = 6 hours/day × 18 days = 108 hours. To complete the same work in 12 days, daily hours needed = 108/12 = 9 hours per day. This is an inverse proportion problem - as days decrease, hours per day must increase proportionally.

Let original boys = x. Work = x×9 (boy-days). When 5 absent: (x-5)×12 = x×9. 12x - 60 = 9x. 3x = 60. x = 20.

Let individual rates be R, S, Ss (Ram, Shyam, Suresh). From statements: R+S=1/18, Ss+R=1/15, S+Ss=1/12 (where all rates are per day). We need combined rate R+S+Ss. Adding all three equations: 2(R+S+Ss) = 1/18 + 1/15 + 1/12. So R+S+Ss = (1/2) × (1/18 + 1/15 + 1/12). But the answer is E (None of these), suggesting the statements together are insufficient or don't give a unique answer.

Let A, B, C be work rates. A+B = 1/18, B+C = 1/24, A+C = 1/36. Add all: 2(A+B+C) = 1/18 + 1/24 + 1/36 = 4/72 + 3/72 + 2/72 = 9/72 = 1/8. So A+B+C = 1/16. Together they complete the work in 16 days. Option C (16 days) is correct.

If A is thrice as efficient as B, then A's work rate is 3x B's rate. Let B take x days, so A takes x-8 days. Since work = rate × time, and work done is the same: 3(x-8) = x, giving x = 12 days. Alternatively, use the formula: if A is k times efficient and takes d days less, then B takes kd/(k-1) = 3×8/2 = 12 days.

6 men in 12 days means 1 man does 1/72 of work per day. 8 women in 18 days means 1 woman does 1/144 of work per day. 18 children in 10 days means 1 child does 1/180 of work per day. Work done by 4 men + 12 women + 20 children in 2 days = 2×(4/72 + 12/144 + 20/180) = 2×(1/18 + 1/12 + 1/9) = 2×(2/36 + 3/36 + 4/36) = 2×(9/36) = 1/2. Remaining work = 1/2. If men complete this in 1 day, we need m men such that m/72 = 1/2, so m = 36 men are required.

8 men in 5 days means total work = 8 × 5 = 40 man-days. So 1 man does 1/40 of work per day. 16 women in 5 days means total work = 16 × 5 = 80 woman-days. So 1 woman does 1/80 of work per day. 2 men + 6 women = 2(1/40) + 6(1/80) = 1/20 + 3/40 = 5/40 = 1/8 of work per day. Days needed = 1 / (1/8) = 8 days. Option C is correct.

If A does thrice as much work as B, then A's work rate is 3 times B's. Let B's rate = 1 unit/day, so A's rate = 3 units/day. Combined rate = 4 units/day. If they complete work in 3 days, total work = 4 × 3 = 12 units. A alone would take 12/3 = 4 days. This is a work-rate problem where efficiency ratios determine individual times.