Rather than calculating a percentage and then adding or subtracting it as a separate step, build a single decimal multiplier:

Example (a): increase \260$by$12\%$: multiplier$= 1.12$, so$\260 \times 1.12 = \291.20$.

Example (b): a \85$jacket is discounted by$20\%$: multiplier$= 0.8$, so$\85 \times 0.8 = \68$.

Example: a phone bought for \420$is sold for$\$357$. Find the percentage loss.

1. Change: \420 - \357 = \63$.
2. Divide by the original value: $\dfrac{63}{420} \times 100\% = 15\%$.
3. Since the value decreased, this is a $15\%$ loss.

where $P$ is the principal (starting amount), $R$ is the interest rate per year, and $T$ is the number of years. If a question asks for the total amount in an account, remember to add the interest back onto the principal: $P + I$.

Unlike simple interest, compound interest is recalculated on the new total each year, not just the original principal — so the amount grows faster over time. Use the multiplier method from earlier, raised to the power of the number of years:

where $P$ is the principal, $R$ is the annual interest rate, and $n$ is the number of years.

Example: \2000$is invested at$4\%$per year compound interest. Find the total value after$3$ years.

1. Multiplier for a $4\%$ increase: $1.04$.
2. Apply it $3$ times (once per year): $2000 \times 1.04^3$.
3. Evaluate: 2000 \times 1.124864 = \2249.73$ (2 d.p.).

• Dividing by the new value instead of the original when finding a percentage change.
• Using a multiplier that's the wrong size, especially for percentages under $10\%$ — always include the leading zero, e.g. $1.05$ not $1.5$.
• Stopping at the interest value $I$ when the question actually asks for the total amount, $P + I$.