Time & Work MCQs for UPSC Prelims

Find rates as fraction of tank per minute. Set X: 20 pipes fill 0.70 in 14 min -> 0.05/min, so one X pipe = 0.0025. Set Y: 10 pipes fill 0.375 in 6 min -> 0.0625/min, one Y pipe = 0.00625. Set Z: 16 pipes empty 0.5 in 20 min -> 0.025/min, one Z pipe = 0.0015625. Now half of X closed = 10 X pipes open: 10 x 0.0025 = 0.025. Half of Y open = 5 Y pipes: 5 x 0.00625 = 0.03125. All Z open (emptying): -0.025. Net = 0.025 + 0.03125 - 0.025 = 0.03125 tank/min. Time to fill 50% (0.5) = 0.5 / 0.03125 = 16 minutes. Answer: (d).

Earnings are shared in proportion to work done, i.e. to the rates of work. Let Q's rate = 1 unit. P works thrice as fast as Q, so P's rate = 3. P and Q together work four times as fast as R: (P + Q) = 4R, so R = (3 + 1)/4 = 1. Thus the rates are P : Q : R = 3 : 1 : 1, which is also the ratio of work done (and hence earnings). Hence (a).

X does one third of the work in 6 days, so X alone takes 18 days and works at 1/18 per day. Y does one third in 8 days, so Y takes 24 days and works at 1/24 per day. Z does three fourths in 12 days, so Z takes 16 days and works at 1/16 per day. Combined rate = 1/18 + 1/24 + 1/16 = (8 + 6 + 9)/144 = 23/144 per day. Y alone finishes the remainder in 8 2/3 = 26/3 days, completing (26/3) x (1/24) = 26/72 = 13/36 of the work. So the three together did 1 - 13/36 = 23/36 = 92/144 of the work in n days: (23/144) x n = 92/144, giving n = 4. The official UPSC key confirms option (b).

Let a man do m units per day and a woman w units per day. The given data yields one equation, 30(24m + 12w) = total work, with two unknown efficiencies. The required time for 12 men and 24 women depends on the sign of (12m + 24w) - (24m + 12w) = 12(w - m), which is unknown: if w = m the answer is exactly 30 days, if w > m it is less, and if w < m it is more. Option (a) assumes equal efficiency, option (b) assumes men are more efficient, and option (c) wrongly excludes the possibility of exactly 30 days. Since no relation between m and w is given, only option (d), data inadequate, is correct.

X completes 20% in 8 days, so X completes the full work in 8 x 5 = 40 days, rate = 1/40 per day. Y completes 25% in 6 days, so Y completes the full work in 6 x 4 = 24 days, rate = 1/24 per day. Combined rate = 1/40 + 1/24 = (3 + 5)/120 = 8/120 = 1/15 of the work per day. Time to complete 40% (= 0.4 work) = 0.4 / (1/15) = 0.4 x 15 = 6 days. The other options come from miscomputing individual full-work times or the combined rate.

Scale each rate to the whole job. W does 25% in 30 days, so the full job takes 30 x 4 = 120 days. X does 1/4 in 10 days, so the full job takes 10 x 4 = 40 days. Y does 40% in 40 days, so the full job takes 40 x (100/40) = 100 days. Z does 1/3 in 13 days, so the full job takes 13 x 3 = 39 days. The smallest time is Z's 39 days, so Z finishes first (just ahead of X's 40 days). Hence Z.

Option (a) is correct. Daily work in 48ths of the job: A does 6, B does 3, C does 4. The rotation A, B, C starts on Monday and repeats every three days. Cumulative work: after day 3 (Wed) 13/48, day 6 (Sat) 26/48, day 9 (Tue) 39/48, day 10 (Wed, A) 45/48. On day 11 B contributes exactly the remaining 3/48, finishing the job at the end of day 11. Counting weekdays from Monday as day 1, day 11 is a Thursday, so statement 1 is correct. Statement 2 is wrong because the work takes 11 days, not 10; after 10 days only 45/48 is done. Hence options (b), (c) and (d) fail.

Total work = M x 6k man-days for M men. To finish the same work in 5k days, the required number of men is M' = (M x 6k) / 5k = 1.2M. The increase is 0.2M, i.e., 20% of the original number of men. Hence 20%.