In what time (in years) will Rs. 1,000 amount to Rs. 1,331 at 10% per annum, if interest is compounded annually?

Introduction / Context:
This question asks for the time required for a given principal to grow to a specific amount at a fixed compound interest rate. Recognising standard powers of (1 + r) can make the solution very quick, especially when the numbers are neat, as in this case.

Given Data / Assumptions:

Principal P = Rs. 1,000.
Amount A = Rs. 1,331.
Rate of interest r = 10% per annum.
Compounding is annual.
We must find time t in years.

Concept / Approach:
For annual compounding: A = P * (1 + r/100)^t. Substitute P, A and r, then solve for t. Here r = 10%, so (1 + r/100) = 1.10. We look for t such that: 1,331 / 1,000 = (1.10)^t. The left side is 1.331, and we check powers of 1.10 to see which matches.

Step-by-Step Solution:
Step 1: Express the equation. 1,331 = 1,000 * (1.10)^t. So (1.10)^t = 1,331 / 1,000 = 1.331. Step 2: Recognise standard values. Compute 1.10^2 = 1.21 (too small). Compute 1.10^3 = 1.331, which matches exactly. Therefore, t = 3 years.

Verification / Alternative check:
Calculate year by year. End of year 1: amount = 1,000 * 1.10 = 1,100. End of year 2: amount = 1,100 * 1.10 = 1,210. End of year 3: amount = 1,210 * 1.10 = 1,331. This confirms that it takes exactly 3 years to reach Rs. 1,331.

Why Other Options Are Wrong:
At 2 years, the amount is only Rs. 1,210. At 4 years, the amount would be 1,000 * 1.10^4 = 1,464.1, which is higher than 1,331. At 5 years, it would be even larger. Thus only t = 3 years produces the given final amount.

Common Pitfalls:
Some students treat the growth as simple interest and solve 1,331 = 1,000 * (1 + 0.10t), which would give a non-integer time. Others use logarithms unnecessarily for such neat numbers. Always check first if the ratio A / P matches a known power of (1 + r/100) before resorting to more complex methods.

Final Answer:
Rs. 1,000 will amount to Rs. 1,331 in 3 years at 10% compound interest per annum.