Percent Change and Growth Factors

Introduction

Percent change questions appear throughout the SAT's Problem-Solving and Data Analysis domain (≈15% of the test). You'll encounter prices rising and falling, populations growing, quantities depreciating — and the SAT loves to chain two or more changes together or ask you to work backward to an original value. Mastering the growth-factor approach makes every one of these problems fast and error-proof.

This note covers percent increase and decrease, multipliers (growth factors), successive percent changes, and reversing a percent change. For the foundational idea of finding the part, whole, or percent from the other two, see the sibling note Percent Basics: Part, Whole, and Percent.

Core Concept

From percent to multiplier

Any percent change can be written as a single multiplication:

Increase by $p\%$ → multiply by $1 + \dfrac{p}{100}$
Decrease by $p\%$ → multiply by $1 - \dfrac{p}{100}$

These multipliers are called growth factors (sometimes scale factors or multipliers). For example:

Situation	Growth factor
5% increase	$1.05$
15% decrease	$0.85$
100% increase (doubling)	$2.00$
125% of original	$1.25$
8% decrease	$0.92$

Key idea
Key idea
Key insight: "125% of original" means growth factor 1.25, not 0.25. Percentages ≥ 100% mean the new value is at least as large as the original.

Successive percent changes — always multiply the factors

If a quantity undergoes two back-to-back percent changes, multiply both growth factors:

\text{Final value} = \text{Original} \times (\text{factor}_1) \times (\text{factor}_2)

Critical trap: You cannot add or subtract the percents. A 15% increase followed by a 10% decrease is not a net 5% increase. The actual net factor is $1.15 \times 0.90 = 1.035$ , a 3.5% net increase.

Reversing a percent change to find the original

If you know the value after a percent change and need the original, divide by the growth factor:

\text{Original} = \frac{\text{New value}}{\text{Growth factor}}

For example, if a price after a 15% increase is $345:

\text{Original} = \frac{345}{1.15} = 300

Do not subtract 15% from the new value — that gives 15% of the new price, not the original.

Key Formulas & Rules

Must memorize (not on the reference sheet):

\boxed{\text{New value} = \text{Original} \times \left(1 \pm \frac{p}{100}\right)}

\boxed{\text{Percent change} = \frac{\text{New} - \text{Original}}{\text{Original}} \times 100}

\boxed{\text{Original} = \frac{\text{New value}}{\text{Growth factor}}}

\boxed{\text{Successive changes: Final} = \text{Original} \times f_1 \times f_2 \times \cdots}

where each $f_i = 1 + \dfrac{p_i}{100}$ (use $-$ for a decrease).

Worked Examples

Example 1

A store's monthly revenue was $4,000 in January. In February, revenue increased by 15% from January's amount. In March, revenue decreased by 10% from February's amount. What was the store's revenue in March?

A) $3,600
B) $4,140
C) $4,200
D) $4,600

Solution

Step 1 — February: apply a 15% increase.

4000 \times 1.15 = 4600

Step 2 — March: apply a 10% decrease to February's revenue.

4600 \times 0.90 = 4140

Verify: $4000 \times 1.15 \times 0.90 = 4000 \times 1.035 = 4140$ ✓

Why each distractor fails:

A) $3,600 — applies only the 10% decrease to the original, ignoring the February increase: $4000 \times 0.90 = 3600$ .
C) $4,200 — adds the percents: $15\% - 10\% = 5\%$ net, then $4000 \times 1.05 = 4200$ . Percents in successive changes must be multiplied, not combined.
D) $4,600 — stops after February's increase: $4000 \times 1.15 = 4600$ .

Answer: B) $4,140

Desmos check

Desmos check: Type 4000 * 1.15 * 0.90 in the expression bar to confirm 4140 instantly.

Example 2

A price increased by 15%, reaching $345. What was the original price?

A) $293.25
B) $300
C) $330
D) $396.75

Solution

The new value equals the original times the growth factor:

\text{Original} \times 1.15 = 345

\text{Original} = \frac{345}{1.15} = 300

Verify: $300 \times 1.15 = 345$ ✓

Why each distractor fails:

A) $293.25 — subtracts 15% of the new price: $345 - 345 \times 0.15 = 345 - 51.75 = 293.25$ . This removes 15% of the wrong base.
C) $330 — subtracts the percent number as a raw dollar amount: $345 - 15 = 330$ .
D) $396.75 — applies another 15% increase to the new price instead of reversing: $345 \times 1.15 = 396.75$ .

Answer: B) $300

Common Mistakes & Traps

Mistake	What it looks like	Correct approach
Adding percents in successive changes	$+15\%$ then $-10\%$ treated as $+5\%$ net	Multiply the factors: $1.15 \times 0.90$
Subtracting % from the new value to reverse	$345 - 0.15(345)$ to find original	Divide: $345 \div 1.15$
Treating "125% of original" as a 125% increase	Multiplying by $2.25$	"125% of" means multiply by $1.25$
Applying a change to the wrong base	Subtracting 10% of January to find March	Always apply the decrease to February's value
Stopping one step early	Reporting the intermediate value	Read the question: it asks for the final value
Treating a percent as a dollar/unit amount	Subtracting 15 dollars for "15% increase"	Convert to decimal, multiply by the whole

Practice Questions

Question 1

A laptop originally cost $X. Its price dropped by 30%, and the new price is $630. What was the original price?

A) $441
B) $660
C) $819
D) $900

Show answer

Answer: D) $900

Set up the equation: $X \times 0.70 = 630$ , so $X = 630 \div 0.70 = 900$ .

Verify: $900 \times 0.70 = 630$ ✓

A) $441 — multiplies the new price by 0.70 again: $630 \times 0.70 = 441$ .
B) $660 — adds $30 (the percent as a number) to 630: $630 + 30 = 660$ .
C) $819 — adds 30% of the new price: $630 \times 1.30 = 819$ (this inflates the new value instead of recovering the original).

Question 2 (Student-produced response)

A car was worth $25,000. It depreciated by 20% in year 1, then by 10% in year 2. What is the car's value (in dollars) at the end of year 2?

Show answer

Answer: 18000

Year 1: $25000 \times 0.80 = 20000$

Year 2: $20000 \times 0.90 = 18000$

Combined: $25000 \times 0.80 \times 0.90 = 25000 \times 0.72 = 18000$ ✓

Desmos check

Desmos check: Enter 25000 * 0.80 * 0.90 to confirm $18,000.

Question 3

A company's year 2 sales were 125% of its year 1 sales. If year 1 sales were $80,000, what were year 2 sales?

A) $20,000
B) $100,000
C) $160,000
D) $180,000

Show answer

Answer: B) $100,000

"125% of $80,000" means multiply by the growth factor 1.25:

80000 \times 1.25 = 100000

Verify: $100000 \div 80000 = 1.25 = 125\%$ ✓

A) $20,000 — takes only the 25% increase portion: $80000 \times 0.25 = 20000$ .
C) $160,000 — doubles the original (confuses 125% with 200%): $80000 \times 2 = 160000$ .
D) $180,000 — adds 125% to the original: $80000 + 80000 \times 1.25 = 180000$ .

Question 4

A quantity grows by 8% each month. Which expression correctly represents its value after 3 months, starting from an initial value of $V$ ?

A) $V + 3(0.08V)$
B) $V(1.08)^3$
C) $V(0.08)^3$
D) $V(1.08 \times 3)$

Show answer

Answer: B) $V(1.08)^3$

Each month the value is multiplied by the growth factor 1.08. After 3 months:

V \times 1.08 \times 1.08 \times 1.08 = V(1.08)^3

A) Adds 8% of the original three times (simple, non-compound growth): $V + 0.24V = 1.24V$ . This underestimates because it doesn't let each month's growth build on the previous month's value.
C) Uses only the rate $0.08$ , not the growth factor $1.08$ .
D) Multiplies the factor by 3 instead of raising it to the 3rd power: $V \times 3.24$ , which vastly overstates the result.

Desmos check

Desmos check: Enter V = 1000 then compute 1000 * 1.08^3 and 1000 + 3*(0.08*1000) to see they differ (1259.71 vs. 1240).

Question 5 (Student-produced response)

A school had 840 students in September. By October, enrollment rose to 882. By what percent did enrollment increase? (Enter a number only; do not enter a percent sign.)

Show answer

Answer: 5

\text{Percent change} = \frac{882 - 840}{840} \times 100 = \frac{42}{840} \times 100 = 0.05 \times 100 = 5

Verify: $840 \times 1.05 = 882$ ✓

Connections

Prerequisite: Percent Basics: Part, Whole, and Percent — you need to be comfortable converting between fractions, decimals, and percents before using growth factors.
Algebra overlap: Percent change equations like $X \times 1.15 = 345$ are linear equations in one variable; the same solve-by-dividing technique appears throughout the Algebra domain.
Exponential growth/decay (Advanced Math): When a growth factor is applied repeatedly over time — $A = P(1.08)^t$ — the skill here is the foundation. The SAT may set up a table or scenario where you must identify the growth factor before writing the exponential model.
Data analysis questions: Percent change appears in two-way tables and data displays where you compare categories; always identify the correct base (denominator) before computing.
Real-world contexts: Roughly 30% of SAT math questions are set in context. Percent change is one of the most common "real-world wrappers" — look for price changes, population growth, survey data, and scientific measurements.

Introduction

Core Concept

From percent to multiplier

Successive percent changes — always multiply the factors

Reversing a percent change to find the original

Key Formulas & Rules

Worked Examples

Example 1

Example 2

Common Mistakes & Traps

Practice Questions

Connections

Figures

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