4.3 Percent Increase, Decrease & Percent Change

4.3 Percent Increase, Decrease & Percent Change\n

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.\n ---\n

1. The Universal Percent Change Formula\n

Whether a quantity grows (percent increase) or shrinks (percent decrease), the fundamental formula for percent change remains identical:\n $\text{Percent Change} = \frac{\text{Amount of Change}}{\text{Original Value}} \times 100\% = \frac{|\text{New Value} - \text{Original Value}|}{\text{Original Value}} \times 100\%$\n

[!IMPORTANT]\n> The denominator of the percent change formula is ALWAYS the Original Value (the starting amount before the change occurred). Using the new value in the denominator is the single most common error on the ACCUPLACER.\n ---\n

2. Percent Increase\n

A percent increase occurs when the new value is greater than the original value ($\text{New} > \text{Original}$).\n $\text{Percent Increase} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$\n

Common Real-World Contexts\n- Retail price markups\n- Salary raises and wage growth\n- Population expansion\n- Inflation and cost-of-living adjustments\n

Worked Example 1: Percent Increase Calculation\nProblem: The price of a textbook increased from $\$80$ to $\$100$. What was the percent increase?\n

1. Identify values: $\text{Original Value} = 80$, $\text{New Value} = 100$.\n2. Calculate Amount of Increase:\n $\text{Amount of Change} = 100 - 80 = 20$\n3. Apply Formula:\n $\text{Percent Increase} = \frac{20}{80} \times 100\% = \frac{1}{4} \times 100\% = 25\%$\n4. Answer: The price increased by $25%$.\n ---\n

3. Percent Decrease\n

A percent decrease occurs when the new value is less than the original value ($\text{New} < \text{Original}$).\n $\text{Percent Decrease} = \frac{\text{Original Value} - \text{New Value}}{\text{Original Value}} \times 100\%$\n

Common Real-World Contexts\n- Store discounts and clearance sales\n- Asset depreciation (e.g., car value loss)\n- Weight loss or reduction in enrollment/attendance\n- Target budget cuts\n

Worked Example 2: Percent Decrease Calculation\nProblem: A winter coat originally priced at $\$150$ is put on sale for $\$105$. What is the percent decrease?\n

1. Identify values: $\text{Original Value} = 150$, $\text{New Value} = 105$.\n2. Calculate Amount of Decrease:\n $\text{Amount of Change} = 150 - 105 = 45$\n3. Apply Formula:\n $\text{Percent Decrease} = \frac{45}{150} \times 100\% = \frac{3}{10} \times 100\% = 30\%$\n4. Answer: The price decreased by $30%$.\n ---\n

4. Finding the New Value Directly\n

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

Worked Example 3: Direct New Value Calculation\nProblem: A town's population was $12,500$ last year. If the population grew by $8%$ this year, what is the new population?\n

1. Convert rate to decimal: $r = 8% = 0.08$.\n2. Calculate multiplier: $1 + r = 1 + 0.08 = 1.08$.\n3. Multiply:\n $\text{New Population} = 12,500 \times 1.08 = 13,500$\n4. Answer: The new population is $13,500$.\n ---\n

5. Finding the Original Value (Working Backward)\n

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

Worked Example 4: Calculating Original Value Before Increase\nProblem: 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?\n

1. Let $P$ = Original Population.\n2. Set up equation:\n $P \times (1 + 0.12) = 56,000 \implies 1.12 P = 56,000$\n3. Solve for $P$:\n $P = \frac{56,000}{1.12} = \frac{5,600,000}{112} = 50,000$\n4. Answer: The original population was $50,000$.\n ---\n

6. Successive (Consecutive) Percent Changes\n

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

[!CAUTION]\n> Percent changes are NEVER additive! A $20%$ increase followed by a $20%$ decrease does NOT return a price to its original starting value.\n

Step-by-Step Demonstration\n

Suppose an item costs $\$100$:\n1. Apply $20%$ Increase:\n $\text{Price after increase} = 100 \times (1 + 0.20) = 100 \times 1.20 = \$120$\n2. Apply $20%$ Decrease to the NEW price ($$120$):\n $\text{Price after decrease} = 120 \times (1 - 0.20) = 120 \times 0.80 = \$96$\n3. Net Result: The final price is $\$96$, which represents an overall $4%$ net decrease from the original $\$100$!\n Why does this happen? The $20%$ decrease is calculated on a larger base ($$120$) than the original $20%$ increase ($$100$).\n ---\n

Summary Reference Table of Formulas\n

| Operation | Given Values | Formula |\n| :--- | :--- | :--- |\n| Percent Increase | Original, New | $\frac{\text{New} - \text{Original}}{\text{Original}} \times 100%$ |\n| Percent Decrease | Original, New | $\frac{\text{Original} - \text{New}}{\text{Original}} \times 100%$ |\n| New Value (Increase) | Original, Rate $r$ | $\text{Original} \times (1 + r)$ |\n| New Value (Decrease) | Original, Rate $r$ | $\text{Original} \times (1 - r)$ |\n| Original Value | New, Rate $r$ | $\frac{\text{New}}{1 \pm r}$ |\n ---\n

ACCUPLACER Exam Traps & Common Errors\n

[!WARNING]\n> Trap 1: Using the New Value in the Denominator\n> If a price increases from $\$80$ to $\$100$, calculating $\frac{20}{100} = 20%$ is WRONG. The original price is $\$80$, so the correct calculation is $\frac{20}{80} = 25%$.\n [!WARNING]\n> Trap 2: Assuming Consecutive Percents Cancel Out\n> A $10%$ markup followed by a $10%$ discount does not leave the price unchanged. For a $\$100$ item, $100 \times 1.10 = 110$, and $110 \times 0.90 = 99$ (a $1%$ net loss).\n [!WARNING]\n> Trap 3: Subtracting Percentages Directly from New Values When Reversing\n> If an item costs $\$110$ after a $10%$ markup, taking $10%$ of $\$110$ ($$11$) and subtracting it gives $\$99$, not the original $\$100$. Always divide by $(1 + r)$: $110 \div 1.10 = 100$.