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SSC MTS Pipes & Cisterns

Study Material — 2 PYQs (2021–2021) · Concept Notes · Shortcuts

SSC MTS Pipes & Cisterns is a frequently tested subtopic — 2 previous year questions from 2021–2021 papers are included below with concept notes, key rules and shortcut tricks.

2 PYQs

2021–2021

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Previous Year Questions

SSC MTS Pipes & Cisterns — Past Exam Questions

2 questions from actual SSC MTS papers · all shown free · click option to reveal solution

Exam Q 12021Previous Year Pattern

A pipe can empty a full cistern in 8 hours. What fraction of the cistern will it empty in 3 hours?

Exam Q 22021Previous Year Pattern

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

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Concept Notes

Pipes & Cisterns— Rules & Concept

💡

Core Concept

Read this first — the foundation of the topic

💡Key Rules

If a pipe fills a tank in 'n' hours, it fills 1/n part in 1 hour. This is the pipe's rate of work. When pipes work together, ADD their rates. For outlet pipes, SUBTRACT their rates from inlet rates.

The combined rate tells you the net filling or emptying speed

🔑Core Formulas

- Pipe Rate = 1/Time taken to fill/empty alone - Combined Rate = Sum of individual rates (+ for inlet, - for outlet) - Time with combined rate = 1/Combined rate - Work done = Rate × Time

📊

Exam Patterns

What examiners ask — read before attempting PYQs

⚡Power Shortcut - The LCM Method

Take LCM of all individual times. This becomes tank capacity. Each pipe's rate becomes LCM/individual time. This eliminates fractions completely and makes calculations super fast

✏️Worked Example 1

1

Find individual rates A's rate = 1/12 per hour B's rate = 1/15 per hour C's rate = -1/20 per hour (negative for outlet)

2

Combined rate = 1/12 + 1/15 - 1/20

3

Take LCM of 12,15,20 = 60 A fills 60/12 = 5 units/hour B fills 60/15 = 4 units/hour C empties 60/20 = 3 units/hour

4

Net rate = 5 + 4 - 3 = 6 units/hour

5

Time = 60/6 = 10 hours Worked Example 2: Two pipes fill tank in 6 hours when opened together. First pipe alone takes 5 hours more than second pipe. Find individual times.

1

Let second pipe take x hours, first takes (x+5) hours

2

Combined rate = 1/6 per hour

3

1/(x+5) + 1/x = 1/6

4

Cross multiply: 6x + 6(x+5) = x(x+5)

5

6x + 6x + 30 = x² + 5x

6

12x + 30 = x² + 5x

7

x² - 7x - 30 = 0

8

(x-10)(x+3) = 0, so x = 10 hours

9

Second pipe = 10 hours, First pipe = 15 hours Shortcut Trick 1 - Efficiency Method: Instead of fractions, use efficiency. If A completes work in 4 hours and B in 6 hours, their efficiencies are in ratio 6:4 = 3:2. Total work = LCM(4,6) = 12 units

⚡Shortcut Trick 2 - Alternating Pipes Formula

If two pipes work alternately starting with first pipe, and their individual times are a and b hours, then cycles needed = Total work/(work in 2 hours) and check remainder. The #1 Most Common Trap: Students forget to make outlet pipe rates NEGATIVE. They add all rates as positive and get wrong answers

💡Always remember

Inlet = +, Outlet = -, then calculate combined rate. Another major mistake is not converting 'part of tank filled' back to actual time. If question asks when tank will be full and you calculated rate, don't forget the final division step.

Key Points to Remember

Inlet pipe rate = +1/time, Outlet pipe rate = -1/time to fill/empty alone
Combined rate = Sum of all individual rates (+ for inlet, - for outlet)
Time with multiple pipes = 1/Combined rate
LCM method eliminates fractions: Tank capacity = LCM of all individual times
Pipe efficiency = LCM/Individual time gives units filled per hour
When pipes opened at different times, calculate work done by each separately
If two pipes together take time T and alone take times in ratio a:b, individual times are T×(a+b)/b and T×(a+b)/a
Alternating pipes: Calculate work done in complete cycles first, then remainder
Three pipes A, B, C with rates 1/a, 1/b, 1/c together take time = abc/(bc+ac-ab) when C is outlet
Always verify: Combined rate positive means filling, negative means emptying

Exam-Specific Tips

If a pipe fills tank in n hours, it fills 1/n part of tank in 1 hour
Two pipes with individual times a and b hours together take ab/(a+b) hours when both are inlet pipes
If inlet pipe fills in a hours and outlet empties in b hours, together they take ab/(b-a) hours to fill
When three pipes A, B (inlet) and C (outlet) work together, time = abc/(bc+ac-ab) hours
Pipe opened for part time contributes: (Part time/Total individual time) × Tank capacity
Two pipes filling together in 6 hours with ratio 2:3 take individual times of 15 and 10 hours
If tank fills in x hours with inlet and empties in y hours with outlet, net time = xy/(y-x) hours

60-Second Revision — Pipes & Cisterns

Remember: Inlet rates are positive, outlet rates are negative always
Formula: Combined rate = Sum of individual rates, Time = 1/Combined rate
Trick: Use LCM method to avoid fractions in complex problems
Trap: Don't forget to subtract outlet pipe rates from total
Pattern: SSC asks 2-3 pipes together and alternating pipe problems most
Quick check: Combined rate positive = filling, negative = emptying
Time-saver: For two pipes a,b hours individually, together take ab/(a+b) hours

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