4.3 Percent Increase, Decrease & Percent Change

4.3 Percent Increase, Decrease & Percent Change

In real-world applications and ACCUPLACER word problems, quantities frequently change over time. Percent change measures the relative growth or decline of a quantity compared to its starting point, expressed as a percentage of that initial value.

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1. The Universal Percent Change Formula

Whether a quantity grows (percent increase) or shrinks (percent decrease), the fundamental formula for percent change remains identical:

$\text{Percent Change} = \frac{\text{Amount of Change}}{\text{Original Value}} \times 100\% = \frac{|\text{New Value} - \text{Original Value}|}{\text{Original Value}} \times 100\%$

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2. Percent Increase

A percent increase occurs when the new value is greater than the original value ($\text{New} > \text{Original}$).

$\text{Percent Increase} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$

Common Real-World Contexts

• Retail price markups
• Salary raises and wage growth
• Population expansion
• Inflation and cost-of-living adjustments

Worked Example 1: Percent Increase Calculation

Problem: The price of a textbook increased from $80 to $100. What was the percent increase?

1. Identify values: $\text{Original Value} = 80$, $\text{New Value} = 100$.
2. Calculate Amount of Increase: $\text{Amount of Change} = 100 - 80 = 20$
3. Apply Formula: $\text{Percent Increase} = \frac{20}{80} \times 100\% = \frac{1}{4} \times 100\% = 25\%$
4. Answer: The price increased by $25\%$.

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3. Percent Decrease

A percent decrease occurs when the new value is less than the original value ($\text{New} < \text{Original}$).

$\text{Percent Decrease} = \frac{\text{Original Value} - \text{New Value}}{\text{Original Value}} \times 100\%$

Common Real-World Contexts

• Store discounts and clearance sales
• Asset depreciation (e.g., car value loss)
• Weight loss or reduction in enrollment/attendance
• Target budget cuts

Worked Example 2: Percent Decrease Calculation

Problem: A winter coat originally priced at $150 is put on sale for $105. What is the percent decrease?

1. Identify values: $\text{Original Value} = 150$, $\text{New Value} = 105$.
2. Calculate Amount of Decrease: $\text{Amount of Change} = 150 - 105 = 45$
3. Apply Formula: $\text{Percent Decrease} = \frac{45}{150} \times 100\% = \frac{3}{10} \times 100\% = 30\%$
4. Answer: The price decreased by $30\%$.

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4. Finding the New Value Directly

If you know the original value and the percent change rate $r$ (in decimal form), you can calculate the new value directly in one step:

$\text{New Value (Increase)} = \text{Original Value} \times (1 + r)$

$\text{New Value (Decrease)} = \text{Original Value} \times (1 - r)$

Worked Example 3: Direct New Value Calculation

Problem: A town's population was $12,500$ last year. If the population grew by $8\%$ this year, what is the new population?

1. Convert rate to decimal: $r = 8\% = 0.08$.
2. Calculate multiplier: $1 + r = 1 + 0.08 = 1.08$.
3. Multiply: $\text{New Population} = 12,500 \times 1.08 = 13,500$
4. Answer: The new population is $13,500$.

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5. Finding the Original Value (Working Backward)

When given the new value after a percent change and asked for the original value, do not take the percent of the new value! Set up an algebraic equation.

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

Worked Example 4: Calculating Original Value Before Increase

Problem: A city's population grew by $12\%$ over five years to reach a current population of $56,000$. What was the original population five years ago?

1. Let $P$ = Original Population.
2. Set up equation: $P \times (1 + 0.12) = 56,000 \implies 1.12 P = 56,000$
3. Solve for $P$: $P = \frac{56,000}{1.12} = \frac{5,600,000}{112} = 50,000$
4. Answer: The original population was $50,000$.

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6. Successive (Consecutive) Percent Changes

A major conceptual trap on the ACCUPLACER involves consecutive percent changes applied one after another.

Step-by-Step Demonstration

Suppose an item costs $100:

1. Apply $20\%$ Increase: $\text{Price after increase} = 100 \times (1 + 0.20) = 100 \times 1.20 = \$120$
2. Apply $20\%$ Decrease to the NEW price ($\text{\textdollar}120$): $\text{Price after decrease} = 120 \times (1 - 0.20) = 120 \times 0.80 = \$96$
3. Net Result: The final price is $96, which represents an overall $4\%$ net decrease from the original $100!

Why does this happen? The $20\%$ decrease is calculated on a larger base ($\text{\textdollar}120$) than the original $20\%$ increase ($\text{\textdollar}100$).

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Summary Reference Table of Formulas

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ACCUPLACER Exam Traps & Common Errors