A pipe problem is a time-and-work problem in which the job is filling or emptying a tank. An inlet has a positive fill rate; an outlet or leak has a negative fill rate. The rate is expressed as a fraction of the whole tank per unit of time. Time, flow units and tank capacity must be consistent.
1. Combine inlet and outlet rates
If inlet A fills a tank in a hours, its rate is 1/a tank/hour. If outlet B empties a full tank in b hours, its rate is −1/b tank/hour. With both open, net rate is 1/a−1/b. If the result is zero, the level stays unchanged; if negative, an initially empty tank will not fill.
Worked example 1. An inlet fills in 6 hours and an outlet empties in 9. Together, their net rate is 1/6−1/9=1/18 tank/hour, so an empty tank fills in 18 hours. This is slower than the inlet alone, as it should be.
For two inlets, add rates. For an inlet and leak, subtract. If a question describes an outlet “emptying the tank in 12 hours,” that time refers to the full tank at its own steady rate, even if the tank is partly filled in the actual scenario.
2. Use LCM units for clean arithmetic
Choose a convenient tank capacity equal to the LCM of the given fill and drain times. For times 8, 12 and 24 hours, let capacity be 24 units. Their rates are 3, 2 and 1 units/hour. Add inlets and subtract drains, then divide capacity by net hourly flow. This is an arithmetic aid; the true tank may have any capacity.
Worked example 2. Two pipes fill a tank in 8 and 12 hours, and a waste pipe empties it in 24. With capacity 24 units, their rates are +3, +2 and −1 units/hour. Net rate is 4 units/hour, so filling takes 24/4=6 hours.
3. Opening and closing in stages
If a pipe is closed midway, calculate volume transferred before the change. The remaining fraction is then handled by the new net rate. A common error is to apply the final rate to the whole tank even though part was already filled.
Worked example 3. A pipe fills a tank in 10 hours. It runs alone for 3 hours, filling 3/10. A second pipe that fills in 15 hours is then opened. Their combined rate is 1/10+1/15=1/6 tank/hour. The remaining 7/10 takes (7/10)/(1/6)=4.2 hours. Total elapsed time is 7.2 hours.
4. Alternate operation and repeating cycles
When pipes work on alternate hours, use one complete cycle to compute the net fraction, then account for the remaining hour in the prescribed order. If an inlet works during odd hours and an outlet during even hours, the tank can finish during an inlet hour before a full cycle is complete. Never automatically round a fractional cycle to a whole cycle.
Worked example 4. An inlet fills 1/4 tank per hour in odd hours; an outlet drains 1/8 tank per hour in even hours. A two-hour cycle adds 1/8 tank. Six full cycles fill 3/4 in 12 hours. The inlet supplies the remaining 1/4 in hour 13, so the tank fills at the end of that hour.
5. Capacity and flow-rate questions
Some questions give litres per minute rather than fill times. Work directly with volume rates: capacity = net litres/minute × minutes. Convert hours to minutes before mixing units. If an inlet supplies 15 L/min and a drain removes 5 L/min, the net is 10 L/min. A 600 L tank fills in 60 minutes from empty.
A partially filled tank changes only the quantity remaining, not the pipe's assumed constant rate. If the tank begins one-third full, the remaining job is two-thirds of a tank. Real pipes may change flow with pressure, but standard exam problems assume constant stated rates unless specified otherwise.
6. Check the direction
With a leak open, the fill time must be longer than with the inlet alone. With an additional inlet, it must be shorter. A net rate of 1/5 tank/hour means five hours for a full tank; a rate of 5 tanks/hour means one-fifth hour. These sanity checks catch reciprocal mistakes.
Recall before practice
1. State the sign of an inlet and an outlet rate. 2. Explain why an outlet can make the fill time longer than either stated pipe time. 3. Describe the step after a pipe is shut at a given hour. 4. Explain how to handle the final incomplete alternating cycle.
Chapter 12 practice — 50 questions
Choose one option for each item. Keep a separate answer list; explanations follow this set.
1. Tap P can fill a tank in 12 hours and tap Q can empty it in 15 hours. If both taps are opened together, in how much time will the empty tank be filled?
A. 60 hours B. 6⅔ hours C. 3 hours D. 27 hours
2. Pipes P and Q together can fill a cistern in 30 hours, while pipe P alone can fill it in 50 hours. In how many hours will pipe Q alone fill the tank?
A. 20 hours B. 80 hours C. 75 hours D. 18¾ hours
3. A pipe can fill a water tank in 32 hours and another pipe can empty the full tank in 80 hours. If both pipes are opened together, in how much time will half (1/2) of the tank be filled?
A. 53⅓ hours B. 26⅔ hours C. 11 3/7 hours D. 16 hours
4. A pipe can fill a tank in 15 hours and another pipe can empty the full tank in 18 hours. If both pipes are opened together, in how much time will two-thirds (2/3) of the tank be filled?
A. 10 hours B. 5 5/11 hours C. 60 hours D. 90 hours
5. A pipe can fill a cistern in 15 hours and another pipe can empty the full tank in 90 hours. If both pipes are opened together, in how much time will half (1/2) of the tank be filled?
A. 6 3/7 hours B. 18 hours C. 7½ hours D. 9 hours
6. 3 pipes of the same size can fill a water tank in 1 hour 40 minutes. How long will 6 such pipes take to fill the same tank?
A. 50 minutes B. 3 hours 20 minutes C. 5 hours D. 1 hour 40 minutes
7. A pipe can fill a tank in 35 hours and another pipe can empty the full tank in 60 hours. If both pipes are opened together, in how much time will two-fifths (2/5) of the tank be filled?
A. 14 hours B. 8 16/19 hours C. 84 hours D. 33⅗ hours
8. 12 pipes of the same size can fill a tank in 2 hours 30 minutes. How long will 5 such pipes take to fill the same tank?
A. 2 hours 30 minutes B. 17 hours 30 minutes C. 1 hour 3 minutes D. 6 hours
9. Three pipes can fill a tank with milk, water and syrup in 10, 9 and 45 minutes respectively. If all three are opened together until the tank is full, what is the ratio of milk, water and syrup in the tank?
A. 45 : 9 : 10 B. 2 : 10 : 9 C. 10 : 9 : 45 D. 9 : 10 : 2
10. Three pipes can fill a cistern with juice, water and soda in 6, 36 and 24 minutes respectively. If all three are opened together until the tank is full, what is the ratio of juice, water and soda in the tank?
A. 3 : 2 : 12 B. 1 : 6 : 4 C. 4 : 6 : 1 D. 12 : 2 : 3
11. A tap can fill a cistern in 9 hours. Because of a leak in the bottom, the full tank empties in 12 hours. If the tap is opened with the leak present, in how much time will the empty tank be filled?
A. 36 hours B. 3 hours C. 21 hours D. 5⅐ hours
12. Tap P can fill a cistern in 18 hours and tap Q can empty it in 20 hours. If both taps are opened together, in how much time will the empty tank be filled?
A. 38 hours B. 2 hours C. 9 9/19 hours D. 180 hours
13. A pipe can fill a tank in 18 hours and another pipe can empty the full tank in 21 hours. If both pipes are opened together, in how much time will three-fifths (3/5) of the tank be filled?
A. 10⅘ hours B. 75⅗ hours C. 126 hours D. 1⅘ hours
14. Tap A can fill a water tank in 30 minutes and tap B can empty it in 60 minutes. If both taps are opened together, in how much time will the empty tank be filled?
A. 60 minutes B. 90 minutes C. 30 minutes D. 20 minutes
15. 2 pipes of the same size can fill a water tank in 2 hours 30 minutes. How long will 4 such pipes take to fill the same tank?
A. 2 hours 30 minutes B. 1 hour 45 minutes C. 1 hour 15 minutes D. 5 hours
16. Pipes A and B can fill a cistern in 45 and 35 hours respectively. All both are opened together, and after 8 hours pipe B is closed. How much more time will pipe A take to fill the remaining tank?
A. 34 5/7 hours B. 26 5/7 hours C. 45 hours D. 11 11/16 hours
17. A pipe can fill a tank in 12 hours and another pipe can empty the full tank in 40 hours. If both pipes are opened together, in how much time will three-fifths (3/5) of the tank be filled?
A. 10 2/7 hours B. 5 7/13 hours C. 17⅐ hours D. 7⅕ hours
18. Pipes P and Q can fill a water tank in 60 and 10 hours respectively. All both are opened together, and after 8 hours pipe Q is closed. How much more time will pipe P take to fill the remaining tank?
A. 4 hours B. 12 hours C. 60 hours D. 4/7 hours
19. Pipes A and B can fill a tank in 14 and 21 hours respectively. All both are opened together, and after 2 hours pipe B is closed. How much more time will pipe A take to fill the remaining tank?
A. 6⅖ hours B. 14 hours C. 12⅔ hours D. 10⅔ hours
20. A pipe can fill a cistern in 10 hours. Due to a leak in the tank, it takes 13 hours to fill it. In how many hours can the leak alone empty the full tank?
A. 43⅓ hours B. 13 hours C. 5 15/23 hours D. 3 hours
21. Two pipes P and Q can fill a tank in 36 hours and 40 hours respectively. Both are opened together, but P must be turned off after some time so that the tank is full in exactly 20 hours. After how many hours should P be turned off?
A. 18 hours B. 10 hours C. 17 7/9 hours D. 2 hours
22. Pipe A can fill a cistern in 9 hours and pipe B in 5 hours. Pipe A is opened at 8:00 a.m. and pipe B at 10:00 a.m. on the same day. At what time will the tank be full?
A. 5:00 p.m. B. 12:30 p.m. C. 11:13 a.m. D. 1:13 p.m.
23. Three pipes can fill a cistern with petrol, diesel and kerosene in 50, 6 and 30 minutes respectively. If all three are opened together until the tank is full, what is the ratio of petrol, diesel and kerosene in the tank?
A. 25 : 3 : 15 B. 5 : 25 : 3 C. 3 : 25 : 5 D. 15 : 3 : 25
24. A tap can fill a water tank in 30 hours. After half the tank is filled, 1 more tap of the same size is opened. What is the total time taken to fill the tank completely?
A. 30 hours B. 22½ hours C. 23½ hours D. 15 hours
25. A pipe can fill a tank in 18 hours and another pipe can empty the full tank in 24 hours. If both pipes are opened together, in how much time will three-fifths (3/5) of the tank be filled?
A. 43⅕ hours B. 10⅘ hours C. 72 hours D. 6 6/35 hours
26. A pipe can fill a tank in 8 hours. Due to a leak in the tank, it takes 10 hours to fill it. In how many hours can the leak alone empty the full tank?
A. 40 hours B. 10 hours C. 4 4/9 hours D. 2 hours
27. A pipe can fill a cistern in 24 hours and another pipe can empty the full tank in 32 hours. If both pipes are opened together, in how much time will half (1/2) of the tank be filled?
A. 96 hours B. 6 6/7 hours C. 48 hours D. 12 hours
28. A pipe can fill a water tank in 20 hours. Due to a leak in the tank, it takes 30 hours to fill it. In how many hours can the leak alone empty the full tank?
A. 12 hours B. 30 hours C. 10 hours D. 60 hours
29. A pipe can fill a tank in 6 hours and another pipe can empty the full tank in 60 hours. If both pipes are opened together, in how much time will two-fifths (2/5) of the tank be filled?
A. 2⅖ hours B. 2⅔ hours C. 2 2/11 hours D. 6⅔ hours
30. A pipe can fill a cistern in 4 hours and another pipe can empty the full tank in 14 hours. If both pipes are opened together, in how much time will half (1/2) of the tank be filled?
A. 2⅘ hours B. 1 5/9 hours C. 2 hours D. 5⅗ hours
31. Two pipes P and Q can fill a water tank in 6 minutes and 18 minutes respectively. Both are opened together, but P must be turned off after some time so that the tank is full in exactly 9 minutes. After how many minutes should P be turned off?
A. 4½ minutes B. 4 minutes C. 3 minutes D. 6 minutes
32. A pipe can fill a cistern in 16 hours. Due to a leak in the tank, it takes 20 hours to fill it. In how many hours can the leak alone empty the full tank?
A. 4 hours B. 20 hours C. 8 8/9 hours D. 80 hours
33. Pipes P and Q together can fill a cistern in 28 hours, while pipe P alone can fill it in 32 hours. In how many hours will pipe Q alone fill one-fourth (1/4) of the tank?
A. 3 11/15 hours B. 15 hours C. 1 hour D. 56 hours
34. A pipe can fill a tank in 14 hours and another pipe can empty the full tank in 15 hours. If both pipes are opened together, in how much time will one-third (1/3) of the tank be filled?
A. 210 hours B. 70 hours C. 2 12/29 hours D. 4⅔ hours
35. A pipe can fill a tank in 12 hours. Due to a leak in the tank, it takes 20 hours to fill it. In how many hours can the leak alone empty the full tank?
A. 20 hours B. 8 hours C. 30 hours D. 7½ hours
36. Three pipes can fill a cistern with milk, water and syrup in 5, 24 and 40 minutes respectively. If all three are opened together until the tank is full, what is the ratio of milk, water and syrup in the tank?
A. 40 : 24 : 5 B. 24 : 5 : 3 C. 5 : 24 : 40 D. 3 : 5 : 24
37. Pipes P and Q together can fill a tank in 20 hours, while pipe P alone can fill it in 36 hours. In how many hours will pipe Q alone fill two-thirds (2/3) of the tank?
A. 30 hours B. 10⅔ hours C. 37⅓ hours D. 8 4/7 hours
38. Three pipes can fill a water tank with petrol, diesel and kerosene in 15, 24 and 16 minutes respectively. If all three are opened together until the tank is full, what is the ratio of petrol, diesel and kerosene in the tank?
A. 15 : 24 : 16 B. 16 : 10 : 15 C. 16 : 24 : 15 D. 15 : 10 : 16
39. A pipe can fill a tank in 14 hours. Due to a leak in the tank, it takes 18 hours to fill it. In how many hours can the leak alone empty the full tank?
A. 18 hours B. 4 hours C. 63 hours D. 7⅞ hours
40. Pipes A and B together can fill a tank in 28 hours, while pipe A alone can fill it in 56 hours. In how many hours will pipe B alone fill three-fourths (3/4) of the tank?
A. 21 hours B. 42 hours C. 63 hours D. 14 hours
41. A leak at the bottom of a water tank can empty the full tank in 8 hours. An inlet pipe that admits 10 litres per minute is opened into the full tank, and now the tank is emptied in 16 hours. What is the capacity of the tank?
A. 19,200 litres B. 9,600 litres C. 3,200 litres D. 4,800 litres
42. Three pipes A, B and C can fill a tank. Pipes A and B together take the same time as pipe C alone. Pipe B takes 3 hours less than A and 1 hour more than C. In how many hours can pipe A alone fill the tank?
A. 3 hours B. 2 hours C. 9 hours D. 6 hours
43. Pipes A, B and C can fill a water tank in 8, 24 and 12 hours respectively. Pipe A is kept open throughout, while B and C are opened for alternate hours starting with B. In how many hours will the tank be full?
A. 8 hours B. 5⅓ hours C. 5⅖ hours D. 4 hours
44. Two pipes P and Q can fill a tank in 9 hours and 40 hours respectively. Both are opened together, but P must be turned off after some time so that the tank is full in exactly 30 hours. After how many hours should P be turned off?
A. 27¾ hours B. 3¼ hours C. 15 hours D. 2¼ hours
45. Two pipes can fill a water tank in 50 hours and 60 hours respectively. Both are opened together. After 1/4 of the tank is filled, a leak develops through which 1/4 of the water supplied by the pipes escapes. What is the total time taken to fill the tank?
A. 34 1/11 hours B. 35 1/11 hours C. 36 4/11 hours D. 27 3/11 hours
46. Two pipes A and B can fill a water tank in 35 minutes and 75 minutes respectively. Both are opened together, but A must be turned off after some time so that the tank is full in exactly 65 minutes. After how many minutes should A be turned off?
A. 4⅔ minutes B. 60⅓ minutes C. 32½ minutes D. 23 19/22 minutes
47. A leak at the bottom of a cistern can empty the full tank in 10 hours. An inlet pipe that admits 6 litres per minute is opened into the full tank, and now the tank is emptied in 13 hours. What is the capacity of the tank?
A. 15,600 litres B. 3,600 litres C. 2,035 litres D. 4,680 litres
48. A leak at the bottom of a water tank can empty the full tank in 8 hours. An inlet pipe that admits 2 litres per minute is opened into the full tank, and now the tank is emptied in 16 hours. What is the capacity of the tank?
A. 960 litres B. 1,920 litres C. 3,840 litres D. 640 litres
49. Pipes P, Q and R can fill a water tank in 9, 30 and 15 hours respectively. All three are opened together, and after 4 hours pipe Q is closed. In how many hours in all is the tank filled?
A. 5⅝ hours B. 4⅞ hours C. 14/19 hours D. ⅞ hours
50. Pipe P can fill a cistern in 6 hours and pipe Q in 4 hours. Pipe P is opened at 7:00 a.m. and pipe Q at 9:00 a.m. on the same day. At what time will the tank be full?
A. 10:36 a.m. B. 9:24 a.m. C. 11:24 a.m. D. 1:00 p.m.
Chapter 12 — Answers and explanations
After checking the key, re-solve any miss without looking at the formula.
1. A. Net rate = 1/12 − 1/15 = (15 − 12)/(12 × 15) = 1/60 of the tank per hour. Time = 60/1 = 60 hours. Formula: ab/(b − a) = 12 × 15/3. (ab/(a + b) = 6⅔ would be right only if both pipes filled.)
APAR26-12-05 | Fill and empty pipes together | Easy
2. C. Rate of Q = 1/30 − 1/50 = 5/150 − 3/150 = 2/150, so Q alone fills the tank in 150/2 = 75 hours. (Taking 50 − 30 = 20 as the answer is the common slip.)
APAR26-12-10 | Second pipe from combined time | Easy
3. B. Net part filled per hour = 1/32 − 1/80 = 3/160. Time for 1/2 of the tank = (1/2) ÷ (3/160) = 160/6 = 26⅔ hours. (The full tank would take 53⅓ hours; forgetting the fraction is the trap.)
APAR26-12-14 | Time to fill a fraction with fill and empty pipes | Easy
4. C. Net part filled per hour = 1/15 − 1/18 = 1/90. Time for 2/3 of the tank = (2/3) ÷ (1/90) = 180/3 = 60 hours. (The full tank would take 90 hours; forgetting the fraction is the trap.)
APAR26-12-03 | Time to fill a fraction with fill and empty pipes | Easy
5. D. Net part filled per hour = 1/15 − 1/90 = 5/90. Time for 1/2 of the tank = (1/2) ÷ (5/90) = 90/10 = 9 hours. (The full tank would take 18 hours; forgetting the fraction is the trap.)
APAR26-12-08 | Time to fill a fraction with fill and empty pipes | Easy
6. A. More pipes ⇒ less time (inverse variation): pipes × time = constant. 3 × 100 min = 6 × t ⇒ t = 3 × 100/6 = 50 minutes = 50 minutes. (Multiplying the other way, 6 × 100/3 = 200 min, is the usual error.)
APAR26-12-09 | Number of pipes — inverse variation | Easy
7. D. Net part filled per hour = 1/35 − 1/60 = 5/420. Time for 2/5 of the tank = (2/5) ÷ (5/420) = 840/25 = 33⅗ hours. (The full tank would take 84 hours; forgetting the fraction is the trap.)
APAR26-12-01 | Time to fill a fraction with fill and empty pipes | Easy
8. D. More pipes ⇒ less time (inverse variation): pipes × time = constant. 12 × 150 min = 5 × t ⇒ t = 12 × 150/5 = 360 minutes = 6 hours. (Multiplying the other way, 5 × 150/12 = 62.5 min, is the usual error.)
APAR26-12-02 | Number of pipes — inverse variation | Easy
9. D. Each liquid's share is proportional to its pipe's rate, i.e. 1/10 : 1/9 : 1/45. Multiply by LCM(10, 9, 45) = 90: 9 : 10 : 2 = 9 : 10 : 2. (Writing the times 10 : 9 : 45 as the ratio is the trap — the faster pipe puts in more, not less.)
APAR26-12-15 | Ratio of liquids from three pipes | Easy
10. D. Each liquid's share is proportional to its pipe's rate, i.e. 1/6 : 1/36 : 1/24. Multiply by LCM(6, 36, 24) = 72: 12 : 2 : 3 = 12 : 2 : 3. (Writing the times 1 : 6 : 4 as the ratio is the trap — the faster pipe puts in more, not less.)
APAR26-12-12 | Ratio of liquids from three pipes | Easy
11. A. In 1 hour: tap fills 1/9, leak empties 1/12; net = 1/9 − 1/12 = 1/36. Time to fill = 36/1 = 36 hours (formula ab/(b − a) = 9 × 12/3). Adding the rates gives 5⅐ h, which is wrong because the leak works against the tap.
APAR26-12-11 | Tap fills, leak empties | Easy
12. D. Net rate = 1/18 − 1/20 = (20 − 18)/(18 × 20) = 1/180 of the tank per hour. Time = 180/1 = 180 hours. Formula: ab/(b − a) = 18 × 20/2. (ab/(a + b) = 9 9/19 would be right only if both pipes filled.)
APAR26-12-07 | Fill and empty pipes together | Easy
13. B. Net part filled per hour = 1/18 − 1/21 = 1/126. Time for 3/5 of the tank = (3/5) ÷ (1/126) = 378/5 = 75⅗ hours. (The full tank would take 126 hours; forgetting the fraction is the trap.)
APAR26-12-13 | Time to fill a fraction with fill and empty pipes | Easy
14. A. Net rate = 1/30 − 1/60 = (60 − 30)/(30 × 60) = 1/60 of the tank per minute. Time = 60/1 = 60 minutes. Formula: ab/(b − a) = 30 × 60/30. (ab/(a + b) = 20 would be right only if both pipes filled.)
APAR26-12-06 | Fill and empty pipes together | Easy
15. C. More pipes ⇒ less time (inverse variation): pipes × time = constant. 2 × 150 min = 4 × t ⇒ t = 2 × 150/4 = 75 minutes = 1 hour 15 minutes. (Multiplying the other way, 4 × 150/2 = 300 min, is the usual error.)
APAR26-12-04 | Number of pipes — inverse variation | Easy
16. B. Total capacity = LCM(45, 35) = 315 units; rates 7, 9 units/h, together 16 units/h. In 8 h: 8 × 16 = 128 units; remaining 187 units. After B closes, rate = 7 units/h, so remaining time = 187/7 = 26 5/7 h.
APAR26-12-28 | One pipe closed midway — remaining time | Medium
17. A. Net part filled per hour = 1/12 − 1/40 = 7/120. Time for 3/5 of the tank = (3/5) ÷ (7/120) = 360/35 = 10 2/7 hours. (The full tank would take 17⅐ hours; forgetting the fraction is the trap.)
APAR26-12-36 | Time to fill a fraction with fill and empty pipes | Medium
18. A. Total capacity = LCM(60, 10) = 60 units; rates 1, 6 units/h, together 7 units/h. In 8 h: 8 × 7 = 56 units; remaining 4 units. After Q closes, rate = 1 units/h, so remaining time = 4/1 = 4 h.
APAR26-12-24 | One pipe closed midway — remaining time | Medium
19. D. Total capacity = LCM(14, 21) = 42 units; rates 3, 2 units/h, together 5 units/h. In 2 h: 2 × 5 = 10 units; remaining 32 units. After B closes, rate = 3 units/h, so remaining time = 32/3 = 10⅔ h.
APAR26-12-16 | One pipe closed midway — remaining time | Medium
20. A. Leak's rate = (pipe alone) − (pipe with leak) = 1/10 − 1/13 = 13/130 − 10/130 = 3/130. So the leak empties the tank in 130/3 = 43⅓ hours. Formula: TT′/(T′ − T) = 10 × 13/3. (The extra 3 hours is not the answer.)
APAR26-12-38 | Leak emptying time from extra fill time | Medium
21. A. Capacity = LCM(36, 40) = 360 units; P gives 10 and Q gives 9 units per hour. Q runs all 20 hours: 9 × 20 = 180 units. P must supply the rest, 360 − 180 = 180 units, in 180/10 = 18 hours. (Equation: x/36 + 20/40 = 1.)
APAR26-12-19 | When to shut a pipe to finish in a given time | Medium
22. B. Capacity = LCM(9, 5) = 45 units; A gives 5, B gives 9 units/h. From 8:00 a.m. to 10:00 a.m. only A runs: 5 × 2 = 10 units; 35 units remain. Together at 14 units/h they take 2½ h = 2 hours 30 minutes. So the tank is full at 10:00 a.m. + 2 hours 30 minutes = 12:30 p.m.
APAR26-12-20 | Clock time when the tank is full | Medium
23. C. Each liquid's share is proportional to its pipe's rate, i.e. 1/50 : 1/6 : 1/30. Multiply by LCM(50, 6, 30) = 150: 3 : 25 : 5 = 3 : 25 : 5. (Writing the times 25 : 3 : 15 as the ratio is the trap — the faster pipe puts in more, not less.)
APAR26-12-31 | Ratio of liquids from three pipes | Medium
24. B. First half: 30/2 = 15 hours with one tap. Second half: 2 taps together fill 2× as fast, so time = (30/2) ÷ 2 = 7½ hours. Total = 15 + 7½ = 22½ hours. (Dividing the whole 30 h by 2 ignores that the extra taps start only at the halfway mark.)
APAR26-12-30 | Extra identical taps after half the tank | Medium
25. A. Net part filled per hour = 1/18 − 1/24 = 1/72. Time for 3/5 of the tank = (3/5) ÷ (1/72) = 216/5 = 43⅕ hours. (The full tank would take 72 hours; forgetting the fraction is the trap.)
APAR26-12-17 | Time to fill a fraction with fill and empty pipes | Medium
26. A. Leak's rate = (pipe alone) − (pipe with leak) = 1/8 − 1/10 = 5/40 − 4/40 = 1/40. So the leak empties the tank in 40/1 = 40 hours. Formula: TT′/(T′ − T) = 8 × 10/2. (The extra 2 hours is not the answer.)
APAR26-12-29 | Leak emptying time from extra fill time | Medium
27. C. Net part filled per hour = 1/24 − 1/32 = 1/96. Time for 1/2 of the tank = (1/2) ÷ (1/96) = 96/2 = 48 hours. (The full tank would take 96 hours; forgetting the fraction is the trap.)
APAR26-12-33 | Time to fill a fraction with fill and empty pipes | Medium
28. D. Leak's rate = (pipe alone) − (pipe with leak) = 1/20 − 1/30 = 3/60 − 2/60 = 1/60. So the leak empties the tank in 60/1 = 60 hours. Formula: TT′/(T′ − T) = 20 × 30/10. (The extra 10 hours is not the answer.)
APAR26-12-23 | Leak emptying time from extra fill time | Medium
29. B. Net part filled per hour = 1/6 − 1/60 = 9/60. Time for 2/5 of the tank = (2/5) ÷ (9/60) = 120/45 = 2⅔ hours. (The full tank would take 6⅔ hours; forgetting the fraction is the trap.)
APAR26-12-40 | Time to fill a fraction with fill and empty pipes | Medium
30. A. Net part filled per hour = 1/4 − 1/14 = 5/28. Time for 1/2 of the tank = (1/2) ÷ (5/28) = 28/10 = 2⅘ hours. (The full tank would take 5⅗ hours; forgetting the fraction is the trap.)
APAR26-12-26 | Time to fill a fraction with fill and empty pipes | Medium
31. C. Capacity = LCM(6, 18) = 18 units; P gives 3 and Q gives 1 units per minute. Q runs all 9 minutes: 1 × 9 = 9 units. P must supply the rest, 18 − 9 = 9 units, in 9/3 = 3 minutes. (Equation: x/6 + 9/18 = 1.)
APAR26-12-39 | When to shut a pipe to finish in a given time | Medium
32. D. Leak's rate = (pipe alone) − (pipe with leak) = 1/16 − 1/20 = 5/80 − 4/80 = 1/80. So the leak empties the tank in 80/1 = 80 hours. Formula: TT′/(T′ − T) = 16 × 20/4. (The extra 4 hours is not the answer.)
APAR26-12-32 | Leak emptying time from extra fill time | Medium
33. D. Rate of Q = 1/28 − 1/32 = 8/224 − 7/224 = 1/224, so Q alone fills the tank in 224/1 = 224 hours. For 1/4 of the tank: 224 × 1/4 = 56 hours. (Taking 32 − 28 = 4 as the answer is the common slip.)
APAR26-12-37 | Second pipe from combined time | Medium
34. B. Net part filled per hour = 1/14 − 1/15 = 1/210. Time for 1/3 of the tank = (1/3) ÷ (1/210) = 210/3 = 70 hours. (The full tank would take 210 hours; forgetting the fraction is the trap.)
APAR26-12-34 | Time to fill a fraction with fill and empty pipes | Medium
35. C. Leak's rate = (pipe alone) − (pipe with leak) = 1/12 − 1/20 = 5/60 − 3/60 = 2/60. So the leak empties the tank in 60/2 = 30 hours. Formula: TT′/(T′ − T) = 12 × 20/8. (The extra 8 hours is not the answer.)
APAR26-12-25 | Leak emptying time from extra fill time | Medium
36. B. Each liquid's share is proportional to its pipe's rate, i.e. 1/5 : 1/24 : 1/40. Multiply by LCM(5, 24, 40) = 120: 24 : 5 : 3 = 24 : 5 : 3. (Writing the times 5 : 24 : 40 as the ratio is the trap — the faster pipe puts in more, not less.)
APAR26-12-22 | Ratio of liquids from three pipes | Medium
37. A. Rate of Q = 1/20 − 1/36 = 9/180 − 5/180 = 4/180, so Q alone fills the tank in 180/4 = 45 hours. For 2/3 of the tank: 45 × 2/3 = 30 hours. (Taking 36 − 20 = 16 as the answer is the common slip.)
APAR26-12-21 | Second pipe from combined time | Medium
38. B. Each liquid's share is proportional to its pipe's rate, i.e. 1/15 : 1/24 : 1/16. Multiply by LCM(15, 24, 16) = 240: 16 : 10 : 15 = 16 : 10 : 15. (Writing the times 15 : 24 : 16 as the ratio is the trap — the faster pipe puts in more, not less.)
APAR26-12-18 | Ratio of liquids from three pipes | Medium
39. C. Leak's rate = (pipe alone) − (pipe with leak) = 1/14 − 1/18 = 9/126 − 7/126 = 2/126. So the leak empties the tank in 126/2 = 63 hours. Formula: TT′/(T′ − T) = 14 × 18/4. (The extra 4 hours is not the answer.)
APAR26-12-35 | Leak emptying time from extra fill time | Medium
40. B. Rate of B = 1/28 − 1/56 = 2/56 − 1/56 = 1/56, so B alone fills the tank in 56/1 = 56 hours. For 3/4 of the tank: 56 × 3/4 = 42 hours. (Taking 56 − 28 = 28 as the answer is the common slip.)
APAR26-12-27 | Second pipe from combined time | Medium
41. B. Inlet's rate = (leak alone) − (leak with inlet) = 1/8 − 1/16 = 1/16 of the tank per hour, so the inlet alone would fill the tank in 16/1 = 16 hours. Inlet supplies 10 × 60 = 600 litres/hour, hence capacity = 600 × 16/1 = 9,600 litres. (Using 16 × 600 = 9,600 ignores the leak.)
APAR26-12-42 | Tank capacity from leak and inlet | Difficult
42. D. Let A take x hours; then B takes (x − 3) and C takes (x − 4). Given 1/x + 1/(x − 3) = 1/(x − 4). Cross-multiplying: (2x − 3)(x − 4) = x(x − 3) ⇒ x² − 8x + 12 = 0 ⇒ (x − 6)(x − 2) = 0. The root x = 6 makes all times positive (6, 3, 2 h; check: 1/6 + 1/3 = 1/2). Answer: 6 hours.
APAR26-12-48 | Pipe times related by differences | Difficult
43. C. Capacity = LCM(8, 24, 12) = 24 units; rates 3, 1, 2 units/h. A 2-hour cycle fills (3 + 1) + (3 + 2) = 9 units. 2 cycles (4 h) fill 18 units. 6 units left: hour 5 (A+B) fills 4, then 2 units at 5 units/h take ⅖ h. Total = 5⅖ hours.
APAR26-12-45 | One pipe constant, two alternating hourly | Difficult
44. D. Capacity = LCM(9, 40) = 360 units; P gives 40 and Q gives 9 units per hour. Q runs all 30 hours: 9 × 30 = 270 units. P must supply the rest, 360 − 270 = 90 units, in 90/40 = 2¼ hours. (Equation: x/9 + 30/40 = 1.)
APAR26-12-47 | When to shut a pipe to finish in a given time | Difficult
45. A. Capacity = LCM(50, 60) = 300 units; inflow = 6 + 5 = 11 units/h. First 1/4 (75 units): 6 9/11 h. After the leak only 3/4 of the inflow stays, i.e. effective rate = 11 × 3/4 = 8.25 units/h, so the remaining 225 units take 27 3/11 h. Total = 6 9/11 + 27 3/11 = 34 1/11 hours.
APAR26-12-43 | Leak develops after a partial fill | Difficult
46. A. Capacity = LCM(35, 75) = 525 units; A gives 15 and B gives 7 units per minute. B runs all 65 minutes: 7 × 65 = 455 units. A must supply the rest, 525 − 455 = 70 units, in 70/15 = 4⅔ minutes. (Equation: x/35 + 65/75 = 1.)
APAR26-12-41 | When to shut a pipe to finish in a given time | Difficult
47. A. Inlet's rate = (leak alone) − (leak with inlet) = 1/10 − 1/13 = 3/130 of the tank per hour, so the inlet alone would fill the tank in 130/3 = 43⅓ hours. Inlet supplies 6 × 60 = 360 litres/hour, hence capacity = 360 × 130/3 = 15,600 litres. (Using 13 × 360 = 4,680 ignores the leak.)
APAR26-12-49 | Tank capacity from leak and inlet | Difficult
48. B. Inlet's rate = (leak alone) − (leak with inlet) = 1/8 − 1/16 = 1/16 of the tank per hour, so the inlet alone would fill the tank in 16/1 = 16 hours. Inlet supplies 2 × 60 = 120 litres/hour, hence capacity = 120 × 16/1 = 1,920 litres. (Using 16 × 120 = 1,920 ignores the leak.)
APAR26-12-44 | Tank capacity from leak and inlet | Difficult
49. B. Total capacity = LCM(9, 30, 15) = 90 units; rates 10, 3, 6 units/h, together 19 units/h. In 4 h: 4 × 19 = 76 units; remaining 14 units. After Q closes, rate = 16 units/h, so remaining time = 14/16 = ⅞ h. Total = 4 + ⅞ = 4⅞ hours.
APAR26-12-46 | One pipe closed midway — total time | Difficult
50. A. Capacity = LCM(6, 4) = 12 units; P gives 2, Q gives 3 units/h. From 7:00 a.m. to 9:00 a.m. only P runs: 2 × 2 = 4 units; 8 units remain. Together at 5 units/h they take 1⅗ h = 1 hour 36 minutes. So the tank is full at 9:00 a.m. + 1 hour 36 minutes = 10:36 a.m.
APAR26-12-50 | Clock time when the tank is full | Difficult