Compound vs Simple Interest — A father divides ₹ 25,000 between his two sons A and B. A invests his share at 8% per annum compounded annually, and B invests his share at 10% per annum simple interest. After 2 years, B's interest exceeds A's interest by ₹ 1,336. Find A's share in the father's property.

Introduction / Context:
This problem mixes compound interest (for A) and simple interest (for B) for the same 2-year period. We are told how much more interest B earns than A, which lets us set up a single linear equation in A's share.

Given Data / Assumptions:

• Total property = ₹ 25,000.
• A invests at 8% p.a. compounded annually for 2 years.
• B invests at 10% p.a. simple interest for 2 years.
• After 2 years, Interest(B) − Interest(A) = ₹ 1,336.

Concept / Approach:
Let A's share be x ⇒ B's share = 25,000 − x. A's 2-year compound interest on x is x * [(1.08)^2 − 1] = 0.1664x. B's 2-year simple interest on (25,000 − x) is (25,000 − x) * 0.10 * 2 = 5,000 − 0.20x. Their difference is given.

Step-by-Step Solution:
(5,000 − 0.20x) − 0.1664x = 1,3365,000 − 0.3664x = 1,3360.3664x = 3,664 ⇒ x = 10,000

Verification / Alternative check:
Interest(A) = 0.1664 * 10,000 = ₹ 1,664. Interest(B) = 0.20 * 15,000 = ₹ 3,000. Difference = ₹ 1,336 ✔

Why Other Options Are Wrong:
Any value other than ₹ 10,000 fails the linear relation and cannot produce a difference of ₹ 1,336 over 2 years with the stated rates.

Common Pitfalls:
Applying simple interest to A's share (instead of compound) or compounding B's share by mistake; both lead to incorrect equations.

Final Answer:
₹ 10,000