Dividing 700 hectares into two parts with a constraint on difference\nA large field of 700 hectares is split into two parts. The difference between the areas equals one-fifth of their average. Find the area of the smaller part (in hectares).

Introduction / Context:
Problems involving sum and difference can be solved using the identities for two numbers: sum S = x + y and difference D = |x − y|, then numbers are (S ± D)/2.

Given Data / Assumptions:

• Total S = 700
• Difference D = (1/5) * average = (1/5) * (S/2) = (1/5) * 350 = 70

Concept / Approach:
Let larger = (S + D)/2 and smaller = (S − D)/2. Compute both and choose the smaller.

Step-by-Step Solution:

Verification / Alternative check:
Average = 350; one-fifth = 70; difference between 385 and 315 is indeed 70, satisfying the condition.

Why Other Options Are Wrong:
Other values do not maintain both the fixed sum of 700 and the required difference of 70 simultaneously.

Common Pitfalls:
Confusing “one-fifth of the average” with “one-fifth of the sum” or misplacing which number is larger vs smaller is common; always compute both explicitly.

Final Answer:
315