Pipes and Cisterns

Combined rate = 1/16 + 1/12 - 1/24 = (3+4-2)/48 = 5/48 tank per hour. Time to fill = 48/5 = 9.6 hours. The third pipe empties, so its contribution is subtracted.

Fill pipe rate: 1/12 per hour. Fill + leak together: 1/15 per hour. Therefore leak alone empties at (1/12 - 1/15) = (5/60 - 4/60) = 1/60 per hour. If the cistern is full, the leak will empty it in 60 hours.

Let I, II, III be rates. I+II = 1/12, II+III = 1/15, and I = 2*III. Substituting: 2*III + II = 1/12 and II + III = 1/15. Solving gives III = 1/60, so I = 2/60 = 1/30, meaning 30 hours.

Original: A fills in 90 hrs (rate = 1/90), B fills in 60 hrs (rate = 1/60). Combined rate = 1/90 + 1/60 = 5/180 = 1/36. Time for original tank = 36 hrs. New capacity = 1.4× original. New rates (80% efficiency): A = 0.8/90 = 1/112.5, B = 0.8/60 = 1/75. Combined new rate = 1/112.5 + 1/75 = 5/450 + 6/450 = 11/450. New time = 1.4 / (11/450) = 1.4 × 450/11 = 630/11 = 57.27... Hmm, not 63. Let me recalculate: Original combined time = 1/(1/90 + 1/60) = 1/(5/180) = 36 hrs. With 80% efficiency, new rates are 0.8× original. New combined rate = 0.8 × (1/90 + 1/60) = 0.8/36 = 1/45. For 1.4× capacity, time = 1.4 × 45 = 63 hrs. Yes! Matches option A exactly.

Tank dimensions: 0.2 km × 0.15 km = 200 m × 150 m = 30000 m². Required volume for 16 m height: 30000 × 16 = 480000 m³. Pipe dimensions: 30 cm × 20 cm = 0.3 m × 0.2 m = 0.06 m². Flow speed: 20 km/hr = 20000 m/hr. Flow rate = 0.06 × 20000 = 1200 m³/hr. Time = 480000 ÷ 1200 = 400 hours. Answer D (400) is correct.

Pipe rates: A = 1/12 = 8.33%/hr, B = 1/15 = 6.67%/hr, C = 1/20 = 5%/hr. From 6:00-9:30 (3.5hr): B fills 3.5 × 6.67% = 23.33%. From 9:30-11:30 (2hr): A+B fills 2 × (8.33% + 6.67%) = 30%. After 11:30: All three fill at 20%/hr. Remaining = 100% - 23.33% - 30% = 46.67%. Time = 46.67% / 20% = 2.33hr = 2hr 20min. Total = 6:00 + 3hr 30min + 2hr + 2hr 20min = 1:50 PM.

Let P, Q, R together fill in 6 hours. Their combined rate = 1/6 per hour. In 2 hours, they fill 2/6 = 1/3 of tank. Remaining = 2/3. P and Q fill 2/3 in 7 hours, so their combined rate = (2/3)/7 = 2/21 per hour. Therefore R's rate = (1/6) - (2/21) = (7/42) - (4/42) = 3/42 = 1/14 per hour. So R alone takes 14 hours.

Pipe A fills in 72 minutes (rate: 1/72 per minute), Pipe B in 108 minutes (rate: 1/108 per minute). Working alternately starting with A: Cycle of 2 minutes fills (1/72 + 1/108) = 5/216 of tank. After 86 minutes (43 complete cycles), tank is 43 × 5/216 = 215/216 full. Remaining 1/216 needs Pipe A (who works 87th minute): (1/72)t = 1/216 gives t = 1/3 minute = 20 seconds. Total time = 86 + 1/3 = 86.33 minutes. The key is finding complete cycles first, then handling the remainder.

Combined rate: 1/18 + 1/40 - 1/20 = 11/360 tank per hour. In 20 hours, tank filled = 20 × 11/360 = 11/18. If 440L is 11/18 of capacity, full capacity = 440 × 18/11 = 720L. Option D is correct.

Work done = (rate × time). All three work together for 10 minutes: 10(1/45+1/72+1/60) = 190/360 tank filled. Then only B and C work for 5 minutes: 5(1/72+1/60) = 55/360 more. Total filled = 245/360. Remaining 115/360 is filled by B alone at 1/72 per minute, taking 115/5 = 23 minutes. Total time = 10+5+23 = 38 minutes.

Pipe A fills at rate 1/18 per hour. Pipe B fills at rate 1/30 per hour. Combined rate = 1/18 + 1/30 = 5/90 + 3/90 = 8/90 = 4/45 per hour. In 5 hours together, they fill 5×4/45 = 20/45 = 4/9 of the tank. Remaining part = 1 - 4/9 = 5/9. Pipe A alone fills at 1/18 per hour, so time for remaining part = (5/9)/(1/18) = 5/9 × 18 = 10 hours.

A+B fill in 24 hours, so combined rate = 1/24. C empties in 120 hours, so C's rate = -1/120. B+C fill in 120 hours, so B's rate - 1/120 = 1/120, giving B's rate = 1/60. Then A's rate = 1/24 - 1/60 = 1/40. So A alone takes 40 hours. Option A (60 hr) incorrectly assigns B's time to A, while C (80 hr) is a miscalculation.

Total work = 1 tank. A+B+C rate = 1/6 per hour. In 2 hours, 1/3 tank filled. Remaining 2/3 filled by A+B in 7 hours, so A+B rate = (2/3)/7 = 2/21. C's rate = 1/6 - 2/21 = 7/42 - 4/42 = 3/42 = 1/14. C alone takes 14 hours.

Let B's time = t hours, so A's time = 3t hours. Together: 1/t + 1/(3t) = 1/3 (combined rate per hour). Solving: 4/(3t) = 1/3, so 3t = 12 and t = 4. Therefore A alone takes 12 hours. Verification: A fills 1/12 per hour, B fills 1/4 per hour, together 1/12 + 3/12 = 4/12 = 1/3 tank per hour, so 3 hours to fill. Standard work-rate problem.

Volume of tank = 200 × 150 × 8 = 240,000 m³. Pipe cross-section = 0.3 × 0.2 = 0.06 m². Flow speed = 20 km/h = 20,000 m/h. Flow rate = 0.06 × 20,000 = 1,200 m³/h. Time = 240,000/1,200 = 200 hours.