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SSC CGL Pipes & Cisterns

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This page covers SSC CGL Pipes & Cisterns with complete concept notes, 5 graded practice MCQs, key points and exam-specific tips. Free to study.

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Concept Notes

Pipes & Cisterns— Rules & Concept

Core ConceptRead this first — the foundation of the topic

Think of it this way

— A FILLING pipe adds water. Its work is POSITIVE. — A DRAINING pipe (called a leak or outlet) removes water. Its work is NEGATIVE. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ KEY RULES / PROPERTIES ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Rule 1: If a pipe fills a tank in 'A' hours, it fills 1/A of the tank in 1 hour

Rule 2

If a pipe empties a tank in 'B' hours, it empties 1/B of the tank in 1 hour

Rule 3

When both pipes work together, combine their rates. Add filling rates. Subtract draining rates

Rule 4

The total work done = 1 (one full tank). Always treat the tank as 1 unit. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Formula BlockMemorise — at least one formula appears in every paper

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FORMULA 1 — Two Filling Pipes (A fills in 'a' hrs, B fills in 'b' hrs):

Time to fill together = (a × b) / (a + b)

FORMULA 2 — One Filling Pipe + One Draining Pipe:

Net Rate = (1/a) - (1/b)

Time to fill = (a × b) / (b - a) [only when b > a, meaning filling is faster]

FORMULA 3 — Three Pipes Together:

Net Rate = (1/a) + (1/b) - (1/c) [c is the drain pipe]

Time = 1 / Net Rate

FORMULA 4 — Pipe opened for part of the time:

Work done by Pipe A in 't' hours = t/a

Remaining work is done by Pipe B alone.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Exam PatternsWhat examiners ask — read before attempting PYQs

— WHAT GETS ASKED ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SSC CGL typically asks: 1. Two pipes filling together — find total time. 2. One fill pipe + one leak — find time or find leak capacity. 3. A pipe is opened after some time — find total time to fill. 4.

Tank is already partially full — find time to fill remaining part. 5. Find the capacity of tank when flow rate is given in litres per hour. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SHORTCUT / TRICK SECTION ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ TRICK 1 — LCM METHOD (Fastest for SSC): Assume tank capacity = LCM of all given times. Convert everything into units per hour instead of fractions. This avoids fractions completely. Example: Pipe A fills in 4 hrs, Pipe B fills in 6 hrs. LCM of 4 and 6 = 12 units (assumed tank capacity) Pipe A fills 12/4 = 3 units/hr Pipe B fills 12/6 = 2 units/hr Together = 5 units/hr Time = 12/5 = 2.4 hours = 2 hours 24 minutes TRICK 2 — LEAK DETECTION SHORTCUT: If a pipe fills tank in 'a' hrs alone, but due to a leak it takes 'b' hrs (b > a), then: Time for leak to empty the full tank = (a × b) / (b - a) TRICK 3 — PARTIAL FILLING SHORTCUT: If tank is already x/y full, remaining work = (1 - x/y). Multiply remaining work by time taken to fill full tank to get answer directly. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Worked ExampleSolve this step-by-step before moving on

1

Step 1

Use LCM method. LCM of 12 and 18 = 36 units.

2

Step 2

Pipe A fills 36/12 = 3 units per hour.

3

Step 3

Pipe B fills 36/18 = 2 units per hour.

4

Step 4

Together = 3 + 2 = 5 units per hour.

5

Step 5

Time = 36/5 = 7.2 hours = 7 hours 12 minutes. Answer: 7 hours 12 minutes. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ WORKED EXAMPLE 2 (Leak Type — Very Common in SSC) ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Question: A pipe fills a tank in 20 hours. Due to a leak at the bottom, it takes 30 hours to fill. How long will the leak take to empty a full tank?

1

Step 1

Filling rate of pipe = 1/20 per hour.

2

Step 2

Net rate (pipe + leak together) = 1/30 per hour.

3

Step 3

Leak rate = Filling rate - Net rate = 1/20 - 1/30.

4

Step 4

LCM of 20 and 30 = 60. 1/20 = 3/60 and 1/30 = 2/60.

5

Step 5

Leak rate = 3/60 - 2/60 = 1/60 per hour.

6

Step 6

Leak empties tank in 60 hours. Answer: 60 hours. Alternate using shortcut: (a × b)/(b - a) = (20 × 30)/(30 - 20) = 600/10 = 60 hours. ✓ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Exam TrapsCommon mistakes students make — avoid these

— THE #1 TRAP ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Most students add times instead of adding rates. WRONG thinking: 'A takes 12 hrs, B takes 18 hrs, together they take 12+18 = 30 hrs or 30/2 = 15 hrs.' THIS IS COMPLETELY WRONG. CORRECT thinking: ALWAYS add RATES (work per hour), NOT times. 1/12 + 1/18 = 3/36 + 2/36 = 5/36. Time = 36/5 hours. Remember: Together is always FASTER than alone. If your answer is more than the smallest given time, you have made an error.

Key Points to Remember

A filling pipe that fills in 'A' hours does 1/A work per hour — this is its RATE, not its time.
A drain pipe that empties in 'B' hours does NEGATIVE work: subtract 1/B from the net rate.
FORMULA: Two pipes filling together → Time = (a × b) / (a + b).
FORMULA: One fill pipe (a hrs) + one leak (b hrs) → Leak empties full tank in (a × b) / (b - a) hours.
LCM TRICK: Assume tank capacity = LCM of all times given; convert to units per hour to avoid fractions.
When pipes work together, the combined time is ALWAYS less than the smallest individual time.
Net Rate = Sum of all filling rates MINUS sum of all draining rates.
If tank is already partially filled (say 1/3 full), remaining work = 2/3; multiply by full fill time to get remaining time.
FORMULA: If a pipe fills in 'a' hrs alone but takes 'b' hrs due to leak → leak time = (a × b) / (b - a).
Always add RATES (1/time), NEVER add times directly — this is the most common error in this topic.

Exam-Specific Tips

If Pipe A fills a tank in 'a' hours and Pipe B empties it in 'b' hours (b > a), net filling time = (a × b) / (b - a) hours.
If a pipe fills a tank in 'x' hours, in 't' hours it fills t/x fraction of the tank — this fraction must equal 1 for a full tank.
The LCM method assigns tank capacity = LCM of all pipe times; each pipe's rate = LCM divided by its individual time.
Two pipes filling in 'a' and 'b' hours together take (a × b)/(a + b) hours — this formula works ONLY for two filling pipes with no leak.
If three pipes A, B (filling) and C (draining) work together, net rate = 1/A + 1/B - 1/C and total time = 1/(net rate).
A pipe filling a tank in 6 hours and a leak emptying in 8 hours together: net rate = 1/6 - 1/8 = 1/24; tank fills in 24 hours.
When a pipe is closed after 't' hours and remaining tank is filled by another pipe, calculate work done = t × (rate of first pipe), then remaining = 1 minus that value.

Practice MCQs

Pipes & Cisterns — Practice Questions

5graded MCQs · easy to hard · full solution & trap analysis

All MCQs →

Practice 1easy

A pipe can empty a full cistern in 8 hours. If the cistern is already 3/4 full, how long will it take for the pipe to empty it completely?

Practice 2easy

A pipe can empty a full cistern in 20 hours. If the cistern is already 3/4 full, how long will it take to empty it completely?

Practice 3medium

Pipe A can fill a cistern in 12 hours, and Pipe B can empty it in 18 hours. If both pipes are opened simultaneously, how long will it take to fill the cistern?

Practice 4medium

Pipe A can fill a cistern in 12 hours, and Pipe B can empty it in 18 hours. If both pipes are opened simultaneously, how long will it take to fill the cistern?

Practice 5hard

Two pipes A and B together can fill a tank in 10 hours. If Pipe A alone takes 6 hours more than Pipe B alone to fill the tank, how long will Pipe B take to fill the tank individually?

60-Second Revision — Pipes & Cisterns

Remember: RATE = 1/Time. Always work with rates, NEVER add times directly.
Formula: Two filling pipes → Together time = (a × b) / (a + b). Quick and direct.
Formula: Fill pipe 'a' hrs + leak discovered, now takes 'b' hrs → Leak empties in (a × b) / (b - a) hrs.
Trick: Use LCM as assumed tank capacity to convert all fractions into whole numbers — saves 40% calculation time.
Trap: Answer must ALWAYS be less than the smallest filling time given. If not, recheck your signs (filling vs draining).
For partial fill questions: Remaining work = (1 - fraction already filled). Then Time = Remaining work / Net rate.
Drain pipe = NEGATIVE rate. Filling pipe = POSITIVE rate. Net rate decides whether tank fills or empties overall.

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