4.3 Percent Increase, Decrease & Percent Change

ACCUPLACER Arithmetic Exam

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1 Chapter 1: Whole Number Operations

1.1 Basic Whole Number Arithmetic 1.2 Order of Operations & Grouping Symbols 1.3 Number Properties, Factors & Prime Factorization 1.4 Whole Number Estimation, Rounding & Applied Word Problems

2 Chapter 2: Fraction Operations

2.1 Fraction Fundamentals & Simplifying 2.2 Addition & Subtraction of Fractions and Mixed Numbers 2.3 Multiplication & Division of Fractions and Mixed Numbers 2.4 Real-World Fraction Applications & Word Problems

3 Chapter 3: Decimal Operations

3.1 Decimal Place Value, Rounding & Comparing Decimals 3.2 Addition & Subtraction of Decimals 3.3 Multiplication & Division of Decimals 3.4 Decimal Word Problems & Financial Calculations

4 Chapter 4: Percent

4.1 Percent Concepts & Conversions 4.2 Percent of a Quantity & Percent Equations 4.3 Percent Increase, Decrease & Percent Change 4.4 Practical Percent Applications

5 Chapter 5: Number Comparisons & Equivalents

5.1 Ordering Mixed Numerical Forms 5.2 Equivalent Numerical Representations & Conversion Strategies 5.3 Ratios, Unit Rates & Proportional Reasoning

4.3 Percent Increase, Decrease & Percent Change

Key Takeaways Explain This Topic AI

Percent change measures relative change compared to the original starting value: Percent Change = (|New - Original| / Original) × 100%.

In all percent change calculations, the denominator MUST be the Original (starting) value, never the new value.

To calculate a new value after a percent increase, multiply the original by (1 + r); for a percent decrease, multiply by (1 - r).

Successive percent changes (e.g., a 20% increase followed by a 20% decrease) are NOT additive and do NOT return a quantity to its original value.

In word problems asking for an original value prior to a percent change, set up the algebraic equation Original × (1 ± r) = New and divide.

Last updated: August 2026

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.

1. The Universal Percent Change Formula

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

Percent Change = Amount of Change Original Value × 100 % = ∣ New Value − Original Value ∣ Original Value × 100 % \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\% Percent Change = Original Value Amount of Change × 100% = Original Value ∣ New Value − Original Value ∣ × 100%

Important

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.

2. Percent Increase

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

Percent Increase = New Value − Original Value Original Value × 100 % \text{Percent Increase} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\% Percent Increase = Original Value New Value − Original Value × 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?

Identify values: Original Value = 80 \text{Original Value} = 80 Original Value = 80 , New Value = 100 \text{New Value} = 100 New Value = 100 .

Calculate Amount of Increase: Amount of Change = 100 − 80 = 20 \text{Amount of Change} = 100 - 80 = 20 Amount of Change = 100 − 80 = 20

Apply Formula: Percent Increase = 20 80 × 100 % = 1 4 × 100 % = 25 % \text{Percent Increase} = \frac{20}{80} \times 100\% = \frac{1}{4} \times 100\% = 25\% Percent Increase = 80 20 × 100% = 4 1 × 100% = 25%

Answer: The price increased by 25 % 25\% 25% .

3. Percent Decrease

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

Percent Decrease = Original Value − New Value Original Value × 100 % \text{Percent Decrease} = \frac{\text{Original Value} - \text{New Value}}{\text{Original Value}} \times 100\% Percent Decrease = Original Value Original Value − New Value × 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?

Identify values: Original Value = 150 \text{Original Value} = 150 Original Value = 150 , New Value = 105 \text{New Value} = 105 New Value = 105 .

Calculate Amount of Decrease: Amount of Change = 150 − 105 = 45 \text{Amount of Change} = 150 - 105 = 45 Amount of Change = 150 − 105 = 45

Apply Formula: Percent Decrease = 45 150 × 100 % = 3 10 × 100 % = 30 % \text{Percent Decrease} = \frac{45}{150} \times 100\% = \frac{3}{10} \times 100\% = 30\% Percent Decrease = 150 45 × 100% = 10 3 × 100% = 30%

Answer: The price decreased by 30 % 30\% 30% .

4. Finding the New Value Directly

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

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

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

Worked Example 3: Direct New Value Calculation

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

Convert rate to decimal: r = 8 % = 0.08 r = 8\% = 0.08 r = 8% = 0.08 .

Calculate multiplier: 1 + r = 1 + 0.08 = 1.08 1 + r = 1 + 0.08 = 1.08 1 + r = 1 + 0.08 = 1.08 .

Multiply: New Population = 12 , 500 × 1.08 = 13 , 500 \text{New Population} = 12,500 \times 1.08 = 13,500 New Population = 12 , 500 × 1.08 = 13 , 500

Answer: The new population is 13 , 500 13,500 13 , 500 .

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.

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

Worked Example 4: Calculating Original Value Before Increase

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

Let P P P = Original Population.

Set up equation: P × ( 1 + 0.12 ) = 56 , 000 ⟹ 1.12 P = 56 , 000 P \times (1 + 0.12) = 56,000 \implies 1.12 P = 56,000 P × ( 1 + 0.12 ) = 56 , 000 ⟹ 1.12 P = 56 , 000

Solve for P P P : P = 56 , 000 1.12 = 5 , 600 , 000 112 = 50 , 000 P = \frac{56,000}{1.12} = \frac{5,600,000}{112} = 50,000 P = 1.12 56 , 000 = 112 5 , 600 , 000 = 50 , 000

Answer: The original population was 50 , 000 50,000 50 , 000 .

6. Successive (Consecutive) Percent Changes

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

Caution

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

Step-by-Step Demonstration

Suppose an item costs $100:

Apply 20 % 20\% 20% Increase: Price after increase = 100 × ( 1 + 0.20 ) = 100 × 1.20 = $ 120 \text{Price after increase} = 100 \times (1 + 0.20) = 100 \times 1.20 = \$120 Price after increase = 100 × ( 1 + 0.20 ) = 100 × 1.20 = $120

Apply 20 % 20\% 20% Decrease to the NEW price ( $ 120 \text{\textdollar}120 $ 120 ): Price after decrease = 120 × ( 1 − 0.20 ) = 120 × 0.80 = $ 96 \text{Price after decrease} = 120 \times (1 - 0.20) = 120 \times 0.80 = \$96 Price after decrease = 120 × ( 1 − 0.20 ) = 120 × 0.80 = $96

Net Result: The final price is $96, which represents an overall 4 % 4\% 4% net decrease from the original $100!

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

Summary Reference Table of Formulas

Operation Given Values Formula

Percent Increase Original, New New − Original Original × 100 % \frac{\text{New} - \text{Original}}{\text{Original}} \times 100\% Original New − Original × 100%

Percent Decrease Original, New Original − New Original × 100 % \frac{\text{Original} - \text{New}}{\text{Original}} \times 100\% Original Original − New × 100%

New Value (Increase) Original, Rate r r r Original × ( 1 + r ) \text{Original} \times (1 + r) Original × ( 1 + r )

New Value (Decrease) Original, Rate r r r Original × ( 1 − r ) \text{Original} \times (1 - r) Original × ( 1 − r )

Original Value New, Rate r r r New 1 ± r \frac{\text{New}}{1 \pm r} 1 ± r New
ACCUPLACER Exam Traps & Common Errors

Warning

Trap 1: Using the New Value in the Denominator If a price increases from $80 to $100, calculating 20 100 = 20 % \frac{20}{100} = 20\% 100 20 = 20% is WRONG. The original price is $80, so the correct calculation is 20 80 = 25 % \frac{20}{80} = 25\% 80 20 = 25% .

Warning

Trap 2: Assuming Consecutive Percents Cancel Out A 10 % 10\% 10% markup followed by a 10 % 10\% 10% discount does not leave the price unchanged. For a $100 item, 100 × 1.10 = 110 100 \times 1.10 = 110 100 × 1.10 = 110 , and 110 × 0.90 = 99 110 \times 0.90 = 99 110 × 0.90 = 99 (a 1 % 1\% 1% net loss).

Warning

Trap 3: Subtracting Percentages Directly from New Values When Reversing If an item costs $110 after a 10 % 10\% 10% markup, taking 10 % 10\% 10% of $110 ( $ 11 \text{\textdollar}11 $ 11 ) and subtracting it gives $99, not the original $100. Always divide by ( 1 + r ) (1 + r) ( 1 + r ) : 110 ÷ 1.10 = 100 110 \div 1.10 = 100 110 ÷ 1.10 = 100 .