Pipes & Cisterns
SSC GD Exam Preparation - Quantitative Aptitude
1️⃣ Introduction
Pipes & Cistern problems are based on the Time & Work concept.
Inlet Pipe
→ Fills a tank
Outlet Pipe
→ Empties a tank
If both work together → Subtract their 1-hour works
2️⃣ Key Formulas
✔ Inlet Pipe
If a pipe fills a tank in x hours,
1-hour work = 1/x
✔ Outlet Pipe
If it empties in y hours,
1-hour work = –1/y
✔ Both Working Together
Net 1-hour work = 1/x - 1/y
✔ Tank filled/emptied in time
Time = 1 / (Net 1-hour work)
3️⃣ Important Points
If filling is faster than emptying → tank fills
If emptying is faster → tank empties
When more than 2 pipes: add all fillings & subtract all emptyings
🧠 15 Practice Questions (With Answers)
Test your pipes & cisterns concepts with these SSC GD level practice questions. Click on "View Answer" to check your understanding.
Q1. A pipe fills a tank in 12 hours. What is its 1-hour work?
View Answer
1/12 part
Q2. An outlet empties the tank in 10 hours. 1-hour work?
View Answer
–1/10 part
Q3. A fills in 6 h, B fills in 3 h. Together?
View Answer
1-hour work = 1/6 + 1/3 = 1/2, Time = 2 hours
Q4. A fills in 8 h, B empties in 12 h. Net time?
View Answer
1-hour work = 1/8 − 1/12 = 1/24, Time = 24 hours
Q5. A fills in 4 h, B empties in 6 h. Net time?
View Answer
1-hour = 1/4 – 1/6 = 1/12, Time = 12 hours
Q6. A pipe fills in 10 h. Another fills in 15 h. Both fill?
View Answer
1-hour = 1/10 + 1/15 = 1/6, Time = 6 hours
Q7. A fills tank in 20 h. Outlet empties in 10 h. Net effect?
View Answer
1-hour = 1/20 – 1/10 = –1/20, Tank empties in 20 hours
Q8. A fills in 30 h. B fills in 20 h. C empties in 60 h. Time?
View Answer
1-hour = 1/30 + 1/20 – 1/60 = 1/15, Time = 15 hours
Q9. A can fill half tank in 6 h. Full tank?
View Answer
Half = 6 h → full = 12 hours
Q10. A fills in 12 h, but leakage empties in 36 h. Time to fill?
View Answer
1-hour = 1/12 – 1/36 = 1/18, Time = 18 hours
Q11. A fills in 18 h. B empties in 54 h. Both operate?
View Answer
1-hour = 1/18 – 1/54 = 1/27, Time = 27 hours
Q12. Two pipes A & B fill a tank in 10 h and 15 h. After 5 h, B closed. Remaining time?
View Answer
Work in 5 h = (1/10 + 1/15) × 5 = 5/6, Remaining = 1/6, A alone = (1/6) ÷ (1/10) = 10/6 = 1.67 hours
Q13. A fills in 25 h. B empties in 50 h. With both, tank fills in?
View Answer
1-hour = 1/25 – 1/50 = 1/50, Time = 50 hours
Q14. A fills in 15 h, B fills in 10 h, C empties in 30 h. Together?
View Answer
1-hour = 1/15 + 1/10 – 1/30 = 1/6, Time = 6 hours
Q15. A pipe can fill in 16 h. A & B together fill in 12 h. B alone?
View Answer
1-hour(B) = 1/12 – 1/16 = 1/48, Time = 48 hours
✅ SSC GD Exam Strategy
🟢 Remember inlet = positive, outlet = negative work
🟢 Practice net effect calculations with multiple pipes
🟢 Master leakage problems (outlet pipes)
🟢 Learn to handle partial work scenarios
🟢 Time yourself - aim for 45-60 seconds per pipes & cistern question
You've completed Pipes & Cisterns Concepts!
SSC GD Tip: Pipes & Cisterns problems are essentially Time & Work problems with the added concept of negative work (emptying). Focus on mastering the basic formula of adding filling rates and subtracting emptying rates. Remember that outlets have negative work values, and the net effect determines whether the tank fills or empties. Practice scenarios with multiple pipes and leakages as they frequently appear in SSC exams. Regular practice with previous year questions will build your speed and accuracy in solving pipes & cisterns problems efficiently.
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