Question: A loan of $\usd{}25\,000$ is charged $4\%$ per annum compound interest for $5$ years. Find the amount owed and the total interest (a) compounded yearly, (b) compounded quarterly.

Step 1. (a) The compound formula with $P = 25000$, $r = 4$, $n = 5$:

$$A = P\left(1 + \dfrac{r}{100}\right)^n = 25000(1.04)^5 = \usd{}30\,416.32$$

Step 2. Interest = final amount − principal:

$$30\,416.32 - 25\,000 = \usd{}5\,416.32$$

Step 3. (b) Quarterly compounding changes both knobs: the rate per period becomes $4 \div 4 = 1\%$, and the number of periods becomes $4 \times 5 = 20$:

$$A = 25000\left(1 + \dfrac{1}{100}\right)^{20} = \usd{}30\,504.75$$

Step 4. Interest $= 30\,504.75 - 25\,000 = \usd{}5\,504.75$ — slightly more than yearly compounding, because interest starts earning interest sooner.

⚠ Watch out: The paired adjustment is the whole trick: divide the rate by the number of periods per year AND multiply $n$ by it — doing only one of the two is the classic error. And $A$ is the amount, not the interest: subtract $P$ at the end. Round money to 2 d.p.

Question: (a) Mary borrowed $\usd{}60\,000$ for $4$ years at simple interest of $8\%$ per year. Find the interest and the total she repays. (b) Andrew borrowed $\usd{}50\,000$ for $3$ years and paid $\usd{}12\,000$ total interest. Find the interest rate.

Step 1. (a) The simple-interest formula, with $P = 60000$, $R = 8$, $T = 4$:

$$I = \dfrac{PRT}{100} = \dfrac{60000 \times 8 \times 4}{100} = \usd{}19\,200$$

Step 2. Total repaid = principal + interest:

$$60\,000 + 19\,200 = \usd{}79\,200$$

Step 3. (b) Same formula, different unknown — substitute what is known and solve:

$$12\,000 = \dfrac{50000 \times R \times 3}{100} = 1500R$$

Step 4.

$$R = \dfrac{12\,000}{1500} = 8\% \text{ per annum}$$

The same rearrangement finds a missing $T$ (loan period) or $P$: one formula, four possible unknowns.

⚠ Watch out: Simple interest is charged on the original principal every year — it never compounds. And keep $R$ as the number of percent ($8$, not $0.08$) when using $\frac{PRT}{100}$; mixing that with the decimal convention double-divides by 100.

Question: $\usd{}25\,000$ is borrowed for $5$ years at $4\%$ per annum. How much more interest does compound interest (compounded yearly) cost than simple interest?

Step 1. Simple interest — the same $4\%$ of $\usd{}25\,000$ every year:

$$I_{\text{simple}} = \dfrac{PRT}{100} = \dfrac{25000 \times 4 \times 5}{100} = \usd{}5\,000$$

Step 2. Compound interest — each year's interest joins the principal:

$$A = 25000(1.04)^5 = \usd{}30\,416.32 \;\Rightarrow\; I_{\text{compound}} = \usd{}5\,416.32$$

Step 3. The difference:

$$5\,416.32 - 5\,000 = \usd{}416.32$$

Step 4. Why compound wins (for the lender): simple interest earns $4\%$ of $25\,000 = \usd{}1\,000$ flat each year, while compound earns $4\%$ of a growing balance — $\usd{}1\,000$, then $\usd{}1\,040$, then $\usd{}1\,081.60$, … The gap widens every extra year.

⚠ Watch out: For one year the two are identical — the divergence only starts in year 2, and "which scheme is better" questions hinge on that. Compare interest with interest, not an amount with an interest; mixing $A$ and $I$ makes compound look absurdly worse.