Q.Find the compound interest on ₹10,000 for 2 years at 10% per annum, compounded yearly.

Compound interest is found by first calculating the amount using $A = P(1+r/100)^t$ and then subtracting the principal.

$A = 10{,}000\left(1+\dfrac{10}{100}\right)^{2} = 10{,}000(1.1)^{2} = 10{,}000 \times 1.21 = 12{,}100$

$CI = A - P = 12{,}100 - 10{,}000 = 2{,}100$

Here $P = ₹10{,}000$, $r = 10\%$ per annum, $t = 2$ years, compounded yearly.

Step 1 — Apply the yearly compound-interest formula.

Step 2 — Evaluate. $(1.1)^2 = 1.21$, so

Step 3 — Find the compound interest.

Year-by-year build-up (verification method): Year 1 interest $= 10\%$ of $10{,}000 = ₹1{,}000$, so the amount after Year 1 is $₹11{,}000$. Year 2 interest is now $10\%$ of the new principal $₹11{,}000$, i.e. $₹1{,}100$, giving an amount of $₹12{,}100$ after Year 2 — the same answer as the direct formula, confirming it. This step-by-step build-up is exactly what the formula $A=P(1+r/100)^t$ compresses into a single calculation.

The most common error is applying simple interest instead of compound interest — i.e. computing $10\%$ of ₹10,000 for both years and getting $₹2{,}000$ instead of $₹2{,}100$. Simple interest charges the same ₹1,000 every year on the original principal; compound interest charges interest on the growing amount, which is why the second year's interest (₹1,100) is larger than the first year's (₹1,000).