# Ratio, Proportion, and Variation

## Identify variations from word problems

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Direct and Inverse Variation

Kendall has to travel 240 miles to the city for a 3:00 pm. medical appointment and must take the bus. The bus travels at a rate of 60 miles per hour and leaves every hour on the hour. How can Kendall figure out which bus to take?

In this concept, you will learn to understand direct and inverse variation.

### Direct Variation

Direct and inverse variation are two different ways that show how one number changes with respect to another number. In direct variation, as one number increases, so does the other number. In other words, ‘\begin{align*}y\end{align*}’ varies directly with ‘\begin{align*}x\end{align*}’(or ‘\begin{align*}x\end{align*}’ and ‘\begin{align*}y\end{align*}’ are directly proportional). For example, there is a direct variation between the age and height. As the age of a child increases in years, the height of the child also increases. This example of direct variation can be expressed using the function \begin{align*}y = k \cdot x\end{align*} such that \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are the age and height of the child and ‘\begin{align*}k\end{align*}’ is called the constant of proportionality. The constant of proportionality tells you how much ‘\begin{align*}y\end{align*}’ will increase for every increase in ‘\begin{align*}x\end{align*}’.

Let’s look at an example.

\begin{align*}y=2x\end{align*}

The increase in \begin{align*}y\end{align*} will equal twice the value of \begin{align*}x\end{align*}. This can be clearly shown by creating a t-table.

\begin{align*}\begin{array}{rcl} y &=& 2x \\ y &=& 2(1) \\ y &=& 2 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} y &=& 2x \\ y &=& 2(2) \\ y &=& 4 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} y &=& 2x \\ y &=& 2(3) \\ y &=& 6 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} y &=& 2x \\ y &=& 2(4) \\ y &=& 8 \end{array}\end{align*}

 \begin{align*}x\end{align*} \begin{align*}y\end{align*} 1 2 2 4 3 6 4 8

Following is a graph of the direct variation equation \begin{align*}y=2x\end{align*}.

Look at the above graph. The line passes through the origin \begin{align*}(0,0)\end{align*}. If you think of the equation of a line written in slope-intercept form as \begin{align*}y=mx+b\end{align*} then the equation of the given graph could be written as \begin{align*}y=2x+0\end{align*}. This means the slope of the graphed line is 2.

Inverse variation is the exact opposite of direct variation. In inverse variation, as one number increases, the other number decreases. If a student skips a lot of Advanced Math Classes then the mark on the exam will be low. The more classes skipped-the lower the mark. This example of inverse variation can be expressed using the function \begin{align*}y=\frac{k}{x}\end{align*} such that \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are the number of classes skipped and the lower math mark and ‘\begin{align*}k\end{align*}’ is still called the constant of proportionality. In the equation \begin{align*}y=\frac{k}{x}\end{align*} the constant ‘\begin{align*}k\end{align*}’ is divided by the value of ‘\begin{align*}x\end{align*}’. As the value of ‘\begin{align*}x\end{align*}’ increases then the value of ‘\begin{align*}y\end{align*}’ decreases.

The above graph shows an inverse variation. Notice how the value of ‘\begin{align*}y\end{align*}’ gets smaller as the value of ‘\begin{align*}x\end{align*}’ gets bigger.

### Examples

#### Example 1

Earlier, you were given a problem about Kendall and her medical appointment. She needs to figure out what time to get on the bus to get to her appointment on time. How can Kendall figure this out?

She can use the direct variation formula to figure out how long it will take her to get to the city on the bus.

First, write the equation for direct variation.

\begin{align*}y = k \cdot x\end{align*}

Next, state the variables.

\begin{align*}y=\end{align*} number of miles to travel

\begin{align*}x=\end{align*} number of hours on the bus

\begin{align*}k=\end{align*} speed of the bus

Next, fill in \begin{align*}k=60\end{align*} and \begin{align*}y=240\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} y &=& k \cdot x \\ 240 &=& 60 \cdot x \\ 240 &=& 60 x \\ \end{array}\end{align*}

Next, divide both sides of the equation by 60 to find the value of \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 240 &=& 60x \\ \frac{240}{60} &=& \frac{\overset{1}{\cancel{60}}}{\cancel{60}}x \\ 4 &=& x \end{array}\end{align*}

The answer is 4 hours on the bus.

If Kendall’s appointment is at 3:00 p.m. and she will be on the bus for 4 hours, she can get on the 11:00 a.m. bus and arrive exactly on time. However, she might want to get on the bus that leaves at 10:00 a.m. to allow an hour for any unexpected delays in travel, like traffic.

#### Example 2

If \begin{align*}y\end{align*} varies directly as \begin{align*}x\end{align*}, and \begin{align*}x=9\end{align*} when \begin{align*}y=15\end{align*}, find \begin{align*}y\end{align*} when \begin{align*}x=33\end{align*}.

First, write the equation for direct variation.

\begin{align*}y=k \cdot x\end{align*}

Next, fill in \begin{align*}x=9\end{align*} and \begin{align*}y=15\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} y &=& k \cdot x\\ 15 &=& k \cdot 9 \\ 15 &=& 9k \end{array}\end{align*}

Next, divide both sides of the equation by 9 to find the value of the constant of proportionality.

\begin{align*}\begin{array}{rcl} 15 &=& 9k \\ \frac{15}{9} &=& \frac{\overset{1}{\cancel{9}}}{\cancel{9}}k \\ \frac{5}{3} &=& k \end{array}\end{align*}

Next, fill in the values \begin{align*}k=\frac{5}{3}\end{align*} and \begin{align*}x=33\end{align*} into the equation for direct variation to calculate the value of \begin{align*}y\end{align*}.

\begin{align*}\begin{array}{rcl} y &=& k \cdot x\\ y &=& \frac{5}{3} (33) \end{array}\end{align*}

Then, simplify the right side of the equation.

\begin{align*}\begin{array}{rcl} y &=& \frac{5}{\cancel{3}} \cancel{\left(\overset{11}{33}\right)} \\ y &=& 5(11) \\ y &=& 55 \end{array}\end{align*}

#### Example 3

The amount of money raised at a fundraiser is directly proportional to the number of people in attendance. If 5 people attending raised 100, then how much money will be raised if 60 people are in attendance? First, write the equation for direct variation. \begin{align*}y=k \cdot x\end{align*} Next, state the variables. \begin{align*}y=\end{align*} amount of money raised \begin{align*}x=\end{align*} number of people in attendance \begin{align*}k=\end{align*} constant of proportionality Next, fill in \begin{align*}x=5\end{align*} and \begin{align*}y=100\end{align*} into the equation. \begin{align*}\begin{array}{rcl} y &=& k \cdot x\\ 100 &=& k \cdot 5 \\ 100 &=& 5k \end{array}\end{align*} Next, divide both sides of the equation by 5 to find the value of \begin{align*}k\end{align*}. \begin{align*}\begin{array}{rcl} 100 &=& 5k \\ \frac{100}{5} &=& \frac{\overset{1}{\cancel{5}}}{\cancel{5}}k \\ 20 &=& k \end{array}\end{align*} Next, fill in the values \begin{align*}k=20\end{align*} and \begin{align*}x=60\end{align*} into the equation for direct variation to calculate the value of \begin{align*}y\end{align*}. \begin{align*}\begin{array}{rcl} y &=& k \cdot x\\ y &=& 20(60) \end{array}\end{align*} Then, simplify the right side of the equation. \begin{align*}\begin{array}{rcl} y &=& 20(60) \\ y &=& 1200 \\ \end{array}\end{align*} The answer is1,200.

#### Example 4

Dr. Burns told Katie the number of cavities she had in her teeth each year was inversely proportional to the number of minutes she spent brushing her teeth each time she brushed. If Katie had four new cavities during the year the time she spent each time brushing was only 30 seconds. How many new cavities will Katie have at the end of next year if she starts brushing for two minutes each time she brushes her teeth?

First, write the equation for inverse variation.

\begin{align*}y=\frac{k}{x}\end{align*}

Next, state the variables.

\begin{align*}y=\end{align*} number of new cavities

\begin{align*}x=\end{align*} number of minutes spent brushing

\begin{align*}k=\end{align*} constant of proportionality

Next, fill in \begin{align*}x=0.5\end{align*} and \begin{align*}y=4\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} y &=& \frac{k}{x} \\ 4 &=& \frac{k}{0.5} \\ \end{array}\end{align*}

Next, multiply both sides of the equation by 0.5 to find the value of \begin{align*}k\end{align*}.

\begin{align*}\begin{array}{rcl} 4 &=& \frac{k}{0.5} \\ (0.5)4 &=& \overset{1}{\cancel{(0.5)}} \frac{k}{\cancel{0.5}}\\ 2 &=& k \end{array}\end{align*}

Next, fill in the values \begin{align*}k=2\end{align*} and \begin{align*}x=2\end{align*} into the equation for inverse variation to calculate the value of \begin{align*}y\end{align*}.

\begin{align*}\begin{array}{rcl} y &=& \frac{k}{x} \\ y &=& \frac{2}{2} \\ y &=& 1 \end{array}\end{align*}

The answer is 1 new cavity.

#### Example 5

Margie was going on a trip and was travelling at a speed of 50 miles per hour. As the number of hours travelling increased so did the distance she travelled. Is this an example of direct or inverse variation? Explain your answer.

As the time travelling increased, the number of miles travelled increased. This is an example of direct variation which says as the value of one quantity increases so does the value of the other quantity

### Review

State whether each situation described is an example of direct variation or inverse variation.

1. The train traveled at a speed of 80 miles per hour. The number of miles increased with every hour that the train was running.

2. The train traveled at a speed of 80 miles per hour, then it increased it’s speed. The number of hours on the train decreased with the increase in speed.

3. Mary is training for a marathon. She runs many hours each week. After a few weeks of running, she noticed that her speed increased.

4. Kevin has been working for the same company for several years. He has received a raise each year that he has been with the company.

5. Joseph has been working for the same company too, but his salary has decreased each year.

6. Kelly spent more hours studying than she ever had before. She was surprised when she received a lower score on this test than she had on any previous tests.

7. Jeff is on a diet. He knows that the number of calories that he burns is directly connected with the number of hours that he exercises.

8. Seth and Sarah spent a lot of time eating while they were on vacation. After vacation was over, they both noticed that they had each gained five pounds.

9. Mary’s training has increased. She has been keeping track of how long it takes her to run 1 mile. She noticed that the she trained, the faster her time became.

10. Over time, the price of a postage stamp has increased a few pennies each year.

Answer each question true or false.

11. In direct variation, one factor increases as the other factor increases.

12. In inverse variation, one factor increases, but the other factor decreases.

13. When you see the letter k in a situation, you can think of a constant variable.

14. If a plane ascends and then quickly descends, this is an example of direct variation.

15. If your exercise increases and you lose weight, then this is an example of direct variation.

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### Vocabulary Language: English

TermDefinition
Direct Variation When the dependent variable grows large or small as the independent variable does.
Inverse Variation Inverse variation is a relationship between two variables in which the product of the two variables is equal to a constant. As one variable increases the second variable decreases proportionally.