ADO - Work Efficiency with Variable Productivity

This topic is an advanced extension of Time and Work. Instead of assuming that a person works at a constant rate, it considers situations where productivity changes over time due to factors like fatigue, learning, incentives, or external conditions. Questions based on this concept require careful tracking of changing work rates across different time intervals.

1. Basic Idea

In standard problems, a person completes a fixed portion of work per day. Here, the efficiency (rate of work) is not constant. It may:

  • Increase over time (learning effect)

  • Decrease over time (fatigue)

  • Change in steps (shift-based work or varying conditions)

Work is still measured in units, and the formula remains:

Work Done = Efficiency × Time

But efficiency is no longer fixed.

2. Types of Variable Productivity

a) Increasing Efficiency

A worker becomes faster over time.

Example:
A person completes 10% more work each day than the previous day.

This forms a progression:
Day 1 = x
Day 2 = 1.1x
Day 3 = 1.21x

Such problems often involve geometric progression.

b) Decreasing Efficiency

A worker slows down due to fatigue.

Example:
A worker’s efficiency reduces by 20% after every 2 days.

You must calculate work done in segments:

  • First 2 days → full efficiency

  • Next 2 days → reduced efficiency

c) Piecewise Constant Efficiency

Efficiency changes at fixed intervals.

Example:

  • First 3 days → 5 units/day

  • Next 4 days → 8 units/day

Total work = (3 × 5) + (4 × 8)

d) Conditional Productivity

Efficiency depends on conditions such as:

  • Working alone vs in a team

  • Availability of tools or resources

  • Time of day (day shift vs night shift)

3. How to Solve These Problems

Step 1: Assume total work (usually LCM or 100 units)
Step 2: Break time into segments based on efficiency change
Step 3: Calculate work done in each segment
Step 4: Add total work done
Step 5: Equate with total work to find unknowns

4. Example Problem

A worker completes 20 units per day for the first 5 days. After that, due to fatigue, his efficiency drops to 15 units per day. If total work is 200 units, how many days will he take?

Solution:

Work in first 5 days = 5 × 20 = 100 units
Remaining work = 200 − 100 = 100 units

Now efficiency = 15 units/day
Days needed = 100 ÷ 15 = 6.67 days

Total time = 5 + 6.67 = 11.67 days

5. Advanced Case: Increasing Productivity

A worker increases efficiency by 5 units every day:
Day 1 = 5
Day 2 = 10
Day 3 = 15

This forms an arithmetic progression.

Total work after n days:
Sum = n/2 × [2a + (n−1)d]

Where:
a = initial efficiency
d = increase per day

6. Important Concepts to Remember

  • Always track efficiency changes carefully

  • Convert the entire problem into work units

  • Use arithmetic or geometric progression when needed

  • Break the timeline into logical parts

  • Avoid assuming constant productivity

7. Common Mistakes

  • Ignoring change in efficiency

  • Treating variable productivity as constant

  • Not breaking the timeline correctly

  • Calculation errors in progression-based problems

8. Exam Relevance

These questions are common in:

  • Data sufficiency

  • Caselet-based problems

  • Advanced time and work questions

They test analytical thinking rather than formula memorization.

If you want, I can give you:

  • Practice questions of this type

  • Shortcut methods for faster solving

  • Previous exam-level questions with solutions