Pipes and Cisterns

Pipe A fills at 1/36 per hour and B at 1/48 per hour. If B is open for t hours, total filled = t/48 + 24/36 = 1. Solving: t/48 = 1 - 2/3 = 1/3, so t = 16 hours. Option C is correct.

First pipe fills in 6 min (rate = 1/6 per minute). Second pipe fills in 7 min (rate = 1/7 per minute). Operating alternately for 1 minute each means in 2 minutes, they fill 1/6 + 1/7 = 13/42 of the cistern. After 6 cycles (12 minutes), cistern is 6 × (13/42) = 78/42 = 39/21 = 13/7 full. In 13th minute (first pipe), it fills 1/6 = 7/42 more. Total = 13/7 + 1/6 = 78/42 + 7/42 = 85/42 = 2 1/42. Remaining = 1 - 85/42 = -43/42. Wait, let me recalculate. After 6 full cycles (12 min): 6 × (1/6 + 1/7) = 6 × (13/42) = 78/42 = 1 + 36/42 = 1 + 6/7. Remaining = 1/7. Next is first pipe's turn (13th minute), fills 1/6. But only 1/7 needed. So time from 13th minute: (1/7) / (1/6) = 6/7 minute. Total = 12 + 6/7 = 90/7 minutes. But answer should be 45/7. Let me reconsider. In 2 minutes: 1/6 + 1/7 = 13/42. After 6 full cycles (12 minutes): 6 × 13/42 = 78/42 = 1 36/42 = 1 6/7. Remaining = 1/7. In 13th minute (first pipe): fills 1/6 per minute, so to fill 1/7 takes (1/7)/(1/6) = 6/7 minute. Total = 12 + 6/7 = 84/7 + 6/7 = 90/7. Hmm. But wait - maybe we should check how much is filled before the last partial minute. After 5 cycles (10 min): 5 × 13/42 = 65/42 = 1 23/42. Remaining = 19/42. 11th minute (first pipe): fills 1/6 = 7/42. Total = 65/42 + 7/42 = 72/42 = 1 30/42 = 1 5/7. Remaining = 2/7. 12th minute (second pipe): fills 1/7 = 6/42. Total = 72/42 + 6/42 = 78/42 = 1 36/42. Remaining = 1/7. 13th minute (first pipe): fills 1/6 per minute, so 1/7 takes 6/7 minute. Total = 12 + 6/7 = 90/7. But answer says 45/7. Let me reconsider. Actually, I think I miscounted. In 2-minute cycles, fill 13/42. After n cycles (2n minutes): 13n/42. We need total >= 1. So 13n/42 >= 1 means n >= 42/13 ≈ 3.23. So after 4 cycles (8 minutes): 4 × 13/42 = 52/42 = 1 10/42. Remaining = -10/42. Wait that means it overflowed. Let me be more careful. After 3 cycles (6 minutes): 3 × 13/42 = 39/42 < 1. Remaining = 3/42 = 1/14. 7th minute (first pipe): fills 1/6 = 7/42. But we only need 3/42. So time from 7th minute: (3/42)/(1/6) = (3/42) × 6 = 18/42 = 3/7 minute. Total = 6 + 3/7 = 45/7 minutes.

The phrase all pipes run alternatively is ambiguous - it could mean hourly rotation, minute-by-minute switching, or some other pattern. Without clarifying what alternatively means in this context, it is impossible to determine whether and when the tank fills or empties. Option E correctly identifies this indeterminacy.

Let first pipe work for t minutes, then only second pipe works for (18-t) minutes. Combined rate of both pipes = 1/20 + 1/30 = 1/12 per minute. Work equation: t/12 + (18-t)/30 = 1. Solving gives t/20 + 3/5 = 1, so t/20 = 2/5, t = 8 minutes. First pipe should be closed after 8 minutes.

Inlet fills in 4 hours, so rate = 1/4 per hour. Outlet empties in 5 hours, so rate = -1/5 per hour. When both open, net rate = 1/4 - 1/5 = 1/20 per hour. When only inlet works, rate = 1/4 per hour. From 9 AM to 5 PM = 8 hours. If outlet closed after x hours, then x/20 + (8-x)/4 = 1. Solving: x/20 + 2 - x/4 = 1, which gives -4x/20 + 2 = 1, so -x/5 = -1, and x = 5 hours. Outlet closed at 9 AM + 5 hours = 2 PM.

Pipe A fills 10 times faster than B, so if B's rate is 1/66 cistern per minute, A's rate is 10/66. Their combined rate is 11/66 = 1/6 cistern per minute, meaning they fill the cistern together in 6 minutes. Option B (5 min) ignores that A alone would take 6.6 minutes, so the combined time cannot be less.

Pipe A fills 1/10 of tank per minute, B fills 1/12, C fills 1/15. Combined rate = 1/10 + 1/12 + 1/15 = 6/60 + 5/60 + 4/60 = 15/60 = 1/4 tank per minute. Time to fill = 1 / (1/4) = 4 minutes. When pipes work together, add their individual rates to get the combined rate, then take the reciprocal.

Fill tap rate = 1/40 per min, empty tap rate = 1/60 per min. Net fill rate with both open = 1/40 - 1/60 = (3-2)/120 = 1/120 per min. Time to fill = 1 / (1/120) = 120 min. The empty tap drains slower than the fill tap fills, so tank still fills, just at reduced rate.

Cistern volume = pi × r² × h = pi × 5² × 2 = 50pi cubic meters = 50,00,000 cubic cm. Pipe flow rate = pi × 10² × 3,00,000 cm/hr = 3pi × 10^7 cubic cm/hr. Time = Volume/Rate = 50,00,000 / (3pi × 10^7/pi) = 50,00,000 / 3,00,000 = 5/3 hours = 1 hour 40 minutes.

Pipe A fills 1/2 tank per hour. From 10-11 AM (1 hour), A fills 1/2 tank. Remaining 1/2 tank is filled by A+B together: combined rate = 1/2 + 1/6 = 2/3 per hour. Time for remaining = (1/2) ÷ (2/3) = 3/4 hour = 45 minutes. Tank fills at 11:45 AM.

Volume of water needed = area of tank × rise in height = 15 × 10 × 3 = 450 m³. Pipe cross-section = 2 × 1.5 = 3 m². Speed = 20 km/hr = 20000 m/hr = 20000/3600 m/sec ≈ 5.56 m/sec. Flow rate = 3 × 5.56 = 16.68 m³/sec. Time = 450/16.68 ≈ 27 seconds.

The rate of drainage is proportional to the cross-sectional area of the pipe, which depends on the square of the radius (or diameter). When diameter doubles from d to 2d, the area increases by factor of 4, so the drainage rate becomes 4 times faster. Therefore, time becomes 40/4 = 10 minutes. The claimed answer is correct.

Let L and S be rates of large and small pipes (pool per hour). L+S = 1/12. Half pool: 4L + 9S = 1/2. Multiplying second by 2: 8L + 18S = 1. From first: L = 1/12 - S. Substituting: 8(1/12 - S) + 18S = 1. 2/3 + 10S = 1, so 10S = 1/3 and S = 1/36. Then L = 1/12 - 1/36 = 1/18. Large pipe alone takes 18 hours.

First pipe fills at 1/10 per hour, second empties at 1/12 per hour. In 2-hour cycle, net fill = 1/10 - 1/12 = 1/60. After 108 hours (54 cycles), tank has 54/60 = 9/10 filled. In 109th hour, first pipe fills another 1/10, completing it. Total time = 108 + 1 = 109 hours. Verification: 109 hours works correctly.

Filling pipe rate: 1/8 per hour. Emptying pipe rate: 1/5 per hour. Net emptying rate when both open: 1/5 - 1/8 = 3/40 per hour. Cistern is 3/4 full, so volume to empty = 3/4. Time = (3/4) ÷ (3/40) = (3/4) × (40/3) = 10 hours. Option B is correct.