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Partial or Negative Work (Leak or Undo Work)

Introduction

Partial or Negative Work problems involve situations where some agent undoes work already done - most commonly a leak in a tank or a person removing completed work. This pattern is important because many Time & Work questions include elements that reduce progress (leaks, theft, cancellation), and treating these as negative rates simplifies the logic.

Key idea: represent every contributor as a rate (work per unit time). Treat an undoing agent (leak/undo) as a negative rate and add all rates algebraically to get the net progress.

Pattern: Partial or Negative Work (Leak or Undo Work)

Pattern

Key concept: Model undoing actions as negative rates; net rate = sum of positive rates - sum of negative rates. In time problems, Time = Total work ÷ Net rate.

Formula summary:
Worker rate = 1 / time_to_complete (positive for filling/doing).
Leak/undo rate = -(1 / time_to_empty_or_undo) (negative).
Net rate = Σ(positive rates) + Σ(negative rates).

Step-by-Step Example

Question

Pipe A fills a tank in 10 hours, Pipe B fills it in 15 hours. A leak can empty the filled tank in 30 hours. If A and B are opened together with the leak, how long will it take to fill the tank?

Solution

  1. Step 1: Identify individual rates (positive and negative):

    A’s filling rate = 1/10 tank/hour. B’s filling rate = 1/15 tank/hour. Leak’s emptying rate = -1/30 tank/hour.

  2. Step 2: Compute net rate:

    Net rate = 1/10 + 1/15 - 1/30. Find LCM = 30 → (3 + 2 - 1)/30 = 4/30 = 2/15 tank/hour.

  3. Step 3: Compute time to complete 1 tank:

    Time = Total work ÷ Net rate = 1 ÷ (2/15) = 15/2 = 7.5 hours.

  4. Final Answer:

    The tank will be filled in 7 hours 30 minutes.

  5. Quick Check:

    In 7.5 hours A does 7.5×(1/10)=0.75, B does 7.5×(1/15)=0.5, leak removes 7.5×(1/30)=0.25. Total = 0.75+0.5-0.25=1.0 (full tank) ✅

Quick Variations

1. Multiple leaks or undoers: subtract the sum of all negative rates from positive rates.

2. Leak starts later or stops earlier: treat in phases - compute work done before/after leak separately.

3. Worker leaves mid-way while leak continues: split into phases and sum contributions.

4. Leak alone emptying time asked: compute leak rate and invert it (time = 1 / leak rate).

Trick to Always Use

  • Step 1 → Convert every actor to a rate (1/time). Use negative sign for undoing actors.
  • Step 2 → Add all rates algebraically to get the net rate.
  • Step 3 → If events happen in phases (start/stop at different times), compute each phase separately and add works: Work = rate × time per phase.
  • Step 4 → Time (for final phase) = Remaining work ÷ net rate of that phase. Always quick-check by summing works to 1.

Summary

Summary

For Partial / Negative Work problems:

  • Convert all participants to rates (positive for doing, negative for undoing).
  • Net progress is the algebraic sum of rates; invert net rate to find time for unit work.
  • Handle phased actions by computing work per phase and summing contributions.
  • Always perform a quick check by summing all positive contributions and subtracting undoing contributions to verify it equals 1 unit of work.

Practice

(1/5)
1. Pipe A can fill a tank in 12 hours, and a leak at the bottom can empty the full tank in 24 hours. How long will it take to fill the tank if the leak is also open?
easy
A. 18 hours
B. 16 hours
C. 24 hours
D. 20 hours

Solution

  1. Step 1: Identify rates:

    Pipe A filling rate = 1/12 tank/hour; Leak emptying rate = -1/24 tank/hour.

  2. Step 2: Compute net rate:

    Net rate = 1/12 - 1/24 = (2 - 1)/24 = 1/24 tank/hour.

  3. Step 3: Find total time:

    Time = 1 ÷ (1/24) = 24 hours.

  4. Final Answer:

    24 hours → Option C.

  5. Quick Check:

    24×(1/12 - 1/24)=24×(1/24)=1 (full tank) ✅
Hint: Subtract leak rate from filler rate to get net rate; invert net rate to get time.
Common Mistakes: Forgetting to subtract the leak's rate (treating it as positive).
2. Pipe A can fill a tank in 10 hours and Pipe B in 15 hours. A leak can empty the tank in 30 hours. All are open together. How long will it take to fill the tank?
easy
A. 8.5 hours
B. 9.25 hours
C. 7.25 hours
D. 7.5 hours

Solution

  1. Step 1: Identify rates:

    Pipe A fills 1/10 of the tank per hour, Pipe B fills 1/15 per hour, and the leak empties 1/30 per hour.

  2. Step 2: Compute net rate:

    Net rate = 1/10 + 1/15 - 1/30 = (3 + 2 - 1)/30 = 4/30 = 2/15 tank/hour.

  3. Step 3: Find total time to fill:

    Time = 1 ÷ (2/15) = 15/2 = 7.5 hours.

  4. Final Answer:

    The tank will be filled in 7.5 hours → Option D.

  5. Quick Check:

    7.5 × (1/10 + 1/15 - 1/30) = 7.5 × (2/15) = 1 ✅
Hint: Treat the leak as a negative rate, add all rates, and take the reciprocal for total time.
Common Mistakes: Using the leak rate as positive instead of negative or forgetting to take LCM correctly.
3. Pipe A can fill a tank in 6 hours. But due to a leak, it takes 8 hours to fill it. How long will the leak alone take to empty the full tank?
easy
A. 24 hours
B. 16 hours
C. 20 hours
D. 18 hours

Solution

  1. Step 1: Identify rates:

    Filling rate (A) = 1/6 tank/hour. Net (with leak) = 1/8 tank/hour.

  2. Step 2: Compute leak rate:

    Leak rate = A - Net = 1/6 - 1/8 = (4 - 3)/24 = 1/24 tank/hour (emptying).

  3. Step 3: Leak alone empties tank in:

    Time = 1 ÷ (1/24) = 24 hours.

  4. Final Answer:

    24 hours → Option A.

  5. Quick Check:

    1/6 - 1/24 = 1/8 (net), which matches the given 8-hour fill time ✅
Hint: Leak rate = filling rate - net rate; invert to get emptying time.
Common Mistakes: Subtracting in the wrong order (net - fill instead of fill - net).
4. Two pipes A and B can fill a tank in 12 and 18 hours respectively. A leak can empty the full tank in 36 hours. If all are opened together, how long will it take to fill the tank?
medium
A. 10 hours
B. 9 hours
C. 8 hours
D. 7.2 hours

Solution

  1. Step 1: Filling and leak rates:

    A = 1/12, B = 1/18, Leak = -1/36 (tank/hour).

  2. Step 2: Net rate:

    1/12 + 1/18 - 1/36 = (3 + 2 - 1)/36 = 4/36 = 1/9 tank/hour.

  3. Step 3: Total time:

    Time = 1 ÷ (1/9) = 9 hours.

  4. Final Answer:

    9 hours → Option B.

  5. Quick Check:

    9×(1/12 + 1/18 - 1/36) = 9×(1/9) = 1 ✅
Hint: Add filler rates, subtract leak rate, then invert to find time.
Common Mistakes: Using wrong LCM or forgetting the negative sign for the leak.
5. A pipe can fill a tank in 8 hours. Because of a leak, the tank is filled in 10 hours. How long will the leak take to empty the tank completely?
medium
A. 40 hours
B. 20 hours
C. 30 hours
D. 24 hours

Solution

  1. Step 1: Filling and net rates:

    Filling rate = 1/8 tank/hour; Net (with leak) = 1/10 tank/hour.

  2. Step 2: Compute leak rate:

    Leak = Filling - Net = 1/8 - 1/10 = (5 - 4)/40 = 1/40 tank/hour (emptying).

  3. Step 3: Leak empties tank in:

    Time = 1 ÷ (1/40) = 40 hours.

  4. Final Answer:

    40 hours → Option A.

  5. Quick Check:

    1/8 - 1/40 = 1/10 (net), which matches the given 10-hour fill time ✅
Hint: Leak time = 1 ÷ (filling rate - net rate).
Common Mistakes: Mixing denominators when subtracting fractional rates.

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