Pipes and Cisterns: Notes और MCQ Practice

आसान भाषा में

एक cistern (tank) में water fill/empty करने के लिए दो या अधिक pipes लगते हैं। सभी pipes का rate (गति) liters per minute में दिया जाता है। कुल time = Volume ÷ Net rate, जहाँ Net rate = Σ (fill rates) − Σ (empty rates).

मुख्य फ़ॉर्मूले / Rules

• Net rate = (fill pipe1 rate + fill pipe2 rate + …) − (empty pipe1 rate + …)
• Time taken = Volume ÷ Net rate
• अगर दो pipes साथ में काम कर रहे हों तो उनका combined rate जोड़ें, अगर एक fill और एक empty है तो घटाएँ.
• Fraction of tank filled in t minutes = (Net rate × t) ÷ Volume

Quick Tricks

1. LCM trick – सभी pipe rates को LCM (liters) में बदलें, फिर काम आसान हो जाता है. Example: 1/4 और 1/6 tank per min → LCM 12 → 3 और 2 parts respectively.
2. एक pipe को 0 मानें – अगर एक pipe बहुत छोटा/बड़ा है, उसे ignore करके approximate answer जल्दी मिल सकता है.

Common Mistakes

1. Fill और empty rates को add करने की बजाय subtract नहीं करना.
2. Net rate को minutes में निकाल कर फिर volume को minutes में divide करना, जबकि rate को पहले same unit (liters/min) में convert न करना.

1. Concept

A pipe (or tap) fills a cistern (tank) while another pipe may empty it. The work done by a pipe is measured in rate = volume per unit time. If a pipe can fill the whole cistern in x hours, its filling rate is 1 ÷ x cistern per hour. Similarly, an emptying pipe that empties the cistern in y hours has a rate of -1 ÷ y cistern per hour. The net rate of several pipes working together is the algebraic sum of their individual rates. The time taken for the cistern to become full (or empty) is 1 ÷ (net rate), provided the net rate is positive for filling and negative for emptying.

2. Important formulas / rules / facts

• Filling rate of a pipe = 1 ÷ (time taken to fill alone) cistern per hour.
• Emptying rate of a pipe = -1 ÷ (time taken to empty alone) cistern per hour.
• Net rate of n pipes = Σ (individual rates).
• Time to fill/empty with net rate = 1 ÷ (net rate).
• If a pipe works for t hours, work done = rate × t.
• When a pipe is closed after some time, recalculate net rate for the remaining period.
• For mixed work (fill + empty), treat emptying pipes as negative rates.
• If the cistern is partially filled initially, subtract the initial fraction from 1 before applying the net rate.

3. Solved examples (exam style)

Example 1

Question: Pipe A can fill a tank in 12 hrs and pipe B can fill the same tank in 15 hrs. Both are opened together. After 3 hrs pipe B is closed. In how many more hours will the tank be completely filled?

Solution:

1. Rate of A = 1 ÷ 12 = 1/12 cistern/hr.
2. Rate of B = 1 ÷ 15 = 1/15 cistern/hr.
3. Combined rate for first 3 hrs = 1/12 + 1/15 = (5+4)/60 = 9/60 = 3/20 cistern/hr.
4. Work done in 3 hrs = 3 × (3/20) = 9/20 of the tank.
5. Remaining work = 1 – 9/20 = 11/20.
6. After B is closed, only A works at 1/12 cistern/hr.
7. Time required = (remaining work) ÷ (rate of A) = (11/20) ÷ (1/12) = (11/20) × 12 = 132/20 = 6.6 hrs = 6 hrs 36 min.

Answer: 6 hrs 36 min.

Example 2

Question: Pipe X can fill a cistern in 8 hrs and pipe Y can empty it in 10 hrs. Both are opened together. How long will it take to fill the cistern?

Solution:

1. Rate of X = 1 ÷ 8 = 1/8 cistern/hr.
2. Rate of Y = -1 ÷ 10 = -1/10 cistern/hr.
3. Net rate = 1/8 – 1/10 = (5‑4)/40 = 1/40 cistern/hr.
4. Time to fill = 1 ÷ (1/40) = 40 hrs.

Answer: 40 hrs.

Example 3

Question: Three pipes A, B and C can fill a tank in 6 hrs, 8 hrs and 12 hrs respectively. Pipe D can empty the tank in 24 hrs. All four pipes are opened together. How long will the tank take to get filled?

Solution:

1. Rates: A = 1/6, B = 1/8, C = 1/12, D = -1/24.
2. Net rate = 1/6 + 1/8 + 1/12 – 1/24.

Find common denominator 24: (4 + 3 + 2 – 1) /24 = 8/24 = 1/3 cistern/hr.

1. Time = 1 ÷ (1/3) = 3 hrs.

Answer: 3 hrs.

Example 4

Question: A cistern is 75% full. Pipe P can fill it in 5 hrs and pipe Q can empty it in 7 hrs. Both are opened together. How long will it take to completely fill the cistern?

Solution:

1. Remaining fraction to fill = 1 – 0.75 = 0.25.
2. Rate of P = 1/5, rate of Q = -1/7.
3. Net rate = 1/5 – 1/7 = (7‑5)/35 = 2/35 cistern/hr.
4. Time = remaining work ÷ net rate = 0.25 ÷ (2/35) = 0.25 × 35/2 = 8.75/2 = 4.375 hrs = 4 hrs 22.5 min.

Answer: 4 hrs 22.5 min.

Example 5

Question: Pipe A fills a tank in 9 hrs. Pipe B empties it in 12 hrs. After the tank is half full, both pipes are opened together. How long will it take to empty the tank completely?

Solution:

1. Rate of A = 1/9, rate of B = -1/12.
2. Net rate = 1/9 – 1/12 = (4‑3)/36 = 1/36 cistern/hr (still positive, i.e., tank continues to fill).
3. Since net rate is positive, the tank will never empty while both are open. To empty, pipe A must be closed. If only B works, emptying rate = -1/12 cistern/hr.
4. Work left to empty = 0.5 cistern.
5. Time = 0.5 ÷ (1/12) = 0.5 × 12 = 6 hrs.

Answer: 6 hrs (after closing pipe A).

4. Shortcuts and tricks

• Use fractions instead of decimals to avoid rounding errors.
• Common denominator method: When adding/subtracting rates, quickly find LCM of denominators.
• “Work done = rate × time” is the core; always write the equation before plugging numbers.
• If a pipe works for a part of the total time, compute its contribution separately and subtract from the total work.
• When the net rate is very small, the total time becomes large; check for sign errors.
• Convert minutes to hours only at the end to keep calculations clean.

5. Common mistakes to avoid

• Forgetting to assign a negative sign to emptying pipes.
• Adding times of pipes instead of adding their rates.
• Ignoring the initial fraction of water already present.
• Using the wrong denominator when finding LCM (e.g., using 60 instead of 120 for 12 and 15 hrs).
• Assuming the tank will fill even when net rate is negative; always verify the sign of the net rate.
• Rounding intermediate results; keep fractions till the final step.