Time and Work Pipes Cisterns Questions for BIHAR POLICE CONSTABLE

Core Concept: Work Rate as Fraction

Think of a pipe's job like a worker's daily output. If Pipe A fills a tank in 10 hours, its work rate is 1/10 of the tank per hour. If Pipe B empties the same tank in 15 hours, its rate is -1/15 per hour (negative because it removes water). When pipes work together, their rates ADD. This is the foundation of every pipes and cisterns problem.

• Inlet pipe (filling): positive rate = 1/time
• Outlet pipe (emptying): negative rate = -1/time
• Combined rate = sum of individual rates
• Time to complete work = 1 / combined rate
• Always express work as a fraction of the tank (total work = 1)

Two Pipes Working Together

When two pipes work simultaneously, add their rates. If Pipe A fills in 6 hours (rate = 1/6) and Pipe B fills in 8 hours (rate = 1/8), together they fill at rate 1/6 + 1/8. Find common denominator (LCM of 6 and 8 = 24): 4/24 + 3/24 = 7/24 per hour. So time = 1 / (7/24) = 24/7 hours. This is the standard two-pipe-filling pattern seen in Bihar Police papers.

• Find LCM of individual times to get common denominator
• Add fractions to get combined rate
• Invert combined rate to get time
• If one is inlet and one is outlet, subtract rates instead
• Watch sign: inlet positive, outlet negative

Three Pipes and Complex Scenarios

When three pipes work together, add all three rates. The method is identical but requires careful arithmetic. Bihar Police often asks: Pipe A fills in 6 hrs, Pipe B in 8 hrs, Pipe C empties in 12 hrs. All open together—time to fill? Add A and B, subtract C. The key is identifying which pipes are inlets (add) and which are outlets (subtract).

• List all pipes with their individual times
• Mark inlet as positive, outlet as negative
• Add all rates algebraically
• Invert to get combined time
• If answer is negative, tank empties (outlet stronger than inlets)

Partial Work and Pipe Closure

A common Bihar Police twist: Pipe A and B fill together for some time, then B closes, and A continues alone. Break the problem into phases. Calculate how much work is done in phase 1 (both pipes), then how much remains. In phase 2, only A works on the remaining portion. This requires two separate rate calculations.

• Phase 1: Both pipes open—calculate work done = combined rate × time
• Remaining work = 1 - work done in phase 1
• Phase 2: One pipe open—time for remaining = remaining work / single pipe rate
• Total time = phase 1 time + phase 2 time
• Always check: total work should equal 1 (full tank)

Efficiency and Work Ratio

If Pipe A fills in 6 hours and Pipe B fills in 9 hours, their efficiencies (work rates) are in ratio 1/6 : 1/9 = 3:2. This means A is 1.5 times faster than B. Bihar Police sometimes asks: if two pipes have efficiency ratio 3:2 and together fill in 6 hours, find individual times. Use the ratio to set up equations. If rates are 3x and 2x, then 3x + 2x = 1/6, so 5x = 1/6, x = 1/30. Pipe A rate = 3/30 = 1/10, so A fills in 10 hours.

• Efficiency ratio = inverse of time ratio
• If A fills in 6 hrs and B in 9 hrs, efficiency ratio is 9:6 = 3:2
• Use ratio to express rates as multiples: 3x and 2x
• Solve for x using combined rate equation
• Convert back to individual times

• Confusing time with rate: If a pipe fills in 6 hours, rate is 1/6, NOT 6. Many aspirants flip this and get answers 36 times too large.
• Forgetting to subtract outlet pipe rate: When one pipe empties, subtract its rate from inlets. Forgetting the minus sign gives wildly wrong answers.
• Arithmetic errors with LCM: Finding LCM of 6, 8, 12 is easy (24), but aspirants often use wrong LCM (like 48) and waste time. Always verify LCM by checking divisibility.
• Not reading 'partial work' carefully: If pipes work together for 2 hours then one closes, many aspirants calculate only the first phase and forget the second. Always break into phases.
• Sign confusion in three-pipe problems: Mixing up which pipe is inlet vs outlet leads to negative time (impossible). Double-check: inlets are positive, outlets negative.

• RATE = 1/TIME: Think 'Rate is the reciprocal.' If tank fills in 10 hours, rate per hour is 1/10. Memorize this as a mantra.
• ADD for together, SUBTRACT for opposite: When pipes work together (both filling or both emptying), ADD rates. When one fills and one empties, SUBTRACT. Inlet minus outlet.
• LCM is your friend: Always find LCM of individual times first. It becomes your common denominator for adding fractions. LCM = least common multiple of times.
• PHASE method for partial work: Break problem into Phase 1 (both pipes) and Phase 2 (one pipe). Calculate work in phase 1, find remainder, solve phase 2 separately. Then add times.

• Bihar Police typically asks 2-3 pipes and cisterns questions per paper, usually in the 'Numerical Ability' section. Difficulty is easy to moderate—straightforward rate calculations, not tricky word problems.
• Most common pattern: Two pipes (one inlet, one outlet) working together, or three pipes with one closure. Rarely asks efficiency ratio problems, but it's good to know.
• Time management: These questions take 2-3 minutes each if you're comfortable with fractions and LCM. Don't spend more than 3 minutes per question; if stuck, move on and return.
• Calculator use: You can use a calculator in Bihar Police exam. Use it for LCM and fraction arithmetic to avoid silly errors. But know the concept without calculator first.
• Recent papers show preference for 'partial work' scenarios (pipe closes after X hours). Practice this pattern heavily. Also watch for 'tank has initial water' variants (less common but possible).