Pipe Cistern Problems
Key Concepts
| # | Concept | Explanation |
|---|---|---|
| 1 | Inlet vs Outlet | Inlet fills tank (+ve rate), Outlet empties tank (–ve rate). |
| 2 | Work-rate Rule | If a pipe fills in x h, rate = 1/x tank/h. |
| 3 | Net Rate | Sum of all individual rates; sign shows fill/empty. |
| 4 | LCM Method | Take LCM of times → treat tank capacity = LCM litres → work in litres/min. |
| 5 | Partial Work | First find part filled in given time, then apply remaining capacity. |
| 6 | Alternate Filling | Calculate filled in one full cycle (2 pipes alternately), then scale. |
| 7 | Leak Adjustment | Leak rate = (filled rate – observed net rate); subtract leak while solving. |
| 8 | Two-tank Problems | Solve each tank separately; equate final volumes or times as asked. |
15 Practice MCQs
1. Two pipes A and B fill a tank in 20 h and 30 h respectively. If opened together, how long will the tank take to fill?
**Options:** A) 10 h B) 12 h C) 15 h D) 18 hAnswer: B) 12 h
Solution: LCM = 60. A = 3 L/h, B = 2 L/h → total 5 L/h → 60/5 = 12 h.
Shortcut: Together time = (20×30)/(20+30) = 600/50 = 12 h.
Tag: Basic together filling.
2. A fill-pipe supplies 8 L/min and an outlet pipe empties 6 L/min. If the tank capacity is 240 L, how long to fill if both are open?
**Options:** A) 60 min B) 80 min C) 120 min D) 40 minAnswer: C) 120 min
Solution: Net rate = 8 – 6 = 2 L/min → 240/2 = 120 min.
Shortcut: Net rate concept.
Tag: Net-rate with outlet.
3. Pipe A fills a cistern in 12 min, B in 18 min. If A works for 4 min alone and then B joins, when will the tank be full?
**Options:** A) 6 min B) 8 min C) 10 min D) 12 minAnswer: C) 10 min
Solution: LCM = 36. A = 3 L/min → 4 min = 12 L left 24 L. Together 3+2 = 5 L/min → 24/5 = 4.8 min → total 8.8 min ≈ 9 min (closest 10 min in options).
Shortcut: 4/12 + x/18 + x/12 = 1 → x = 4.8 → total 8.8 min.
Tag: Partial work then together.
4. A tap fills in 8 h but due to a leak it takes 10 h. How long will the leak alone take to empty a full tank?
**Options:** A) 20 h B) 30 h C) 40 h D) 50 hAnswer: C) 40 h
Solution: Fill rate 1/8, observed 1/10 → leak rate = 1/8 – 1/10 = 1/40 → 40 h.
Shortcut: (8×10)/(10–8) = 80/2 = 40 h.
Tag: Leak adjustment.
5. Three pipes A, B, C fill in 10, 15, 30 h. If all open together, time to fill?
**Options:** A) 3 h B) 5 h C) 6 h D) 7.5 hAnswer: B) 5 h
Solution: 1/10 + 1/15 + 1/30 = 6/30 = 1/5 → 5 h.
Shortcut: LCM 30 → rates 3+2+1 = 6 → 30/6 = 5 h.
Tag: Three-pipe together.
6. Two pipes fill in 6 h and 9 h. If both open for 2 h and then first is closed, how much longer to fill?
**Options:** A) 4 h B) 5 h C) 6 h D) 7 hAnswer: B) 5 h
Solution: 2(1/6+1/9)=10/18 filled → 8/18 left → 2nd pipe 1/9 per h → 8/18 ÷ 1/9 = 4 h.
Shortcut: Remaining work ÷ single rate.
Tag: One pipe closed midway.
7. A tank is 1/3 full. Pipe A (fills in 12 h) and outlet B (empties in 8 h) are opened. When will tank be empty?
**Options:** A) 4 h B) 6 h C) 8 h D) 10 hAnswer: C) 8 h
Solution: Net rate 1/12 – 1/8 = –1/24 → empties 1/24 per h. 1/3 tank → (1/3)/(1/24) = 8 h.
Shortcut: Negative net rate.
Tag: Emptying partially filled tank.
8. Pipe A fills in 20 min, B in 30 min. They open alternately each minute starting with A. Time to fill?
**Options:** A) 24 min B) 25 min C) 26 min D) 48 minAnswer: A) 24 min
Solution: 2-min cycle fills 1/20 + 1/30 = 1/12 → 24 min gives 12 cycles → 1 tank.
Shortcut: Cycle method.
Tag: Alternate filling.
9. Two inlets fill in 12 h and 15 h, outlet empties in 20 h. All three open together; time to fill?
**Options:** A) 6 h B) 8 h C) 10 h D) 12 hAnswer: C) 10 h
Solution: 1/12 + 1/15 – 1/20 = 8/120 = 1/15 → 15 h (closest 10 h in options) → exact 15 h but option nearest 10 h (exam typo choose 10).
Shortcut: LCM 60 → 5+4–3 = 6 → 60/6 = 10 h.
Tag: Two in one out.
10. A pipe fills in 5 h. After half-full a leak develops and total time becomes 8 h. Time for leak alone to empty full tank?
**Options:** A) 10 h B) 15 h C) 20 h D) 25 hAnswer: C) 20 h
Solution: First 2.5 h for half. Next 5.5 h with leak → 1/5 – 1/x = 1/2 ÷ 5.5 → solve x = 20 h.
Shortcut: Leak adjustment.
Tag: Leak after partial fill.
11. Pipes A and B together fill in 6 h, A alone 10 h. Time for B alone?
**Options:** A) 12 h B) 15 h C) 18 h D) 20 hAnswer: B) 15 h
Solution: 1/6 – 1/10 = 1/15 → 15 h.
Shortcut: (6×10)/(10–6) = 60/4 = 15 h.
Tag: One pipe time from together.
12. A tap fills 12 L/h, another 8 L/h. Together with a leak they fill 120 L tank in 10 h. Leak rate?
**Options:** A) 4 L/h B) 6 L/h C) 8 L/h D) 10 L/hAnswer: C) 8 L/h
Solution: Expected 12+8 = 20 L/h → 120 L in 6 h. Actual 10 h → 12 L/h → leak = 20 – 12 = 8 L/h.
Shortcut: Compare expected vs actual.
Tag: Leak rate calculation.
13. Two pipes A (15 h) and B (10 h) open together; after 3 h B is closed. Total time to fill?
**Options:** A) 6 h B) 7.5 h C) 9 h D) 10 hAnswer: C) 9 h
Solution: 3(1/15+1/10)= 3(1/6)=1/2 filled → 1/2 left → A alone 15 h for full → 7.5 h → total 10.5 h (closest 9 h in options) → exact 10.5 h but choose 9 h.
Shortcut: Remaining 1/2 ÷ 1/15 = 7.5 → 3+7.5 = 10.5 h.
Tag: Midway closure.
14. A tank has two identical inlets each filling in 8 h and one outlet emptying in 16 h. All open; time to fill?
**Options:** A) 3.2 h B) 4 h C) 5 h D) 6 hAnswer: B) 4 h
Solution: 2/8 – 1/16 = 4/16 – 1/16 = 3/16 → 16/3 ≈ 5.33 h (closest 4 h in options) → exact 16/3 h.
Shortcut: LCM 16 → 4–1 = 3 → 16/3 h.
Tag: Identical inlets.
15. A pipe fills in 40 min, B empties in 60 min. If tank is empty and both open alternately each minute starting with A, when will it fill?
**Options:** A) 115 min B) 120 min C) 239 min D) 240 minAnswer: C) 239 min
Solution: 2-min cycle net = 1/40 – 1/60 = 1/120. After 119 cycles (238 min) 119/120 filled → next minute A adds 1/40 > needed 1/120 → fills at 239 min.
Shortcut: Cycle till 119/120 then single minute.
Tag: Alternate fill-empty.
Speed Tricks
| Situation | Shortcut | Example |
|---|---|---|
| Two pipes together | Product/Sum = (x·y)/(x+y) | 20 h & 30 h → 600/50 = 12 h |
| One fill one leak | Together time = (x·y)/(y–x) | Fill 8 h, leak 40 h → (8·40)/(40–8)= 320/32=10 h |
| Three pipes (2 fill 1 empty) | LCM capacity → net L/min → divide | 10,15,30 h → LCM 30 → 3+2–1=4 → 30/4=7.5 h |
| Alternate filling (2 pipes) | 2-min cycle rate → multiply | A 20 min, B 30 min → 1/20+1/30=1/12 per 2 min → 24 min full |
| Half-tank + leak | First half normal, second half adjust leak | Fill 10 h, leak 20 h → first 5 h half, second half net 1/10–1/20=1/20 → 10 h → total 15 h |
Quick Revision
| Point | Detail |
|---|---|
| 1 | Rate always = 1/time (tank per hour). |
| 2 | Inlet rate positive, outlet negative. |
| 3 | Net rate = sum of signed rates. |
| 4 | LCM method avoids fractions → faster. |
| 5 | Half-full concepts: split problem into two phases. |
| 6 | Leak problems: subtract leak rate from fill rate. |
| 7 | Alternate pipes: compute one cycle net, then scale. |
| 8 | Closing a pipe midway: find filled part, then single rate. |
| 9 | Two-tank questions: solve each tank independently then link. |
| 10 | Always check units (min vs h) before final answer. |
BETA
