Proportional Reasoning and Financial Mathematics

Unit Rates and Complex Fractions help us compare quantities in a simple and useful way. In many real-life situations, we want to know how much there is for 1 unit of something else. That is called a unit rate.

For example, if 3 notebooks cost $6, then the unit rate is the cost for 1 notebook. We divide:

$$\frac{6\text{ dollars}}{3\text{ notebooks}}=2\text{ dollars per notebook}$$

In this lesson, you will learn how to find unit rates even when the numbers are fractions. When a fraction contains another fraction, it is called a complex fraction.

For example,

$$\frac{\tfrac{3}{4}\text{ mile}}{\tfrac{1}{2}\text{ hour}}$$

is a complex fraction. It tells us a rate: miles per hour. We will learn how to simplify it to a unit rate.

1. What is a Unit Rate?

A rate compares two quantities with different units, such as miles and hours, or dollars and pounds. A unit rate is a rate with a denominator of 1.

Examples of unit rates include:

•  dollars per 1 sandwich
• 60 miles per 1 hour
• 15 words per 1 minute

To find a unit rate, divide the first quantity by the second quantity.

$$\text{Unit rate}=\frac{\text{amount}}{\text{number of units}}$$

2. What is a Complex Fraction?

A complex fraction is a fraction where the numerator, the denominator, or both are also fractions.

Examples:

$$\frac{\tfrac{5}{6}}{\tfrac{2}{3}}, \quad \frac{\tfrac{3}{4}\text{ cup}}{\tfrac{1}{8}\text{ serving}}$$

Complex fractions often appear in unit rate problems when the quantities are less than 1 or measured in fractional parts.

To simplify a complex fraction, remember this rule:

$$\frac{a}{b}=a\div b$$

So,

$$\frac{\tfrac{3}{4}}{\tfrac{1}{2}}=\tfrac{3}{4}\div \tfrac{1}{2}$$

Then divide by multiplying by the reciprocal:

$$\tfrac{3}{4}\div \tfrac{1}{2}=\tfrac{3}{4}\times \tfrac{2}{1}=\tfrac{6}{4}=\tfrac{3}{2}=1.5$$

3. Steps for Finding a Unit Rate with Fractions

1. Write the rate as a fraction.
2. Identify what should become 1. Usually, the denominator should become 1.
3. Divide numerator by denominator.
4. Simplify the answer.
5. Include units. Units tell what the answer means.

If the rate is

$$\frac{\tfrac{2}{3}\text{ gallon}}{\tfrac{1}{4}\text{ hour}}$$

then we divide:

$$\tfrac{2}{3}\div \tfrac{1}{4}=\tfrac{2}{3}\times 4=\tfrac{8}{3}=2\tfrac{2}{3}$$

So the unit rate is

$$2\tfrac{2}{3}\text{ gallons per hour}$$

4. Why This Matters

Unit rates help us compare prices, speed, earnings, and many other situations. In financial math, unit rates are especially useful for finding:

• cost per item
• pay per hour
• price per pound
• fuel use per mile

When the values are fractions, complex fractions let us still find the basic “for 1” comparison.

5. Worked Examples

Example 1: Simple Fractional Rate

A cyclist travels \(\tfrac{3}{4}\) mile in \(\tfrac{1}{2}\) hour. What is the speed in miles per hour?

Step 1: Write the rate.

$$\frac{\tfrac{3}{4}\text{ mile}}{\tfrac{1}{2}\text{ hour}}$$

Step 2: Divide.

$$\tfrac{3}{4}\div \tfrac{1}{2}=\tfrac{3}{4}\times 2=\tfrac{6}{4}=\tfrac{3}{2}$$

Step 3: Write as a mixed number or decimal.

$$\tfrac{3}{2}=1\tfrac{1}{2}=1.5$$

Answer: The cyclist’s speed is 1.5 miles per hour.

Example 2: Price Per Pound

\(\tfrac{5}{8}\) pound of almonds costs \(\$\tfrac{15}{4}\). What is the cost per pound?

Step 1: Set up the unit rate.

$$\frac{\tfrac{15}{4}\text{ dollars}}{\tfrac{5}{8}\text{ pound}}$$

Step 2: Divide.

$$\tfrac{15}{4}\div \tfrac{5}{8}=\tfrac{15}{4}\times \tfrac{8}{5}$$

Simplify:

$$\tfrac{15}{4}\times \tfrac{8}{5}=\tfrac{15\cdot 8}{4\cdot 5}$$

$$=\tfrac{120}{20}=6$$

Answer: The almonds cost $6 per pound.

Example 3: Earnings Rate

A student earns \(\$\tfrac{21}{2}\) for working \(\tfrac{3}{2}\) hours. What is the hourly pay rate?

Step 1: Write the complex fraction.

$$\frac{\tfrac{21}{2}\text{ dollars}}{\tfrac{3}{2}\text{ hours}}$$

Step 2: Divide by multiplying by the reciprocal.

$$\tfrac{21}{2}\div \tfrac{3}{2}=\tfrac{21}{2}\times \tfrac{2}{3}$$

Step 3: Simplify.

$$\tfrac{21}{2}\times \tfrac{2}{3}=\tfrac{21}{3}=7$$

Answer: The student earns $7 per hour.

Example 4: Interpreting a More Difficult Rate

A machine uses \(\tfrac{7}{10}\) liter of fuel in \(\tfrac{1}{5}\) hour. How many liters does it use per hour?

Step 1: Write the rate.

$$\frac{\tfrac{7}{10}\text{ liter}}{\tfrac{1}{5}\text{ hour}}$$

Step 2: Divide.

$$\tfrac{7}{10}\div \tfrac{1}{5}=\tfrac{7}{10}\times \tfrac{5}{1}$$

$$=\tfrac{35}{10}=\tfrac{7}{2}=3.5$$

Answer: The machine uses 3.5 liters per hour.

6. A Helpful Shortcut

Whenever you see a complex fraction like

$$\frac{\tfrac{a}{b}}{\tfrac{c}{d}}$$

you can rewrite it as

$$\tfrac{a}{b}\div \tfrac{c}{d}=\tfrac{a}{b}\times \tfrac{d}{c}$$

This means:

• Keep the first fraction the same.
• Change division to multiplication.
• Flip the second fraction.

This is one of the most important skills for solving unit rates with fractions.

7. Checking if Your Answer Makes Sense

After finding a unit rate, ask yourself:

• Are the units correct?
• Did I divide in the right order?
• Does the answer seem reasonable?

For example, if less than 1 pound costs more than $3, then the cost per full pound should be more than $3. This kind of thinking helps catch mistakes.

8. Common Mistakes to Avoid

• Forgetting the units. Always say miles per hour, dollars per pound, and so on.
• Flipping the wrong fraction. Only the second fraction is flipped when dividing.
• Dividing in the wrong direction. To find “per 1 hour,” divide by the number of hours.
• Not simplifying. Reduce fractions or write decimals when needed.

9. Practice Ideas

Try these on your own:

• \(\tfrac{1}{2}\) mile in \(\tfrac{1}{4}\) hour
• \($\tfrac{9}{2}\) for \(\tfrac{3}{4}\) pound of cheese
• \(\tfrac{2}{5}\) gallon in \(\tfrac{1}{10}\) hour

For each one, write the complex fraction, divide, simplify, and include units.

10. Summary

A unit rate tells how much there is for 1 unit of another quantity. A complex fraction is a fraction that contains fractions, and it often appears in rate problems with fractional values.

To find a unit rate from a complex fraction, divide the numerator by the denominator. When dividing fractions, multiply by the reciprocal:

$$\frac{\tfrac{a}{b}}{\tfrac{c}{d}}=\tfrac{a}{b}\times \tfrac{d}{c}$$

With practice, you can use this skill to compare speed, prices, wages, and other real-world situations clearly and accurately.

Constants of Proportionality help us describe situations where two quantities change together in a perfectly consistent way.

When one quantity is always a constant multiple of another, the relationship is proportional. The number that connects them is called the constant of proportionality.

This idea is important in maths and in real life. It appears in unit prices, wages, speed, currency exchange, recipes, and many financial situations.

In this lesson, you will learn:

• what a constant of proportionality is,
• how to find it from a table, graph, equation, or word problem,
• how to tell whether a relationship is proportional,
• and how to use the constant to solve problems.

1. What is a constant of proportionality?

If two quantities, usually called \(x\) and \(y\), are proportional, then they follow the rule

Here, \(k\) is the constant of proportionality.

This means that for every value of \(x\), the value of \(y\) is found by multiplying by the same number \(k\).

You can also find \(k\) by dividing:

as long as \(x \ne 0\).

So the constant of proportionality is really the unit rate or the multiplier that connects the two quantities.

Example: If apples cost \(\$3\) per kilogram, then

The constant of proportionality is \(3\), because the cost is always 3 times the number of kilograms.

2. How to recognize a proportional relationship

A relationship is proportional if:

• the ratio \(\frac{y}{x}\) is the same for all pairs of values,
• it can be written in the form \(y = kx\),
• and its graph is a straight line that passes through the origin, \((0,0)\).

If any of these are not true, then the relationship is not proportional.

Important: Not every straight line is proportional. A line must go through the origin to represent a proportional relationship.

For example:

• \(y = 4x\) is proportional because it matches \(y = kx\).
• \(y = 4x + 2\) is not proportional because of the extra \(+2\).

3. Finding the constant from a table

When you are given a table, divide each \(y\)-value by the matching \(x\)-value. If the answer is always the same, that shared value is the constant of proportionality.

Worked Example 1: Finding \(k\) from a table

Find \(\frac{y}{x}\) for each pair:

The ratio is always 12, so the relationship is proportional.

The constant of proportionality is

This means the person earns \(\$12\) per hour.

The equation is

4. Finding the constant from an equation

If an equation is already written in the form \(y = kx\), then the constant of proportionality is the number multiplying \(x\).

Example: In

the constant of proportionality is \(7.5\).

This could mean \(\$7.50\) per item, 7.5 metres per second, or any other unit rate depending on the context.

5. Finding the constant from a graph

On a graph of a proportional relationship, the line goes through the origin. To find the constant of proportionality, choose any point \((x,y)\) on the line and calculate

The value of \(k\) is also the rate of change of the line in this special case.

For example, if the graph passes through \((4,20)\), then

So the equation is

6. Interpreting the constant in real-life situations

The constant of proportionality is not just a number. It has meaning.

It tells us how much of one quantity there is for every 1 unit of the other quantity.

• If \(y\) is cost and \(x\) is number of items, then \(k\) is the price per item.
• If \(y\) is distance and \(x\) is time, then \(k\) is the speed.
• If \(y\) is money earned and \(x\) is hours worked, then \(k\) is the hourly pay rate.

Always pay attention to the units. They help you understand what the constant means.

Worked Example 2: Financial context

A streaming service charges the same amount each month. After 3 months, the total cost is \(\$27\). After 8 months, the total cost is \(\$72\).

Let \(x\) be the number of months and \(y\) be the total cost.

Find the constant of proportionality:

The ratio is constant, so the relationship is proportional.

The constant of proportionality is

This means the service costs \(\$9\) per month.

The equation is

If you want the cost for 12 months:

So 12 months cost \(\$108\).

7. Using the constant to solve missing-value problems

Once you know \(k\), you can find any missing value using \(y = kx\).

Worked Example 3: Solve for a missing value

At a market, oranges cost \(\$2.50\) per kilogram. What is the cost of 6 kilograms?

Here, the constant of proportionality is

Use the equation

So 6 kilograms cost \(\$15\).

Now work backwards: if someone spent \(\$22.50\), how many kilograms did they buy?

They bought 9 kilograms.

8. When a relationship is not proportional

Sometimes two quantities are related, but not proportionally.

This happens when the ratio \(\frac{y}{x}\) changes, or when there is a starting amount that is not zero.

Example: A taxi charges a \(\$4\) starting fee plus \(\$2\) per kilometre.

The equation is

This is not proportional because it does not have the form \(y = kx\).

There is an extra \(+4\), so the graph would not pass through the origin.

Worked Example 4: Decide if it is proportional

Check the ratios:

All the ratios are equal, so the relationship is proportional.

The constant of proportionality is

This means the fuel costs \(\$1.50\) per litre.

The equation is

9. A step-by-step method

When you are asked to find a constant of proportionality, use these steps:

1. Identify the two related quantities.
2. Decide which is \(x\) and which is \(y\).
3. Compute \(\frac{y}{x}\).
4. Check whether the ratio stays the same.
5. If it does, that common ratio is the constant of proportionality \(k\).
6. Write the equation as \(y = kx\).
7. Interpret what \(k\) means in the situation.

10. Common mistakes to avoid

• Mixing up the order: If the rule is \(y = kx\), then \(k = \frac{y}{x}\), not \(\frac{x}{y}\).
• Forgetting units: A constant usually represents something like dollars per item or kilometres per hour.
• Assuming every line is proportional: The graph must pass through \((0,0)\).
• Ignoring a starting amount: If there is a fixed fee or starting value, the relationship may not be proportional.

11. Quick practice ideas

Try these on your own:

• If 5 notebooks cost \(\$20\), what is the constant of proportionality?
• If \(y = 13x\), what is \(k\)?
• If a graph passes through \((0,0)\) and \((6,18)\), what is the constant of proportionality?
• Is \(y = 4x + 1\) proportional? Why or why not?

12. Summary

A constant of proportionality is the fixed number that connects two proportional quantities.

In a proportional relationship, the equation has the form

and the constant is found using

If the ratio stays the same for all pairs of values, then the relationship is proportional.

Understanding constants of proportionality helps you solve problems involving prices, rates, wages, and other real-life comparisons.

Direct and Inverse Variation are two ways to describe how one quantity changes when another quantity changes.

In direct variation, two variables change together at a constant rate. If one doubles, the other also doubles. If one is cut in half, the other is cut in half.

In inverse variation, two variables change in opposite ways. If one doubles, the other is cut in half. If one gets larger, the other gets smaller.

These ideas are useful in maths, science, and finance because they help us model real situations with equations.

1. Direct Variation

Two variables, usually written as \(x\) and \(y\), are in direct variation if they follow the rule

Here, \(k\) is called the constant of variation.

This means the ratio \(\frac{y}{x}\) stays the same, as long as \(x \neq 0\).

So if a relationship is direct variation, then

How to recognize direct variation:

• The equation can be written as \(y = kx\).
• The graph is a straight line passing through the origin, \((0,0)\).
• The ratio \(\frac{y}{x}\) is constant.

Example of direct variation in real life: If each notebook costs \(\$3\), then the total cost \(C\) varies directly with the number of notebooks \(n\).

If you buy twice as many notebooks, the cost doubles.

2. Finding the constant of variation for direct variation

If you know one pair of values for \(x\) and \(y\), you can find \(k\) using

Then substitute that value back into \(y = kx\).

Worked Example 1: Direct Variation

Suppose \(y\) varies directly with \(x\), and \(y=20\) when \(x=5\). Find the equation.

Step 1: Use \(y = kx\).

Step 2: Substitute the known values.

Step 3: Solve for \(k\).

Step 4: Write the equation.

Check: If \(x=5\), then \(y=4(5)=20\), which is correct.

3. Inverse Variation

Two variables are in inverse variation if they follow the rule

Again, \(k\) is the constant of variation.

This means the product \(xy\) stays constant.

How to recognize inverse variation:

• The equation can be written as \(y = \frac{k}{x}\).
• The product \(xy\) is constant.
• As \(x\) increases, \(y\) decreases.
• The graph is a curve, not a straight line.

Example of inverse variation in real life: If 4 workers can finish a job in 6 hours, then increasing the number of workers decreases the time needed. The number of workers and the time vary inversely.

4. Finding the constant of variation for inverse variation

If you know one pair of values, use

Then substitute into \(y = \frac{k}{x}\).

Worked Example 2: Inverse Variation

Suppose \(y\) varies inversely with \(x\), and \(y=3\) when \(x=8\). Find the equation.

Step 1: Use \(y = \frac{k}{x}\).

Step 2: Find \(k\).

Step 3: Write the equation.

Check: If \(x=8\), then \(y=\frac{24}{8}=3\), which is correct.

5. Comparing direct and inverse variation

• Direct variation: variables move in the same direction.
• Inverse variation: variables move in opposite directions.
• Direct variation equation: \(y = kx\)
• Inverse variation equation: \(y = \frac{k}{x}\)
• Direct variation constant: \(\frac{y}{x}\)
• Inverse variation constant: \(xy\)

6. Understanding the graphs

For direct variation, the graph is a line through the origin. This shows that when \(x=0\), then \(y=0\).

For inverse variation, the graph is a curve. It gets closer and closer to the axes but does not touch them.

This behavior is called asymptotic behavior.

For the equation \(y = \frac{k}{x}\):

• As \(x\) gets very large, \(y\) gets very close to 0.
• As \(x\) gets very close to 0, \(y\) becomes very large or very small.

This is important because it shows that inverse variation does not behave like a line. The variables change quickly at first and then more slowly.

7. Worked Example 3: Decide whether the relationship is direct or inverse

A table shows the values:

• \(x=2, y=10\)
• \(x=4, y=20\)
• \(x=6, y=30\)

Check for direct variation by finding \(\frac{y}{x}\):

The ratio is constant, so this is direct variation.

The equation is

Now compare with another table:

• \(x=2, y=12\)
• \(x=3, y=8\)
• \(x=4, y=6\)

Check for inverse variation by finding \(xy\):

The product is constant, so this is inverse variation.

The equation is

8. Worked Example 4: Word problem

A car rental company charges a cost that varies directly with the number of days rented. If renting a car for 3 days costs \(\$120\), what is the cost for 7 days?

Step 1: Write the direct variation equation.

Here, \(C\) is cost and \(d\) is number of days.

Step 2: Use the given information to find \(k\).

Step 3: Write the equation.

Step 4: Find the cost for 7 days.

The cost for 7 days is \(\$280\).

9. Common mistakes to avoid

• Do not confuse \(y = kx\) with \(y = \frac{k}{x}\).
• For direct variation, check whether \(\frac{y}{x}\) is constant, not \(xy\).
• For inverse variation, check whether \(xy\) is constant, not \(\frac{y}{x}\).
• Remember that direct variation graphs are straight lines through the origin.
• Remember that inverse variation graphs are curves with asymptotic behavior.

10. Quick steps for solving problems

1. Decide whether the relationship is direct or inverse.
2. Write the correct formula: \(y=kx\) or \(y=\frac{k}{x}\).
3. Use the given values to find \(k\).
4. Substitute \(k\) back into the equation.
5. Use the equation to answer the question.

Summary

In direct variation, variables change together and follow the form \(y = kx\). The ratio \(\frac{y}{x}\) stays constant, and the graph is a line through the origin.

In inverse variation, variables change in opposite directions and follow the form \(y = \frac{k}{x}\). The product \(xy\) stays constant, and the graph is a curve that shows asymptotic behavior.

When solving problems, first decide which type of variation you have, then find the constant of variation, write the equation, and use it to solve.

Cross-Multiplication Justification

When students solve proportions, they are often told to “cross-multiply.” This method works, but in math it is important to understand why it works, not just memorize a trick.

In this lesson, you will learn the algebra behind cross-multiplication. We will use ideas you already know about equal fractions, multiplying both sides of an equation by the same number, and clearing denominators.

A proportion is an equation that says two ratios or fractions are equal. For example,

is a proportion because both sides represent the same value.

A common form of a proportion is

where the denominators are not zero. That means \(b \neq 0\) and \(d \neq 0\).

The cross-multiplication rule says that if

then

But this is not magic. It comes from multiplying both sides by a common denominator.

Why does cross-multiplication work?

Start with the proportion

If we multiply both sides by \(bd\), we do not change the equality, because we are doing the same thing to both sides.

Now simplify each side.

On the left side, the \(b\) in the denominator cancels with the \(b\) in \(bd\):

On the right side, the \(d\) in the denominator cancels with the \(d\) in \(bd\):

So the equation becomes

This is the reason cross-multiplication works. It is really just a shortcut for multiplying both sides by the common denominator.

Important condition: This only makes sense when the denominators are not zero. A fraction with denominator 0 is undefined.

Another way to think about it

If two fractions are equal, then they name the same number. Multiplying both sides by the product of the denominators removes the fractions. The result is an equation without denominators, and that equation is exactly the cross-products being equal.

Worked Example 1: Check a true proportion

Show why

is true using cross-multiplication.

Step 1: Multiply the numerator on the left by the denominator on the right.

Step 2: Multiply the numerator on the right by the denominator on the left.

Step 3: Compare the products.

Since the cross-products are equal, the proportion is true.

Justification: This works because if we multiply both sides of

by \(5 \cdot 20=100\), we get

Worked Example 2: Solve for a missing value

Solve

Step 1: Multiply both sides by the common denominator \(4 \cdot 10=40\).

Step 2: Simplify.

Step 3: Solve for \(x\).

So,

You could also say this as cross-multiplication:

Worked Example 3: A proportion in a financial situation

A notebook costs \(\$6\) for 3 notebooks. At the same rate, how much would 5 notebooks cost?

We can write a proportion using cost per number of notebooks:

Step 1: Multiply both sides by \(3 \cdot 5=15\).

Step 2: Simplify.

Step 3: Solve.

So 5 notebooks cost \(\$10\).

Check: \(\$6\) for 3 notebooks means \(\$2\) per notebook. For 5 notebooks, \(5 \cdot 2=10\). The answer makes sense.

Worked Example 4: When a proportion is not true

Determine whether

is a true proportion.

Step 1: Find the cross-products.

Step 2: Compare them.

Since the cross-products are not equal, the fractions are not equal. So this is not a true proportion.

What students often misunderstand

• Cross-multiplication is not a separate rule from algebra. It comes from multiplying both sides by the product of the denominators.
• You must start with two fractions set equal to each other. Cross-multiplication is for equations like \(\frac{a}{b}=\frac{c}{d}\), not for adding or subtracting fractions.
• Denominators cannot be zero. If a denominator is 0, the fraction is undefined.
• Keep track of which numbers are multiplied. The products are across the equal sign: top-left with bottom-right, and bottom-left with top-right.

A quick algebra proof with variables

Suppose

with \(m \neq 0\) and \(n \neq 0\).

Multiply both sides by \(mn\):

Simplify:

This shows that cross-multiplication is really just clearing the denominators.

How this connects to proportional reasoning

In 9th Grade math, proportions are used to compare quantities that change at the same rate. Cross-multiplication helps you solve for unknown values in situations involving prices, unit rates, scale drawings, speed, tax, discounts, and other financial or real-world problems.

Still, it is best to remember the reason behind it: when two fractions are equal, multiplying both sides by a common denominator leads to equal cross-products.

Summary

• A proportion is an equation with two equal fractions.
• If \(\frac{a}{b}=\frac{c}{d}\), with \(b \neq 0\) and \(d \neq 0\), then multiplying both sides by \(bd\) gives \(ad=bc\).
• This is why cross-multiplication works.
• It is not just a trick. It is an algebra step based on multiplying both sides of an equation by the same nonzero value.

Successive Percent Change means applying one percent change after another.

This is common in real life. For example, a price may increase by 10% one month and then decrease by 5% the next month. A population may grow by 3% each year. A bank balance may earn interest again and again.

The most important idea is this: successive percent changes are multiplied, not added.

Many students make the mistake of adding the percentages. For example, they may think that a 20% increase followed by a 10% increase is a 30% increase. That is not correct, because the second change is based on the new amount, not the original amount.

1. Turning a percent change into a multiplier

A percent change can be written as a decimal multiplier.

• An increase of 15% means multiply by \(1.15\).
• An increase of 8% means multiply by \(1.08\).
• A decrease of 20% means multiply by \(0.80\).
• A decrease of 7% means multiply by \(0.93\).

In general:

For an increase of \(p\%\):

For a decrease of \(p\%\):

2. How to find a successive percent change

Start with the original amount. Then multiply by each percent-change multiplier in order.

If an amount starts at \(A\), then changes by multipliers \(m_1, m_2, m_3\), the final amount is:

This works because each new percent change is applied to the current amount.

3. Why you cannot usually add the percentages

Suppose a shirt costs \(\$100\). It increases by 20%, then increases by 10%.

If you incorrectly add the percentages, you might say the final change is 30%, so the new price is \(\$130\).

But the correct calculation is:

The final price is \(\$132\), not \(\$130\).

The extra \(\$2\) happens because the 10% increase is taken on the already increased price.

4. A useful shortcut for the overall percent change

After multiplying the change factors, compare the final multiplier to 1.

• If the final multiplier is greater than 1, there is an overall increase.
• If the final multiplier is less than 1, there is an overall decrease.

For example, if the combined multiplier is \(1.188\), then:

So the overall change is an increase of \(18.8\%\).

If the combined multiplier is \(0.92\), then:

So the overall change is a decrease of \(8\%\).

5. Increase followed by decrease

An increase and a decrease of the same percent do not cancel out.

This is a very important fact.

For example, increase by 10%, then decrease by 10%:

The final multiplier is \(0.99\), which means a 1% decrease overall.

So if a value goes up by 10% and then down by 10%, it ends lower than where it started.

Worked Example 1: Two increases

A phone costs \(\$250\). Its price increases by 12%, then increases by 5%. Find the final price.

Step 1: Write each percent change as a multiplier.

• 12% increase \(\rightarrow 1.12\)
• 5% increase \(\rightarrow 1.05\)

Step 2: Multiply.

Final answer: The final price is \(\$294\).

Step 3: Find the overall percent change if needed.

So the price increased by 17.6% overall.

Worked Example 2: Increase followed by decrease

A jacket costs \(\$80\). The price increases by 25%, then later decreases by 20%. Find the final price and the overall percent change.

Step 1: Write the multipliers.

• 25% increase \(\rightarrow 1.25\)
• 20% decrease \(\rightarrow 0.80\)

Step 2: Multiply.

Final price: \(\$80\)

Step 3: Find the overall change.

The combined multiplier is \(1\), so there is no overall change.

This example shows that different percent changes can sometimes balance each other, but this does not happen just because the numbers look close.

Worked Example 3: Same percent up and down

A bicycle costs \(\$300\). The price goes up by 15%, then goes down by 15%. Find the final price.

Step 1: Write the multipliers.

• 15% increase \(\rightarrow 1.15\)
• 15% decrease \(\rightarrow 0.85\)

Step 2: Multiply.

Final answer: The final price is \(\$293.25\).

Step 3: Interpret the result.

So there is an overall 2.25% decrease.

Even though the price went up 15% and down 15%, it did not return to the original price.

Worked Example 4: Repeated growth

A savings account starts with \(\$500\) and grows by 4% each year for 3 years. Find the amount after 3 years.

Each year, a 4% increase means multiply by \(1.04\).

So after 3 years:

Rounded to the nearest cent, the amount is \(\$562.43\).

This repeated multiplying is called compounding.

6. Step-by-step method to use every time

1. Write the starting amount.
2. Change each percent into a multiplier.
3. Multiply the starting amount by all the multipliers.
4. If needed, compare the final multiplier to 1 to find the overall percent change.

7. Common mistakes to avoid

• Do not add percentages unless the problem clearly means a single change from the original amount.
• Do not subtract percentages directly for an increase followed by a decrease.
• Be careful with decreases: a 30% decrease means multiply by \(0.70\), not \(0.30\).
• Use the new amount each time, because each percent change happens after the previous one.

8. Quick check questions

Try these on your own:

• A price of \(\$60\) increases by 10%, then by 20%. What is the final price?
• A value of \(200\) decreases by 5%, then decreases by 10%. What is the final value?
• An amount rises by 30% and then falls by 30%. Is the final amount the same as the starting amount?

Answers:

• \(60 \times 1.10 \times 1.20 = 79.2\), so \(\$79.20\)
• \(200 \times 0.95 \times 0.90 = 171\)
• No. \(1.30 \times 0.70 = 0.91\), so there is a 9% decrease overall.

Summary

Successive percent change means applying percent changes one after another.

To solve these problems, turn each percent change into a multiplier and multiply. This is why successive changes are multiplicative, not additive.

Always remember: an increase and a decrease of the same percent do not usually cancel out, because the second change is based on a different amount.

Simple and Compound Interest Modeling

When people save money in a bank or borrow money as a loan, interest is often involved. Interest is money added to an account or money paid for borrowing.

In this lesson, you will learn how to model simple interest and compound interest. Modeling means writing a math rule or formula that shows how money changes over time.

These ideas are important in real life because they help you understand savings accounts, loans, credit cards, and investments.

1. Key vocabulary

• Principal: the starting amount of money
• Interest: the extra money earned or paid
• Rate: the percent charged or earned each year
• Time: how long the money is invested or borrowed
• Amount: the total money after interest is included
• Compound: interest is added to the balance, and future interest is calculated on the new balance

2. Simple interest

With simple interest, interest is calculated only on the original principal. This means the same amount of interest is added each year.

Because the increase is the same each year, simple interest creates a linear pattern.

The formula for simple interest is

where:

• \(I\) = interest earned or paid
• \(P\) = principal
• \(r\) = annual interest rate written as a decimal
• \(t\) = time in years

To find the total amount, use

Since \(I = Prt\), we can combine the formulas:

This formula shows that the total amount grows at a constant rate over time.

3. Compound interest

With compound interest, interest is calculated on the current balance, not just the original principal.

That means after interest is added once, the next interest amount is a little bigger. This creates exponential growth, not linear growth.

If interest is compounded once per year, the amount after \(t\) years is

If interest is compounded more than once per year, use the general compound interest formula:

where:

• \(A\) = total amount after interest
• \(P\) = principal
• \(r\) = annual interest rate as a decimal
• \(n\) = number of times interest is compounded each year
• \(t\) = time in years

Common values for \(n\):

• \(n = 1\): yearly
• \(n = 2\): semiannually
• \(n = 4\): quarterly
• \(n = 12\): monthly

4. Simple interest vs. compound interest

Both types of interest use a principal, a rate, and time. But they grow differently.

• Simple interest: adds the same amount each year
• Compound interest: adds larger and larger amounts over time

This means compound interest usually grows faster than simple interest, especially over a long time.

5. Turning percent into decimal

Before using any interest formula, change the percent rate to a decimal.

• \(5\% = 0.05\)
• \(3.2\% = 0.032\)
• \(12\% = 0.12\)

You can do this by dividing the percent by 100.

6. Worked Example 1: Simple interest

Problem: A student deposits \(\$800\) into an account that earns \(4\%\) simple interest per year for 3 years. How much interest is earned, and what is the final amount?

Step 1: Identify the values.

• \(P = 800\)
• \(r = 0.04\)
• \(t = 3\)

Step 2: Use the simple interest formula.

The interest earned is \(\$96\).

Step 3: Find the total amount.

Answer: The account earns \(\$96\) in interest and ends with \(\$896\).

7. Worked Example 2: Simple interest as a linear model

Problem: Write a model for the total amount in an account with principal \(\$500\) at \(6\%\) simple interest. Then find the amount after 5 years.

For simple interest, use

Substitute the known values:

This is the model.

Now use \(t = 5\):

Answer: The model is \(A = 500(1 + 0.06t)\), and after 5 years the amount is \(\$650\).

Notice that with simple interest, the account gains the same amount each year:

So the balance increases by \(\$30\) per year.

8. Worked Example 3: Compound interest compounded yearly

Problem: A savings account starts with \(\$1{,}200\) and earns \(5\%\) interest compounded yearly for 4 years. What is the final amount?

Since the interest is compounded yearly, use

Substitute the values:

Now evaluate:

Answer: The final amount is about \(\$1{,}458.61\).

Compare this to simple interest with the same principal, rate, and time:

Compound interest gives more money because each year's interest also earns interest.

9. Worked Example 4: Compound interest with monthly compounding

Problem: A bank account has \(\$2{,}000\) deposited at \(3.6\%\) annual interest compounded monthly for 2 years. Find the final amount.

Step 1: Identify the values.

• \(P = 2000\)
• \(r = 0.036\)
• \(n = 12\)
• \(t = 2\)

Step 2: Use the compound interest formula.

Step 3: Evaluate.

Answer: The final amount is about \(\$2{,}149.12\).

10. How to decide which formula to use

Ask yourself: Is the interest simple or compound?

• If it says simple interest, use \(I = Prt\) or \(A = P(1 + rt)\).
• If it says compounded, use \(A = P\left(1 + \frac{r}{n}\right)^{nt}\).

Also check the compounding period:

• yearly: \(n=1\)
• semiannually: \(n=2\)
• quarterly: \(n=4\)
• monthly: \(n=12\)

11. Common mistakes to avoid

• Forgetting to change percent to decimal
  Example: use \(0.07\), not \(7\).
• Using the wrong formula
  Simple interest and compound interest are not the same.
• Forgetting the compounding number \(n\)
  Monthly compounding must include \(n=12\).
• Mixing up total amount and interest only
  \(A\) is the full balance, while \(I\) is just the extra money.
• Not checking time units
  If the rate is annual, time should be in years.

12. Why these models matter

Simple and compound interest are both ways to describe how money changes over time.

Simple interest is useful for situations where the increase stays constant. Compound interest is more realistic for many savings accounts and investments because growth builds on itself.

Understanding the difference helps you compare financial choices. For example, two accounts may have similar rates, but the one that compounds more often may grow more.

13. Quick check for understanding

1. What is the difference between principal and amount?
2. Why is simple interest linear?
3. Why does compound interest grow faster over time?
4. What value of \(n\) would you use for quarterly compounding?

14. Lesson summary

Simple interest adds interest only to the original principal, so it grows in a linear way. Its formulas are

Compound interest adds interest to the current balance, so it grows exponentially. Its formula is

When solving problems, identify the principal, rate, time, and compounding period. Then choose the correct formula, substitute carefully, and check whether you are finding interest only or the total amount.

Amortization and Depreciation Basics

In financial math, two important ideas are amortization and depreciation. Both describe change over time, but they apply to different situations.

Amortization is about paying off a loan over time with regular payments. Each payment usually covers some interest and some of the amount borrowed, called the principal.

Depreciation is about how an item, such as a car or computer, loses value over time. As the item gets older or more used, it is often worth less money.

These ideas connect to proportional reasoning because both involve repeated change. In amortization, a balance changes as payments are made. In depreciation, a value changes over time, often by a fixed amount or a fixed percent.

1. Amortization: Paying Off a Loan

When a person borrows money, they usually agree to repay it in parts over time. This is called a loan. Examples include car loans, home loans, and some student loans.

The original amount borrowed is called the principal. The extra money charged for borrowing is called interest.

A regular loan payment is often split into two parts:

• money that pays the interest
• money that reduces the principal

As the principal gets smaller, the interest charged usually gets smaller too. This means that over time, more of each payment goes toward the principal.

A simple way to think about one payment is:

Then the new loan balance is:

If the interest rate is given per time period, the interest for one period can be found with:

For example, if the balance is \(\$1000\) and the monthly interest rate is \(2\%\), then:

So \(\$20\) of that month's payment goes to interest first.

Important idea: In an amortized loan, the balance should go down over time, as long as the payments are large enough to cover the interest and reduce the principal.

2. Depreciation: Losing Value Over Time

Depreciation means a decrease in value. Many items lose value as they age. A new car, for example, is usually worth less after one year than when it was first bought.

There are two common basic ways to model depreciation:

• Linear depreciation: the value drops by the same amount each time period
• Percent depreciation: the value drops by the same percent each time period

Linear depreciation is like subtracting the same number again and again.

The formula is:

where:

• \(V\) = value after time
• \(V_0\) = starting value
• \(r\) = amount of value lost each time period
• \(t\) = number of time periods

Percent depreciation means the item keeps a certain percent of its value each time period.

The formula is:

where:

• \(V\) = value after time
• \(V_0\) = starting value
• \(r\) = depreciation rate written as a decimal
• \(t\) = number of time periods

For example, a yearly depreciation rate of \(15\%\) means \(r = 0.15\), so the item keeps:

That means after each year, the item is worth \(85\%\) of what it was worth the year before.

3. Comparing Amortization and Depreciation

• Amortization is about a loan balance going down as payments are made.
• Depreciation is about an asset value going down over time.
• Amortization includes interest and payments.
• Depreciation focuses on decreasing value.

Both ideas involve careful tracking of change over time, which is why tables can be very helpful.

4. Using a Table to Track Change

Tables help organize what happens each month or each year.

For an amortization table, you might include:

• starting balance
• interest charged
• payment amount
• amount applied to principal
• new balance

For a depreciation table, you might include:

• time period
• starting value
• amount lost
• new value

Worked Example 1: One Loan Payment

A student borrows \(\$500\). The monthly interest rate is \(1\%\). The monthly payment is \(\$60\). Find the interest for the month, the amount paid toward principal, and the new balance.

Step 1: Find the interest.

So \(\$5\) of the payment goes to interest.

Step 2: Find how much of the payment goes to principal.

So \(\$55\) reduces the loan balance.

Step 3: Find the new balance.

Answer: Interest is \(\$5\), principal paid is \(\$55\), and the new balance is \(\$445\).

Worked Example 2: Two Months of Amortization

A loan starts at \(\$300\). The monthly interest rate is \(2\%\). The monthly payment is \(\$80\). Find the balance after two months.

Month 1

Interest:

Principal paid:

New balance:

Month 2

Now use the new balance of \(\$226\).

Interest:

Principal paid:

New balance:

Answer: After two months, the balance is \(\$150.52\).

Notice that the second month's interest is smaller than the first month's interest. That is because the balance is smaller.

Worked Example 3: Linear Depreciation

A laptop is bought for \(\$900\). It loses \(\$120\) in value each year. What is its value after 4 years?

Use the linear depreciation formula:

Substitute the values:

Answer: After 4 years, the laptop is worth \(\$420\).

Worked Example 4: Percent Depreciation

A car is worth \(\$12{,}000\) and depreciates by \(10\%\) each year. What is its value after 3 years?

Use the percent depreciation formula:

Substitute the values:

Answer: After 3 years, the car is worth \(\$8{,}748\).

Notice that the car does not lose the same dollar amount each year. Instead, it loses the same percent each year.

5. Common Mistakes to Avoid

• Mixing up principal and interest: The principal is the amount borrowed or still owed. Interest is the extra charge for borrowing.
• Forgetting to change a percent to a decimal: For example, \(8\% = 0.08\).
• Subtracting the full payment from the loan balance before finding interest: Interest must be found from the current balance first.
• Using linear depreciation when the problem describes percent loss: “Loses 15% each year” means use percent depreciation, not subtracting the same dollar amount.
• Using the original value every time in percent depreciation: Percent depreciation uses the current value each period.

6. Quick Check for Understanding

1. A loan balance is \(\$800\), the monthly interest rate is \(1.5\%\), and the payment is \(\$100\). How much interest is charged that month?
2. If an item is worth \(\$700\) and loses \(\$50\) each year, what is its value after 5 years?
3. If a phone costs \(\$600\) and depreciates by \(20\%\) each year, what multiplier is used each year?

Answers:

1. \(800 \times 0.015 = 12\), so the interest is \(\$12\).
2. \(700 - 50(5) = 450\), so the value is \(\$450\).
3. \(1 - 0.20 = 0.80\), so the multiplier is \(0.80\).

Summary

Amortization and depreciation both describe change over time, but they are used in different financial situations. Amortization shows how a loan is paid off through regular payments that cover interest and reduce principal.

Depreciation shows how an asset loses value over time. If the value drops by the same amount each period, use linear depreciation. If the value drops by the same percent each period, use percent depreciation.

By identifying whether a problem is about a loan balance or an item's value, and whether the change is by a fixed amount or a percent, you can choose the correct method and solve it step by step.

Misleading Proportionality happens when someone treats a multiplicative situation as if it were additive.

In simple words, this means they add the same amount when they should be multiplying by the same factor.

This mistake is very common in ratio, rate, scale, percentage, and financial questions. Learning to spot it will help you avoid wrong answers and understand proportional reasoning more clearly.

Key idea: In a proportional relationship, if one quantity is multiplied by a number, the other quantity must also be multiplied by that same number.

If two quantities are proportional, they can be written like this:

$$y = kx$$

Here, 7k is the constant of proportionality. It tells us how much of one quantity there is for each 1 unit of the other.

For example, if apples cost \(\$2\) each, then the total cost \(C\) for \(a\) apples is

$$C = 2a$$

This is proportional because the ratio \(\frac{C}{a} = 2\) stays the same.

Where the mistake happens

Students sometimes think like this:

• “If 3 items cost \(\$12\), then 6 items cost \(\$15\) because I added 3 items, so I add \(\$3\).”

That is misleading proportionality. The number of items doubled from 3 to 6, so the cost should also double from \(\$12\) to \(\$24\).

They used additive reasoning instead of multiplicative reasoning.

Additive reasoning focuses on differences.

For example, “I added 3 more items, so I add 3 more dollars.”

Multiplicative reasoning focuses on scale factors.

For example, “The number of items was multiplied by 2, so the cost must also be multiplied by 2.”

In proportional situations, multiplicative reasoning is the correct method.

How to tell whether a situation is proportional

A situation is proportional if:

• the ratio between the two quantities stays constant,
• the graph would be a straight line through the origin,
• the relationship can be written as \(y = kx\).

Some common proportional situations are:

• cost per item when each item has the same price,
• distance traveled at a constant speed,
• pay earned at a constant hourly rate,
• currency conversion with a fixed exchange rate,
• simple scaling in recipes or maps.

Important warning: Not every situation with two changing numbers is proportional.

For example, a taxi fare might have a starting fee plus a cost per kilometer. That is not proportional because it does not start at 0.

How to avoid misleading proportionality

1. Ask: Is this relationship proportional?
2. Look for a constant ratio, not a constant difference.
3. Find the scale factor between the known values.
4. Multiply both quantities by the same factor.
5. Check your answer by comparing unit rates if possible.

Method 1: Use a unit rate

If 4 notebooks cost \(\$10\), then one notebook costs

$$\frac{10}{4} = 2.5$$

So 7 notebooks cost

$$7 \times 2.5 = 17.5$$

So the total is \(\$17.50\).

Method 2: Use a scale factor

If 4 notebooks become 8 notebooks, the number of notebooks is multiplied by 2.

So the cost must also be multiplied by 2.

If 4 notebooks cost \(\$10\), then 8 notebooks cost

$$10 \times 2 = 20$$

Worked Example 1: Basic cost problem

Three movie tickets cost \(\$27\). How much do 5 tickets cost?

Step 1: Decide if the situation is proportional.

If each ticket costs the same amount, then yes, it is proportional.

Step 2: Find the cost of 1 ticket.

$$\frac{27}{3} = 9$$

So one ticket costs \(\$9\).

Step 3: Multiply by 5 tickets.

$$5 \times 9 = 45$$

Answer: 5 tickets cost \(\$45\).

Common wrong idea: Some students notice that 5 is 2 more than 3 and then add \(\$2\) to \(\$27\). That gives \(\$29\), which is wrong because the price is based on equal groups, not equal increases of 1 dollar per ticket.

Worked Example 2: Wages and hours

A student earns \(\$48\) for 4 hours of work. How much would the student earn for 7 hours?

Step 1: Find the hourly rate.

$$\frac{48}{4} = 12$$

So the student earns \(\$12\) per hour.

Step 2: Multiply by 7 hours.

$$7 \times 12 = 84$$

Answer: The student earns \(\$84\).

Why additive reasoning fails: Going from 4 hours to 7 hours is an increase of 3 hours, but you cannot just add \(\$3\) to the pay. Each extra hour adds \(\$12\), not \(\$1\).

Worked Example 3: Percentage increase and scaling

A small backpack costs \(\$20\). A larger version is 1.5 times the price of the small backpack. What is the price of the larger backpack?

Step 1: Use multiplicative thinking.

The large backpack is not \(\$1.50\) more. It is 1.5 times as much.

Step 2: Multiply.

$$20 \times 1.5 = 30$$

Answer: The larger backpack costs \(\$30\).

Connection to percentages: A factor of \(1.5\) means 150% of the original price, or a 50% increase.

Worked Example 4: Spotting a non-proportional situation

A gym charges a \(\$25\) sign-up fee and \(\$15\) per month. Is the total cost proportional to the number of months?

Step 1: Write a rule.

$$C = 15m + 25$$

Step 2: Compare with the form \(y = kx\).

This rule has an added 25, so it is not proportional.

Why this matters: If the number of months doubles, the total cost does not exactly double because the sign-up fee stays the same.

For example:

• 1 month costs \(15(1)+25=40\)
• 2 months costs \(15(2)+25=55\)

If it were proportional, doubling from 1 month to 2 months would double the cost from 40 to 80. But it becomes 55, not 80.

So this is a good example of a situation where proportional thinking would be misleading.

Comparing additive and multiplicative thinking

• Additive: “How much bigger is it?”
• Multiplicative: “How many times as big is it?”

For proportional relationships, ask “how many times?”

Example:

If a recipe uses 2 cups of flour for 1 batch, then 6 cups of flour makes how many batches?

Since 6 is 3 times 2, the number of batches is also 3 times 1.

So it makes 3 batches.

Financial mathematics connection

Misleading proportionality appears often in money problems.

Here are some common cases:

• Shopping: If the number of items changes, total cost changes by multiplication when the unit price is fixed.
• Hourly pay: If time worked changes, earnings change by multiplication when the hourly rate is fixed.
• Interest or growth: Amounts often change by a factor, not by adding the same amount each time.
• Discounts: A 20% discount means multiply by \(0.8\), not subtract 20 dollars unless the original price is \(\$100\).

Example of a discount mistake

A jacket costs \(\$60\) and is on sale for 25% off.

A wrong method is to subtract 25 and say the sale price is \(\$35\).

That mistake treats 25% like 25 dollars.

The correct method is:

$$0.25 \times 60 = 15$$

Then subtract the discount:

$$60 - 15 = 45$$

So the sale price is \(\$45\).

Quick checklist for students

• Does the situation have a constant rate or unit price?
• Can I write it as \(y = kx\)?
• Am I comparing with multiplication, not just addition?
• Did I use the same scale factor for both quantities?
• Does my answer make sense with the unit rate?

Practice thinking

When you see a ratio or rate problem, pause and ask:

“Should I add, or should I multiply?”

If the relationship is proportional, the answer is multiply.

Summary

Misleading proportionality happens when additive reasoning is used in a situation that needs multiplicative reasoning.

In proportional relationships, the ratio stays constant, and both quantities change by the same scale factor.

To avoid mistakes, check whether the relationship is truly proportional, use unit rates or scale factors, and be careful in financial situations like shopping, wages, and discounts.