Pipes AND Cisterns FOR Placement

Meta Description: Master Pipes & Cisterns for 2026 placement drives. Learn formulas, leak tricks & 23 solved TCS, Infosys, Wipro questions to ace your quantitative aptitude.

Introduction

Pipes and Cisterns is a high-yield quantitative aptitude topic that appears consistently in campus placement drives across IT, consulting, and core engineering companies. It is essentially an applied extension of the Time & Work concept, but with a fluid/tank context that tests your ability to handle positive (filling) and negative (emptying) rates simultaneously. In 2026 placement cycles, candidates can expect 1 to 3 questions per exam, typically carrying a weightage of 2-4% of the quantitative section. Major recruiters like TCS NQT, Infosys, Wipro, Accenture, and Cognizant frequently include this topic to evaluate logical reasoning, rate conversion skills, and accuracy under time pressure. Whether it’s calculating combined filling times, determining leak drainage rates, or solving partial-work scenarios, mastering Pipes and Cisterns gives you a decisive edge in clearing the aptitude cutoff. This guide by PapersAdda covers every sub-topic, provides 23 fully solved problems, and equips you with company-level shortcuts for 2026.

Key Formulas & Concepts

Understanding the rate-based approach is crucial. Treat the cistern/tank capacity as 1 unit or LCM of given times for faster computation.

1. Filling Pipe Rate:
If a pipe fills a tank in x hours, its filling rate = 1/x tank per hour.

2. Emptying Pipe Rate:
If a pipe empties a tank in y hours, its emptying rate = 1/y tank per hour (negative work).

3. Combined Work Rate:
When multiple pipes operate together:
Net Rate = (Sum of filling rates) − (Sum of emptying rates)
Time to fill/empty = 1 / |Net Rate|

4. Leak Impact Formula:
If a pipe fills a tank in x hours alone, but takes y hours due to a leak (y > x):
Leak Rate = 1/x − 1/y
Time taken by leak alone to empty full tank = 1 / Leak Rate = (x × y) / (y − x)

5. Partial Work & Time Relation:
Work Done = Rate × Time
Remaining Work = 1 − Work Done
Time to finish remaining work = Remaining Work / Active Rate

6. Alternating Pipes:
When pipes open alternately (e.g., A for 1 hr, B for 1 hr), calculate work in 2-hour cycles, find full cycles, then handle the remaining fraction.

Solved Examples (Basic Level)

1. A pipe fills a tank in 8 hours. What fraction of the tank is filled in 1.5 hours?
Solution: Rate = 1/8 per hour. In 1.5 hrs = 3/2 hrs, Work = (1/8) × (3/2) = 3/16.

2. Pipe A fills a cistern in 12 hours and Pipe B in 15 hours. How long will they take if opened together?
Solution: Combined rate = 1/12 + 1/15 = (5+4)/60 = 9/60 = 3/20 per hour. Time = 1 ÷ (3/20) = 20/3 = 6 hrs 40 mins.

3. A tap fills a tank in 10 hours, but a leak empties it in 20 hours. If both are open, how long to fill?
Solution: Net rate = 1/10 − 1/20 = (2−1)/20 = 1/20 per hour. Time = 20 hours.

4. Pipe P fills a tank in 6 hours. Due to a leak, it takes 9 hours. How long will the leak take to empty the full tank?
Solution: Leak rate = 1/6 − 1/9 = (3−2)/18 = 1/18. Time for leak alone = 18 hours.

5. A cistern is 1/4 full. Pipe X fills it in 5 hours. How much more time to fill completely?
Solution: Remaining work = 1 − 1/4 = 3/4. Rate = 1/5. Time = (3/4) ÷ (1/5) = 15/4 = 3.75 hours = 3 hrs 45 mins.

Practice Questions (Medium Level)

6. Two pipes A (10h) and B (15h) fill a tank. A third pipe C (20h) empties it. If all open together, find fill time.
Solution: Net rate = 1/10 + 1/15 − 1/20 = (6+4−3)/60 = 7/60. Time = 60/7 ≈ 8.57 hours.

7. Pipe X fills in 12h. Pipe Y fills in 16h. Both open for 4h, then X is closed. How long more for Y to finish?
Solution: Combined 4h work = 4×(1/12+1/16) = 4×(7/48) = 7/12. Remaining = 5/12. Y’s rate = 1/16. Time = (5/12)÷(1/16) = 20/3 = 6.67 hours.

8. A cistern has a leak that drains 50 liters/hr. A pipe fills it in 4 hours, but with leak takes 6 hours. Find capacity.
Solution: Leak empties in T = 1/(1/4−1/6) = 12 hours. Capacity = 12 × 50 = 600 liters.

9. Pipes A(8h), B(10h), C(12h) open together for 2 hours. A is closed. How much more time to fill?
Solution: 2h work = 2×(1/8+1/10+1/12) = 2×(44/120) = 11/15. Remaining = 4/15. B+C rate = 1/10+1/12 = 11/60. Time = (4/15)÷(11/60) = 16/11 ≈ 1.45 hours.

10. Pipe A is twice as fast as B. Together they fill in 18 hours. Find individual times.
Solution: Let B = x hrs, A = x/2. 1/x + 2/x = 3/x = 1/18 → x = 54. B = 54h, A = 27h.

11. A tap fills 1/3 tank in 2 hours. Another empties 1/4 tank in 1.5 hours. Both open, find net fill time.
Solution: Fill rate = (1/3)/2 = 1/6. Empty rate = (1/4)/1.5 = 1/6. Net rate = 0. Tank never fills (or stays constant).

12. Three pipes: A(15h), B(20h), C(30h). A opens at 8 AM, B at 9 AM, C at 10 AM. When is it full?
Solution: 8-9 AM: A does 1/15. 9-10 AM: A+B do 1/15+1/20=7/60. Total by 10 AM = 4/60+7/60=11/60. Remaining=49/60. All 3 rate = 1/15+1/20+1/30=13/60. Time=(49/60)/(13/60)=49/13≈3.77h=3h46m. Full at 1:46 PM.

13. Pipe P fills 40L/min. Pipe Q empties 30L/min. Tank capacity 2400L. Both open, time to fill?
Solution: Net = 10 L/min. Time = 2400/10 = 240 mins = 4 hours.

Advanced Questions

14. A cistern has two filling pipes (A: 12h, B: 18h) and one leak. With leak, it fills in 10h. Find leak’s emptying time.
Solution: Combined fill = 1/12+1/18=5/36. Net with leak = 1/10. Leak rate = 5/36−1/10= (25−18)/180=7/180. Time = 180/7 ≈ 25.71h.

15. Pipe A fills in 10h. After 4h, Pipe B (emptying, 15h) opens. How long more to fill?
Solution: Work done in 4h = 4/10=2/5. Remaining=3/5. Net rate=1/10−1/15=1/30. Time=(3/5)/(1/30)=18h.

16. Two pipes fill in 20h and 30h. Both open, but at 1/4 filled, a leak starts draining at 1/6 tank/h. Find total time.
Solution: Time to 1/4: Both rate=1/12. Time=(1/4)/(1/12)=3h. Remaining=3/4. New net=1/12−1/6=−1/12. Tank drains! Never fills. (Trick: leak > combined fill)

17. Pipe X fills 1/5 tank in 3h. Pipe Y empties 1/3 in 2h. Both open for 1h, then Y closed. Find remaining time for X.
Solution: X rate=1/15, Y rate=−1/6. 1h net=1/15−1/6=−1/10. Tank loses 1/10. Start=0, now=−1/10. X must fill 1+1/10=11/10. Time=(11/10)/(1/15)=16.5h.

18. A tank has inlet A(12h), inlet B(16h). Outlet C empties full tank in 24h. If all open, how long to fill 3/4 tank?
Solution: Net=1/12+1/16−1/24=(4+3−2)/48=5/48. Time for 3/4 = (3/4)/(5/48)=36/5=7.2h.

Common Mistakes to Avoid

• Ignoring sign conventions: Emptying pipes must be subtracted, not added. Always assign negative rates to outlets/leaks.
• Mixing rates and times: Never add hours directly (e.g., 10h + 15h ≠ 25h). Convert to fractions/LCM first.
• Forgetting leak direction: A leak always increases filling time. If your calculated time decreases, you’ve added instead of subtracted.
• Assuming linear work in alternating pipes: Alternating pipes create cycles. Calculate full cycles first, then handle the remainder separately.
• Confusing capacity with rate: If liters/hour are given, find capacity first before switching to unit rates. Keep units consistent throughout.
• Rounding too early: Fractions preserve accuracy. Convert to decimals/hours-minutes only in the final step.

Shortcut Tricks

1. LCM Efficiency Method: Take LCM of given hours as tank capacity. Convert each pipe’s rate to liters/hour. Add/subtract directly.
Example: A(6h), B(8h), C(12h empty). LCM=24L. A=4L/h, B=3L/h, C=−2L/h. Net=5L/h → 24/5=4.8h.

2. Direct Leak Formula: Leak Time = (Normal Time × Delayed Time) / (Delayed Time − Normal Time)
Example: 8h alone, 12h with leak → (8×12)/(12−8)=24h emptying time.

3. Two-Pipe Combined Shortcut: Time together = (A × B) / (A + B)
Example: 12h & 15h → (12×15)/(12+15)=180/27=20/3h. Works only for two filling pipes.

4. Partial Work Multiplier: If W work takes T time, remaining (1−W) takes T × (1−W)/W.
Example: 2/5 filled in 4h → Remaining 3/5 takes 4 × (3/5)/(2/5) = 6h.

Previous Year Questions

1. TCS NQT 2024: Pipe A fills in 20 min, B in 30 min. Both open. After 5 min, A closes. Total time?
Solution: LCM=60 units. A=3u/m, B=2u/m. 5 min work=5×(5)=25u. Remaining=35u. B alone=2u/m → 35/2=17.5 min. Total=22.5 min.

2. Infosys Campus 2023: A cistern fills in 12h. A leak at bottom empties it in 24h. Time to fill?
Solution: Net=1/12−1/24=1/24. Time=24 hours.

3. Wipro Elite 2024: Two taps fill in 10h & 15h. A leak drains 6L/h. With leak, fills in 18h. Find capacity.
Solution: Fill rate=1/10+1/15=1/6. Net with leak=1/18. Leak=1/6−1/18=1/9. 1/9 capacity=6L → Capacity=54L.

4. Accenture APAC 2025: Pipe P fills 1/3 in 4h. Pipe Q empties 1/2 in 5h. Both open 2h, then Q closed. Remaining time?
Solution: P=1/12, Q=−1/10. 2h net=2×(5−6)/60=−1/30. Tank loses 1/30. Need to fill 1+1/30=31/30. P alone: (31/30)/(1/12)=12.4h.

5. Cognizant GenC 2023: A(8h), B(12h), C(24h empty). Open together for 2h, then C closed. Total time?
Solution: Net=1/8+1/12−1/24= (3+2−1)/24=4/24=1/6. 2h work=1/3. Remaining=2/3. A+B=1/8+1/12=5/24. Time=(2/3)/(5/24)=16/5=3.2h. Total=5.2h.

Quick Revision

• Filling rate = 1/time, Emptying rate = −1/time
• Net rate = Σfill − Σempty
• Time = 1 / Net Rate
• Leak Time = (x·y)/(y−x) where x=normal, y=delayed
• Use LCM method for whole-number efficiency
• Negative net rate → tank never fills
• Convert partial work → remaining fraction → divide by active rate
• Always verify units (hours/minutes/liters) before computing
• TCS, Infosys, Wipro, Accenture prefer combined + leak + partial scenarios
• Practice 15-20 mixed questions before 2026 placement day