Percent Increase
and Decrease

Master the art of calculating how much a value grows or shrinks using percent change: a core skill for the SSAT and everyday math.

SSAT Middle Level

What Is a Percent?

A percent means "per hundred." Writing $ 25\% $ is the same as writing the fraction $ \frac{25}{100} $ or the decimal $ 0.25 $. Percents let us describe parts of a whole using a familiar, consistent scale of 100.

What Is Percent Change?

When a value goes up or down, we can describe how much it changed as a percent of the original value. This is called percent change.

Percent Increase

The new value is larger than the original. The price went up, the population grew, or you scored higher.

Percent Decrease

The new value is smaller than the original. The price dropped, the temperature fell, or your time improved.

Why Does Percent Change Matter?

Raw numbers can be misleading. A $10 increase on a $20 item is huge, but on a $1,000 item it's tiny. Percent change gives you a fair comparison by measuring change relative to the starting point.

Shopping Sales

"$30\%$ off" tells you exactly how much you save per dollar.

Test Scores

"Improved by $15\%$" shows growth regardless of total points.

SSAT Problems

Percent change questions appear in nearly every quantitative section.

The Core Formulas

Visualizing the Formula

Increase vs. Decrease

	Percent Increase	Percent Decrease
Direction	Value goes up	Value goes down
Numerator	New $-$ Original	Original $-$ New
Denominator	Original	Original
Example	$50 \to 65$: $30\%$	$80 \to 60$: $25\%$
Key check	$ \text{New} > \text{Original} $	$ \text{New} < \text{Original} $

Finding the New Value

Sometimes the problem gives you the percent change and asks you to find the new value. Use these shortcuts:

How to Solve Percent Change

1

Identify the original and new values. The original is the starting or "before" number.

2

Find the change. Subtract: $\text{Change} = |\text{New} - \text{Original}|$.

3

Divide by the original. $\frac{\text{Change}}{\text{Original}}$.

4

Multiply by 100. Convert the decimal to a percent.

5

Label it. State whether it is an increase or decrease.

Example 1: Price Increase

A video game costs $40. The price rises to $50. What is the percent increase?

Example 2: Score Decrease

A student scored 90 on a quiz last week and 72 this week. What is the percent decrease?

1

Identify values: Original = 90, New = 72

2

Find the change: $90 - 72 = $ 18

3

Divide by original: $\frac{18}{90} = 0.2$

4

Multiply by 100: $0.2 \times 100 = $ $20\%$

5

Label: This is a $20\%$ decrease

Example 3: Finding the New Value

A bicycle costs $120. It is on sale for $30\%$ off. What is the sale price?

Example 4: Finding the Original

After a $20\%$ increase, a number becomes 60. What was the original number?

1

Set up the equation: $\text{Original} \times 1.20 = 60$

2

Solve for Original: $\text{Original} = \frac{60}{1.20}$

3

Calculate: $\frac{60}{1.20} = $ 50

Common Mistakes to Avoid

✕ Wrong Denominator

Dividing the change by the new value instead of the original. Always use the starting number.

✕ Reversing the Operation

To find the original after an increase, divide by the multiplier: don't subtract the percent from the new value.

✕ Forgetting to Multiply by 100

Getting $0.25$ is correct math, but the answer is $25\%$. Don't forget the final step.

Quick Reference Summary

Task	Formula	Example
Find % increase	$\frac{\text{New} - \text{Orig}}{\text{Orig}} \times 100$	$\frac{65-50}{50} \times 100 = 30\%$
Find % decrease	$\frac{\text{Orig} - \text{New}}{\text{Orig}} \times 100$	$\frac{80-60}{80} \times 100 = 25\%$
New after increase	$\text{Orig} \times (1 + \frac{p}{100})$	$200 \times 1.15 = 230$
New after decrease	$\text{Orig} \times (1 - \frac{p}{100})$	$200 \times 0.85 = 170$
Find original	$\frac{\text{New}}{\text{Multiplier}}$	$\frac{60}{1.20} = 50$

Question 1 of 5

A shirt was priced at $40 and is now $52. What is the percent increase?

The change is $52 - 40 = 12$. Divide by the original: $\frac{12}{40} = 0.30$. Multiply by 100 to get $30\%$.

Question 2 of 5

A town's population dropped from 5,000 to 4,000. What is the percent decrease?

The change is $5{,}000 - 4{,}000 = 1{,}000$. Divide by the original: $\frac{1{,}000}{5{,}000} = 0.20$. That gives $20\%$ decrease.

Question 3 of 5

A jacket costs $80 and is on sale for $25\%$ off. What is the sale price?

$25\%$ of $\$80$ is $80 \times 0.25 = \$20$. The sale price is $\$80 - 20 = \mathbf{\$60}$. Or, using the multiplier: $\$80 \times 0.75 = \mathbf{\$60}$.

Question 4 of 5

After a $40\%$ increase, a number becomes 70. What was the original number?

A 40% increase means the multiplier is $1.40$. So, $\text{Original} \times 1.40 = 70$, which gives $\text{Original} = \frac{70}{1.40} = \mathbf{50}$.

Question 5 of 5

A store raises a $60 item by $10\%$, then puts it on sale for $10\%$ off the new price. What is the final price?

After 10% increase: $60 \times 1.10 = \$66$. After 10% decrease: $66 \times 0.90 = \mathbf{\$59.40}$. A 10% increase followed by a 10% decrease does NOT return to the original: the base changed!

Task	Formula	Example
Find % increase	\(\frac{\text{New} - \text{Orig}}{\text{Orig}} \times 100\)	\(\frac{65-50}{50} \times 100 = 30\%\)
Find % decrease	\(\frac{\text{Orig} - \text{New}}{\text{Orig}} \times 100\)	\(\frac{80-60}{80} \times 100 = 25\%\)
New after increase	\(\text{Orig} \times (1 + \frac{p}{100})\)	\(200 \times 1.15 = 230\)
New after decrease	\(\text{Orig} \times (1 - \frac{p}{100})\)	\(200 \times 0.85 = 170\)
Find original	\(\frac{\text{New}}{\text{Multiplier}}\)	\(\frac{60}{1.20} = 50\)

Percent Increaseand Decrease

What Is a Percent?

What Is Percent Change?

Why Does Percent Change Matter?

The Core Formulas

Visualizing the Formula

Increase vs. Decrease

Finding the New Value

How to Solve Percent Change

Example 1: Price Increase

Example 2: Score Decrease

Example 3: Finding the New Value

Example 4: Finding the Original

Common Mistakes to Avoid

Quick Reference Summary

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Percent Increase
and Decrease