3.4 Percentage Change, Discounts, Successive Changes, and Reverse Percentages

Many LANTITE Numeracy questions involve quantities that change over time—such as school enrolments, bookshop discounts, Goods and Services Tax (GST), and attendance rates. Mastering single-step multipliers allows you to handle forward changes, successive changes, and reverse percentage problems without falling into common additive traps.

Percentage Increase and Decrease Using Single-Step Multipliers

Treat the original value as $100\%$ (or $1$). Increasing a value by $r\%$ means you now have $(100 + r)\%$ of the original, while decreasing (discounting) a value by $r\%$ leaves $(100 - r)\%$ of the original:

$$\text{New Value} = \text{Original Value} \times \left(1 \pm \frac{r}{100}\right)$$

• Increase of $r\%$: Multiply by $1 + \frac{r}{100}$ (a $15\%$ increase uses $1.15$). For example, if a school's $4,500 library technology budget is increased by $12\%$, the new budget is $4{,}500 \times 1.12 = 5{,}040$ ($5,040).
• Discount (decrease) of $r\%$: Multiply by $1 - \frac{r}{100}$ (a $15\%$ discount uses $0.85$). For example, if an $850 excursion bus charter receives an $8\%$ early-booking discount, the payable cost is $850 \times 0.92 = 782$ ($782).
• Adding $10\%$ Australian GST: Multiply the pre-tax price by $1 + 0.10 = 1.10$.

Calculating Relative Percentage Change

When given the initial and final values of a quantity, find the percentage increase or decrease using:

$$\text{Percentage Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$$

Critical Rule: Always divide the change by the original (starting) value, never by the final value. For example, if a school's annual energy bill drops from $25,000 to $21,500, the decrease is $3,500, and the percentage decrease is:

$$\frac{3500}{25000} \times 100\% = \frac{35}{250} \times 100\% = 14\%$$

Similarly, if student borrowings from the school library rise from $1{,}250$ books in Term 1 to $1{,}475$ books in Term 2, the increase is $1{,}475 - 1{,}250 = 225$ books, and the percentage increase is $\frac{225}{1250} \times 100\% = 18\%$.

Successive Percentage Changes

When two percentage changes occur one after another, the second percentage acts on the intermediate updated value, not on the original starting value. You must multiply the individual multipliers rather than adding or subtracting the percentages:

$$m_{\text{net}} = m_1 \times m_2$$

Trap 1: Equal Percentage Increase and Decrease Do Not Cancel Out

Suppose an excursion venue raises its hire fee by $20\%$ during peak season and then offers schools a $20\%$ off-peak discount on that higher fee. The combined multiplier is:

$$1.20 \times 0.80 = 0.96$$

Because $0.96 = 96\%$ of the original fee, the result is a net $4\%$ decrease ($100\% - 96\% = 4\%$), not $0\%$ change. The $20\%$ discount is calculated on the larger $120\%$ amount ($0.20 \times 120\% = 24\%$), pulling the price $4\%$ below the starting value.

Trap 2: Successive Discounts Do Not Add Directly

If an educational supplier advertises $20\%$ off storewide and offers teachers an additional $10\%$ off the already reduced price, the combined multiplier is:

$$0.80 \times 0.90 = 0.72$$

Paying $72\%$ of the original price corresponds to a total discount of $100\% - 72\% = 28\%$, not $30\%$.

Reverse Percentages: Finding the Original Value

In a reverse percentage problem, you are given the final value after an increase or discount has already been applied, and you must recover the unknown original value. Because the forward operation multiplied the original value by $(1 \pm \frac{r}{100})$, the reverse operation divides the final value by that same multiplier:

$$\text{Original Value} = \frac{\text{Final Value}}{1 \pm \frac{r}{100}}$$

Worked Example 1 (Reverse Discount): A school purchases a teacher laptop for $1,190 after receiving a $15\%$ educational discount. What was the original retail price before the discount?

• A $15\%$ discount means the school paid $100\% - 15\% = 85\%$ ($0.85$) of the original price.
• Divide the discounted price by $0.85$:

$$\text{Original Price} = \frac{1{,}190}{0.85} = 1{,}400\text{ dollars}$$

Why multiplying $1{,}190 \times 1.15 = 1{,}368.50$ is wrong: The $15\%$ discount was $15\%$ of the original price ($1,400), not $15\%$ of the reduced price ($1,190).

Worked Example 2 (Reverse Increase and GST): If a school's current enrolment of $690$ students represents a $15\%$ increase over its previous census count, then $690$ equals $115\%$ ($1.15$) of the original count, so the original enrolment was $\frac{690}{1.15} = 600$ students. Likewise, if an invoice total including $10\%$ GST is $770, the pre-GST price is $\frac{770}{1.10} = 700$, or $700 (and the GST component is $70, which is $\frac{1}{11}$ of $770, not $10\%$ of $770).

Percentage Points vs. Relative Percentage Change

When comparing two figures that are already percentages (such as attendance rates or literacy proficiency rates), distinguish carefully between two phrases:

• Percentage point change (absolute difference): Subtract the initial percentage from the final percentage ($\text{New Rate} - \text{Original Rate}$).
• Relative percentage change: Divide the percentage point change by the original percentage rate ($\frac{\text{New Rate} - \text{Original Rate}}{\text{Original Rate}} \times 100\%$).

Worked Example: If a school's attendance rate rises from $80\%$ in Term 1 to $88\%$ in Term 2:

• Percentage point increase: $88\% - 80\% = 8\text{ percentage points}$.
• Relative percentage increase: $\frac{8}{80} \times 100\% = 10\%$.