The Inverse Relationship Principle
In the context of competitive exams like UPSC CSAT or SSC, the most fundamental concept to master is the inverse proportion between time and efficiency. If a person is more efficient, they naturally require less time to complete a given unit of work. Conversely, if a person is slower, their time requirement increases proportionally.
Think of it as a balance scale: as efficiency increases, time decreases. This relationship is mathematically represented as E ∝ 1/T, where E is efficiency and T is time. When solving multi-worker scenarios, we often normalize the total work to a constant value—usually the Least Common Multiple (LCM) of the individual times provided—to make calculations seamless and avoid complex fractions.
Calculating Total Work and Efficiency
When dealing with problems involving multiple workers, avoid assuming the total work is always ‘1 unit’. Instead, assign a specific value to the total work that is easily divisible by the time taken by each worker. This is known as the LCM Method. For instance, if Person A takes 10 days and Person B takes 15 days, assume the total work is 30 units (the LCM of 10 and 15).
Efficiency is not just about speed; it is about the capacity to perform a specific fraction of the total work within a standard unit of time.
Once the total work is set, calculating individual efficiency becomes trivial. Person A’s efficiency would be 30/10 = 3 units/day, and Person B’s would be 30/15 = 2 units/day. When they work together, you simply add their efficiencies (3 + 2 = 5 units/day) to determine how quickly the collective task will be finished.
The Concept of Man-Days (M-D-H Formula)
For more complex scenarios involving groups of workers, hours per day, and varying difficulty levels of tasks, we utilize the Chain Rule, commonly referred to as the M-D-H formula. This formula accounts for the number of men (M), the number of days (D), and the number of hours (H) spent working.
The standard formula is expressed as: (M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂. Here, W represents the work done. This is particularly useful in questions where the amount of work changes or the number of people is adjusted midway through a project. It ensures that the ratio of input (men, time) to output (work) remains constant.
Handling Efficiency Ratios
Competitive exams frequently present problems where one worker is described as being “X times as efficient” as another. In these cases, you should immediately convert the efficiency ratio into a time ratio. Remember that the ratio of time taken is always the inverse of the ratio of efficiency.
- If A is twice as efficient as B (Efficiency Ratio A:B = 2:1), then the Time Ratio A:B will be 1:2.
- This allows you to quickly assign variable values (like 2x and x) to efficiency, simplifying algebraic setups.
- Always check if the problem mentions “working together” or “working alternatively,” as these require different logical approaches.
Important Facts and Formulas
| Concept | Formula/Relationship |
|---|---|
| Basic Efficiency | Efficiency = Work / Time |
| Combined Efficiency | (A+B)’s Efficiency = Efficiency(A) + Efficiency(B) |
| Inverse Relationship | Time ∝ 1 / Efficiency |
| Chain Rule | (M₁D₁H₁) / W₁ = (M₂D₂H₂) / W₂ |
| Alternative Days | Work = (Eff A + Eff B) per cycle of 2 days |
Previous Year Question Hints
- Scenario 1: If A can finish a work in 20 days and B in 30 days, how long if they work together? Hint: Use LCM(20, 30) = 60. A’s eff = 3, B’s eff = 2. Total eff = 5. Time = 60/5 = 12 days.
- Scenario 2: A is 50% more efficient than B. If B takes 12 days, how long does A take? Hint: Efficiency A = 1.5 * Efficiency B. Time A = Time B / 1.5 = 12 / 1.5 = 8 days.
Quick Revision Summary
- Total Work = Efficiency × Time.
- Always use the LCM of time periods to define ‘Total Work’ for easier arithmetic.
- The ratio of time taken is the reciprocal of the ratio of efficiency.
- When workers work together, add their individual efficiency rates.
- Use the M-D-H formula for problems involving multiple variables (Men, Days, Hours).
- For ‘alternative day’ problems, calculate the work done per cycle (usually 2 days) rather than per day.
- Always verify if the question asks for the total time or the ‘remaining work’ time.
- Efficiency is constant; only the time spent or the work assigned changes across scenarios.
