Comparing wages by speed: Worker 1 can finish in 26 days and Worker 2 in 39 days. If they work together on the same job, by what percentage are the wages of the first worker more than those of the second (assuming pay proportional to work done)?

Introduction / Context:
When two workers collaborate and are paid by contribution, their wage ratio equals their rate ratio. Since rate is the reciprocal of days to finish alone, a quicker worker earns more. The question asks for the percentage by which the first worker’s wages exceed the second’s when both work the same time on the same job.

Given Data / Assumptions:

• Worker 1 alone: 26 days ⇒ rate r1 = 1/26.
• Worker 2 alone: 39 days ⇒ rate r2 = 1/39.
• Wages proportional to rate (same time worked).

Concept / Approach:
Wage ratio r1 : r2 = 1/26 : 1/39 = 39 : 26, which simplifies to 3 : 2. The percent by which the first exceeds the second is ((3 − 2)/2) * 100 = 50% more.

Step-by-Step Solution:

Verification / Alternative check:
If second earns ₹ 200, first earns ₹ 300; the increase is ₹ 100 over ₹ 200 ⇒ 50%.

Why Other Options Are Wrong:
25%, 35%, 15%, and 20% do not match the exact 3:2 ratio.

Common Pitfalls:
Computing percentage of the total rather than relative to the second worker’s wage.

Final Answer:
50%