Percentages

Introduction
Percentages are a fundamental concept in mathematics, widely used in everyday life and
critical for success on standardized tests like the SAT. Percentages allow us to express parts
of a whole, calculate increases or decreases, analyze data, and solve financial problems like
interest and taxes. This chapter explores all essential subtopics related to percentages,
emphasizing SAT-style problem-solving and real-world applications. The goal is to ensure
you understand percentages conceptually, solve problems efficiently, and apply them
accurately in various contexts.

1. Understanding Percentages
Concept in a Nutshell

A percentage represents a number as a fraction of 100. For example:

- 25% means 25 out of 100 or 1/4.
- 50% means 50 out of 100 or 1/2.

Percentages are often used to describe proportions, discounts, interest rates, and changes in
values. Imagine you score 75% on a test. This means you answered 75 questions correctly
out of every 100 questions.

Concept Explanation

Percentages can be converted between fractions and decimals:

- To convert a percentage to a decimal, divide by 100:

25% = 25 ÷ 100 = 0.25.

- To convert a percentage to a fraction, write it over 100 and simplify:

25% = 25/100 = 1/4.

- To convert a decimal to a percentage, multiply by 100:

0.25 = 0.25 × 100 = 25%.
Percentages are commonly used to express data comparisons, proportions, and
probabilities.

Solved Examples

Example 1: Converting Percentages

Problem: Convert 80% to a fraction and a decimal.
Solution:
As a fraction:
80% = 80/100 = 4/5.
As a decimal:
80% = 80 ÷ 100 = 0.8.
Final Answer: 4/5 and 0.8.

Example 2: Real-Life Application

Problem: If 70% of a class of 50 students passed a test, how many students passed?
Solution:
70% of 50 = 0.7 × 50 = 35.
Final Answer: 35 students passed.

2. Calculating Percentages of a Quantity

Concept in a Nutshell

Finding a percentage of a quantity is essential for problems involving discounts, taxes, or

proportions. For example, if an item costs $200 and you need to calculate 15% sales tax, the
tax amount is:
15% of 200 = 0.15 × 200 = $30.

Concept Explanation

The formula to find a percentage of a quantity is:

Percentage Amount = (Percentage ÷ 100) × Total Quantity

This skill is crucial for solving SAT problems involving percentage-based calculations in
word problems or charts.

Solved Examples
Example 1: Sales Tax
Problem: A laptop costs $800. If the sales tax rate is 8%, what is the total cost including
tax?
Solution:
Tax = 8% of 800 = (8 ÷ 100) × 800 = 64.
Total Cost = 800 + 64 = $864.
Final Answer: $864.

Example 2: Discount
Problem: A store offers a 20% discount on a jacket originally priced at $120. What is the
discounted price?
Solution:
Discount = 20% of 120 = (20 ÷ 100) × 120 = $24.
Discounted Price = 120 - 24 = $96.
Final Answer: $96.

3. Percentage Increase and Decrease

Concept in a Nutshell

Percentage increase or decrease measures how much a value changes relative to its original
value. For example:

- If a product’s price increases from $50 to $60, the percentage increase is:
Increase = (New Value - Original Value) ÷ Original Value × 100 = (60 - 50) ÷ 50 × 100 =
20%.

- Similarly, a decrease from $50 to $40 represents a 20% decrease.

Concept Explanation
The formula for percentage change is:
Percentage Change = (Change ÷ Original Value) × 100
where Change = New Value - Original Value.
- Use this formula for both increases and decreases.
This is particularly useful for solving word problems related to price changes, population
growth, or percentage loss.

Solved Examples

Example 1: Price Increase

Problem: The price of a concert ticket increased from $80 to $100. What is the percentage
increase?
Solution:
Change = 100 - 80 = 20.
Percentage Increase = (20 ÷ 80) × 100 = 25%.
Final Answer: 25%.

Example 2: Percentage Decrease

Problem: A car’s value dropped from $20,000 to $16,000. What is the percentage decrease?
Solution:
Change = 20,000 - 16,000 = 4,000.
Percentage Decrease = (4,000 ÷ 20,000) × 100 = 20%.
Final Answer: 20%.

4. Finding the Original Value

Concept in a Nutshell

Sometimes, you’re given the percentage and the new value, and you need to find the
original value. For example, if a product is discounted by 25% and now costs $75, you can
calculate the original price.

Concept Explanation

The formula to find the original value is:

Original Value = New Value ÷ (1 ± Percentage Change)

- Use "+" for percentage increase.

- Use "−" for percentage decrease.

Solved Examples
Example 1: Reverse Discount
Problem: A shirt is on sale for $60 after a 20% discount. What was the original price?
Solution:
Let the original price be x.
New Price = Original Price × (1 - Discount).
60 = x × (1 - 0.2).
60 = x × 0.8.
x = 60 ÷ 0.8 = 75.
Final Answer: $75.

Example 2: Reverse Percentage Increase

Problem: After a 15% increase, a property’s value is $230,000. What was its original value?
Solution:
Let the original value be x.
New Value = Original Value × (1 + Percentage Increase).
230,000 = x × (1 + 0.15).
230,000 = x × 1.15.
x = 230,000 ÷ 1.15 = 200,000.
Final Answer: $200,000.

5. Successive Percentage Changes

Concept in a Nutshell

When an item undergoes multiple percentage changes, the total effect is not the simple sum
of percentages. For example, if a price increases by 10% and then decreases by 20%, the
final change is calculated using successive percentages.

Concept Explanation

The formula for successive percentage changes is:

New Value = Original Value × (1 ± First Change) × (1 ± Second Change)

Solved Examples

Example 1: Successive Increase and Decrease

Problem: The price of a product increases by 25% and then decreases by 10%. If the
original price was $100, what is the final price?
Solution:
Step 1: Apply the 25% increase.
New Price = 100 × 1.25 = $125.
Step 2: Apply the 10% decrease.
Final Price = 125 × 0.9 = $112.50.
Final Answer: $112.50.

Example 2: Two Successive Increases

Problem: A house’s value increases by 10% in the first year and 15% in the second year. If
the original value was $300,000, what is the final value?
Solution:
Step 1: Apply the 10% increase.
New Value = 300,000 × 1.1 = $330,000.
Step 2: Apply the 15% increase.
Final Value = 330,000 × 1.15 = $379,500.
Final Answer: $379,500.

6. Applications of Percentages
Concept in a Nutshell

Percentages are essential for understanding taxes, interest, profit margins, and growth
rates. For example:
- A $50 item with 6% sales tax costs $50 + ($50 × 0.06) = $53.
- A $1,000 investment with 5% annual interest grows to $1,050 after one year.

Concept Explanation

Percentages are applied in real-world scenarios such as:

- Financial contexts (discounts, taxes, interest rates).
- Data analysis (population changes, test scores).
- Profit and loss (business revenue).

Solved Examples

Example 1: Profit Calculation

Problem: A business buys an item for $200 and sells it for $250. What is the percentage
profit?
Solution:
Profit = Selling Price - Cost Price = 250 - 200 = $50.
Percentage Profit = (Profit ÷ Cost Price) × 100 = (50 ÷ 200) × 100 = 25%.
Final Answer: 25%.

Example 2: Compound Interest

Problem: An investment of $2,000 earns 5% annual interest compounded yearly. What is
its value after 3 years?
Solution:
Compound Interest Formula:
Value = Principal × (1 + Rate)^Time.
Value = 2,000 × (1 + 0.05)^3 = 2,000 × 1.157625 = $2,315.25.
Final Answer: $2,315.25.

Conclusion
Percentages are a cornerstone of mathematical understanding and are frequently tested on
the SAT due to their practical importance and versatility. They help us analyze changes,
compare quantities, and solve complex real-world problems involving discounts, taxes,
profit margins, and growth rates. By mastering percentages, you gain a skill set applicable
not just in exams but in everyday life. Advanced problems often combine percentages with
ratios, proportions, and multi-step calculations, so practicing higher-level questions is
essential for excellence. Remember to break problems into smaller, manageable steps, use
formulas effectively, and check for logical consistency in your solutions.

Solved Examples of Higher Level of

Difficulty
Example 1: Multi-Step Percentage Problem
Problem: A store offers a 20% discount on a jacket. During a special promotion, an
additional 10% discount is applied to the already discounted price. If the jacket originally
costs $150, what is the final price?

Solution:
Step 1: Calculate the price after the first discount.
First Discount = 20% of $150 = 0.2 × 150 = $30.
Price after first discount = 150 - 30 = $120.

Step 2: Apply the second discount to the new price.

Second Discount = 10% of $120 = 0.1 × 120 = $12.
Price after second discount = 120 - 12 = $108.

Final Answer: The final price of the jacket is $108.

Example 2: Reverse Percentage Problem with Successive Changes

Problem: A stock’s value increased by 10% in the first year and then decreased by 15% in
the second year. If the final value of the stock is $229.50, what was the original value?

Solution:
Step 1: Let the original value be x. Apply the successive percentage changes in reverse.

After the first year, the value of the stock was:

x × 1.1.

After the second year, the value became:

(x × 1.1) × 0.85 = $229.50.

Step 2: Solve for x.

x × 1.1 × 0.85 = 229.50
x × 0.935 = 229.50
x = 229.50 ÷ 0.935
x = 245.31

Final Answer: The original value of the stock was $245.31.

Example 3: Compound Interest Problem

Problem: A savings account earns 4% annual interest compounded yearly. If the balance
after 2 years is $10,816, what was the initial deposit?

Solution:
Compound Interest Formula:
Final Value = Principal × (1 + Rate)^Time.

Let the initial deposit be P.

10,816 = P × (1 + 0.04)^2
10,816 = P × 1.0816

Solve for P:
P = 10,816 ÷ 1.0816
P = 10,000

Final Answer: The initial deposit was $10,000.

Example 4: Percentage and Proportion Combined

Problem: In a class of 60 students, 40% are boys. If 25% of the boys and 10% of the girls
participate in a science competition, how many students are participating?

Solution:
Step 1: Calculate the number of boys and girls.
Number of boys = 40% of 60 = 0.4 × 60 = 24.
Number of girls = Total students - Boys = 60 - 24 = 36.

Step 2: Calculate the number of participants.

Boys participating = 25% of 24 = 0.25 × 24 = 6.
Girls participating = 10% of 36 = 0.1 × 36 = 3.

Total participants = Boys participating + Girls participating = 6 + 3 = 9.

Final Answer: 9 students are participating in the science competition.

Example 5: Successive Percentage Change in Area

Problem: The length of a rectangle is increased by 20%, and its width is decreased by 10%.
What is the percentage change in the area of the rectangle?

Solution:
Step 1: Let the original dimensions of the rectangle be L and W. The original area is:
Area = L × W.

Step 2: Calculate the new dimensions.

New Length = L × 1.2.
New Width = W × 0.9.
New Area = (L × 1.2) × (W × 0.9) = 1.08 × (L × W).

Step 3: Calculate the percentage change in area.

Percentage Change = ((New Area - Original Area) ÷ Original Area) × 100
= ((1.08LW - LW) ÷ LW) × 100
= (0.08LW ÷ LW) × 100
= 8%.

Final Answer: The area increases by 8%.

Example 6: Advanced Real-Life Problem with Percentages and Unit Conversion

Problem: A car travels 150 miles on 5 gallons of gas. Another car travels 240 kilometers on
8 liters of gas. Which car has better fuel efficiency? (1 mile = 1.609 kilometers, 1 gallon =
3.785 liters)

Solution:
Step 1: Convert both distances to kilometers and fuel to liters.
Car 1 Distance = 150 × 1.609 = 241.35 kilometers.
Car 1 Fuel = 5 × 3.785 = 18.925 liters.
Car 2 Distance = 240 kilometers (no conversion needed).
Car 2 Fuel = 8 liters.

Step 2: Calculate fuel efficiency for both cars.

Car 1 Efficiency = 241.35 ÷ 18.925 ≈ 12.75 km/l.
Car 2 Efficiency = 240 ÷ 8 = 30 km/l.

Step 3: Compare the two efficiencies.

Car 2 has better fuel efficiency.

Final Answer: Car 2 is more fuel-efficient.

Practice Exercise

question option1 option2 option3 option4

In a class of 40 students, 10% are
left-handed. How many students 2 3 4 5
are left-handed?
A car’s value depreciates by 15%
each year. If its current value is
$18,000 $19,000 $17,000 $17,000
$20,000, what will be its value after
one year?
In a local election, a candidate
received 1200 votes, which was
150% of the votes they received in
800 800 1000 1100
the previous election. How many
votes did the candidate receive in
the previous election?
In a class, 60% of students passed
the exam. If there are 30 students
12 students 18 students 10 students 15 students
in the class, how many students did
not pass the exam?
A restaurant found that 70% of its
customers prefer spicy food. If 40%
of the customers are first-time
28% 40% 70% 42%
visitors, what percentage of
first-time visitors are likely to
prefer spicy food?
What percentage is 990 greater
500 550 500 600
than 165?
The sales increased from $90 to
$540 over a quarter. Find the 500 550 600 650
percentage increase.
The value of a stock rose from $45
to $270 in a week. What is the 500 500 600 700
percentage increase?