Threshold Year under Compound Interest — At 10% per annum compounded annually, in which first year will a sum become more than double its original amount?

Introduction / Context:
The question asks for the earliest integer year n such that the compound-interest growth factor (1.10)^n exceeds 2. This is a threshold or crossing-point problem under geometric growth, common in time-value-of-money topics.

Given Data / Assumptions:

• Growth per year g = 1 + 10% = 1.10.
• We seek smallest integer n with (1.10)^n > 2.

Concept / Approach:
Either compute successive powers or use logarithms. Powers are quick here: 1.1^7 ≈ 1.9487, still below 2; 1.1^8 ≈ 2.1436, above 2. Hence the first year exceeding double is year 8.

Step-by-Step Solution:
1.1^5 ≈ 1.610511.1^6 ≈ 1.771561.1^7 ≈ 1.948721.1^8 ≈ 2.14359 > 2 ⇒ threshold crossed at n = 8

Verification / Alternative check:
Log method: n > ln(2) / ln(1.1) ≈ 0.6931 / 0.09531 ≈ 7.27 ⇒ smallest integer n = 8.

Why Other Options Are Wrong:
6th and 7th years are below the threshold; “Data inadequate” is incorrect because g is fully specified; 9th year is not the first crossing.

Common Pitfalls:
Confusing “at least double” with “more than double,” which changes the inequality if a power equaled 2 exactly (it does not here).

Final Answer:
8th year