Pipes and Cisterns – Complete Short Notes for Competitive Exams

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1. Basic Concept

Pipes and cisterns problems are direct applications of Time and Work concepts.

👉 A pipe fills or empties a tank/cistern — so work = filling or emptying.

• Inlet pipe: Fills the tank → Positive work

• Outlet pipe: Empties the tank → Negative work

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2. Fundamental Formulas

Case	Formula
Time to fill the tank	Work / Filling rate
Time to empty the tank	Work / Emptying rate
Net rate (if both open)	(1/x) ± (1/y)
Time to fill/empty jointly	1 / (1/x ± 1/y)

+ sign when both fill; − sign when one fills and the other empties.

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3. Key Relationship

$$\text{Work (Tank filled)}=\text{Rate of Filling}\times \text{Time}$$

If a pipe fills in x hours → in 1 hour, it fills 1/x part of tank.
If it empties in y hours → in 1 hour, it empties 1/y part of tank.

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Example:

A pipe fills tank in 10 hrs, another empties in 15 hrs.
Net 1 hr work = (1/10 − 1/15) = 1/30
⇒ Tank filled in 30 hrs

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4. When Two Pipes Fill Together

If one pipe fills in x hrs and another in y hrs,

$$\text{Time together}=\frac{xy}{x+\mathrm{y}}$$

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Example:

Pipe A fills in 12 hrs, Pipe B fills in 15 hrs
Together = (12×15)/(12+15) = 180/27 = 6 hrs 40 min

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5. When One Pipe Fills and Other Empties

If fill pipe takes x hrs, and outlet empties in y hrs,

$$\text{Time to fill tank}=\frac{xy}{y-x}(y>x)$$

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Example:

Inlet fills in 10 hrs, outlet empties in 15 hrs
Time = (10×15)/(15 − 10) = 150/5 = 30 hrs

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6. When Both Are Outlets

If two pipes empty the tank in x and y hrs respectively:

$$\text{Together empty time}=\frac{xy}{x+\mathrm{y}}$$

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7. Three Pipes (Two Fillers, One Emptier)

If two inlets fill in x, y hrs and one outlet empties in z hrs:

$$\text{Effective rate}=\frac{1}{x}+\frac{1}{y}-\frac{1}{\mathrm{z}}$$$$\text{Time to fill tank}=\frac{1}{(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})}$$

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Example:

A fills in 12 hr, B fills in 15 hr, C empties in 20 hr
Rate = (1/12 + 1/15 − 1/20) = (5 + 4 − 3)/60 = 6/60
⇒ Time = 60/6 = 10 hr

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8. Tank with Leakage

If a pipe fills a tank in x hrs, but due to leakage it takes t hrs more:

$$\text{Leak rate} = \frac{1}{x} - \frac{1}{x + t}$$

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Example:

Pipe fills in 10 hrs but due to leakage takes 12 hrs
Leak rate = (1/10 − 1/12) = 1/60
⇒ Leak can empty full tank in 60 hrs

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9. Leakage + Outlet Together

If a pipe fills in x hrs, leakage empties in y hrs,
Net time to fill tank = 1 / (1/x − 1/y)

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10. Fraction of Tank Filled

If both pipes are open for t hours:

$$\text{Fraction filled}=t\times (\frac{1}{x}\pm \frac{1}{y})$$

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Example:

Pipe A (fills in 6 hrs), Pipe B (empties in 9 hrs).
Both open for 3 hrs:
= 3 × (1/6 − 1/9) = 3 × 1/18 = 1/6 → 1/6 of tank filled

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11. Efficiency & Ratio Relation

Work done ∝ Time × Rate
If Pipe A fills in 6 hr, B fills in 8 hr →
Efficiency ratio = 8 : 6 = 4 : 3

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12. Alternate Hours Work

If pipes operate alternately every hour,
Find work done in 2 hours = (1/x ± 1/y)
Divide total work (1) by that value accordingly.

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Example:

A fills in 4 hr, B empties in 6 hr, open alternately.
2-hour work = (1/4 − 1/6) = 1/12
After 12 such hours → work done = 12 × 1/12 = 1 → tank full in 12 hours

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13. Tank Filling Half or Partially

If asked time to fill k fraction of tank:

$$\text{Time}=k\times \text{Total time}$$

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Example:

Tank full in 18 hr → Half = 9 hr, One-third = 6 hr.

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14. Inlet Closed After Some Time

If inlet is closed after t₁ hours and outlet continues:
Work done = (t₁/x − t₂/y) = 1 (solve for unknown)

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Example:

A fills in 12 hr, B empties in 20 hr.
If both opened and A closed after 4 hr, tank full in 10 hr total.
1 = (4/12 + 6/20) → verify & solve for consistency.

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15. Important Ratios

Type	Ratio Relation
Time	∝ 1 / Efficiency
Work Done	∝ Efficiency × Time
Efficiency	1 / Time

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16. Trick Table

Condition	Formula	Remark
Two inlets	(xy)/(x + y)	Both fill
Inlet + Outlet	(xy)/(y − x)	y > x
Two outlets	(xy)/(x + y)	Both empty
3 pipes (2 fill, 1 empty)	1/(1/x + 1/y − 1/z)	z empties
Leakage	1/x − 1/(x + t)	x = normal time, t = extra time

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17. Quick Examples

Q1. A fills in 8 hr, B fills in 12 hr, together with C empties in 24 hr → find total time.
= 1/(1/8 + 1/12 − 1/24) = 1/(3/24 + 2/24 − 1/24) = 1/(4/24) = 6 hr

Q2. Pipe A can fill a tank in 10 hr, but due to leakage tank fills in 12 hr → leak alone empties in?
= 1/(1/10 − 1/12) = 60 hr

Q3. Pipe A fills 1/3 of tank in 4 hr → full tank in 12 hr.

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18. Common Tricks for Exams

✅ Always take total work = 1 (or full tank)
✅ Filling = +, emptying = −
✅ Use 1 hr work method for simplicity
✅ If values are fractional, convert to LCM of times for easy ratio
✅ Leakage questions always use difference of reciprocals

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19. Shortcut Summary

Formula	Meaning
1/x	Inlet fills in x hr
1/y	Outlet empties in y hr
1/x ± 1/y	Combined 1 hr work
xy/(x + y)	Time (both fill)
xy/(y − x)	Time (fill + empty)
1/x − 1/(x + t)	Leak rate

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20. Real-life Concept Applications

• Water tank filling & leakage

• Drainage and pumping systems

• Industrial tank operations

• Civil water distribution design

• Pressure & flow-rate problems

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✅ In One Line Summary

“Pipes and Cisterns = Time & Work with Positive (inlet) and Negative (outlet) flow.”