Earnings, Simple and Compound Interest

Earnings questions usually combine a basic hourly rate with overtime at a higher rate.

Wage = hours × rate per hour.

Worked. "Maria earns $15 per hour. She works 38 hours in a week. Find her wage."

• $38 \times 15 = 570$, so wage is $570.

Overtime. Usually at $1.5\times$ ("time and a half") or $2\times$ ("double time") the basic rate.

Worked. "Tom earns $12 per hour basic, with overtime at time-and-a-half for hours over 40. He works 46 hours. Find his wage."

• Basic pay: $40 \times 12 = 480$, so $480.
• Overtime pay: 6 hours at $1.5 \times 12 = 18$ per hour → $6 \times 18 = 108$, so $108.
• Total: $588.

Salary is an annual figure; divide by $52$ for weekly pay or $12$ for monthly.

Income tax / deductions. If a percentage is deducted, use the multiplier method. "$25\%$ tax on wages" → take-home is $\times 0.75$ of gross.

Simple interest is interest paid on the original principal only — every year, you earn the same amount.

$I = \dfrac{P \times R \times T}{100}$

where $P$ is the principal (initial amount), $R$ is the annual interest rate (as a percentage), and $T$ is the time in years.

The TOTAL amount after $T$ years is $A = P + I = P\left(1 + \dfrac{RT}{100}\right)$.

Worked. "Sara invests $2,000 at $4\%$ simple interest for $5$ years. Find the interest and the total."

• $I = \dfrac{2000 \times 4 \times 5}{100} = 400$, so interest is $400.
• Total: $2000 + 400 = 2400$, so $2,400.

Mid-year periods. If $T$ is given in months, convert to a fraction of a year. "$8$ months" → $T = 8/12 = 2/3$.

Compound interest earns interest on interest. After $n$ years, the running total is

$A = P\left(1 + \dfrac{r}{100}\right)^{n}$

where $r$ is the annual rate (%).

The interest portion is $I = A - P$.

Worked. "Ahmed invests $2,000 at $4\%$ compound interest for $5$ years."

• $A = 2000 \times (1.04)^{5} = 2000 \times 1.21665 \approx 2433.31$, so $2,433.31.
• Interest earned: $433.31.

Compare with simple interest above ($400): compound earns about $33 more over the same period.

Why the multiplier? Each year's balance is multiplied by $(1 + r/100)$. Repeating $n$ times gives the formula above. The multiplier method we used in fractions/decimals is the same idea applied to time.

Compound on shorter intervals. If interest is compounded $m$ times per year (e.g. monthly $\Rightarrow m = 12$), the formula becomes $A = P\left(1 + \dfrac{r}{100m}\right)^{m n}.$ Cambridge usually states the interval — annually unless told otherwise.

Depreciation. A LOSS in value over time — a "negative" compound rate. $A = P\left(1 - \dfrac{r}{100}\right)^{n}.$

Worked. "A car is bought for $24,000 and depreciates by $15\%$ each year. Find its value after $4$ years."

• $A = 24000 \times (0.85)^{4} = 24000 \times 0.5220 \approx 12529.13$, so $12,529.13.

Profit and loss.

$\text{Profit} = \text{Selling price} - \text{Cost price}.$ $\text{Profit \%} = \dfrac{\text{Profit}}{\text{Cost price}} \times 100.$

Always express profit as a percentage of the COST price, not the selling price.

Worked. "Tariq buys a watch for $80 and sells it for $100. Find the profit %."

• Profit: $100 - 80 = 20$, so $20.
• Profit %: $\dfrac{20}{80} \times 100 = 25\%$ profit.

For a loss, the calculation is the same but $\text{Selling} < \text{Cost}$.

Discount. A percentage off the marked price.

Worked. "A jacket marked at $160 has a $20\%$ discount."

• Discount amount: $0.20 \times 160 = 32$, so $32 off.
• Sale price: $160 - 32 = 128$, or $160 \times 0.80 = 128$. Either way $128.

Reverse problems. "After a $20\%$ discount, a jacket sells for $128. Find the marked price."

• Multiplier: $0.80$. Marked price: $128 / 0.80 = 160$, so $160.