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Here you will learn about direct and inverse variation, including the concepts of direct proportion and inverse proportion. You will look at solving some real-life word problems involving these different proportional relationships.
Students will first learn about direct and inverse variation as part of ratios and proportional relationships in 7th grade.
Direct and inverse variation are two different types of proportional relationships that describe how quantities are related to each other.
For example,
The cost of a banana is 70 cents. As the number of bananas increases, so does the cost; 3 bananas would cost 3 times the cost of one banana (\$2.10).
If y is directly proportional to x (y\propto{x}), then y=kx where k is the constant of proportionality.
Step-by-step guide: Directly proportional
For example, it takes 1 worker 9 hours to dig a hole. As the number of workers increases, the number of hours it takes to dig the same hole decreases. 3 workers would take a third of the time ( 3 hours).
To calculate inverse variation problems, you need to appreciate that multiplication and division are inverse operations of each other.
If y is inversely proportional to x (y\propto\cfrac{1}{x}), then y=\cfrac{k}{x} where k is the constant of proportionality.
Step-by-step guide: Inversely proportional
How does this relate to 7th grade math and high school math?
Use this worksheet to check your 7th and 8th grade studentsβ understanding of direct variation. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREEUse this worksheet to check your 7th and 8th grade studentsβ understanding of direct variation. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREEIn order to answer word problems involving direct and inverse variation:
A t-shirt costs \$4. How much do 5 t-shirts cost?
As the number of t-shirts increases, so does the cost. This is an example of direct variation, or direct proportion.
2Calculate the constant of proportionality, \textbf{k} .
For direct proportion, the constant of proportionality k is the cost of one t-shirt.
As this is a given value (a t-shirt costs \$4 ) you can say k=4 and so y=4x where the y value would represent the total cost of x number of t-shirts.
3Calculate the unknown value.
Substituting x=5 into y=4x, you have y=4\times{5}=20.
4Write the solution.
The cost of 5 t-shirts is \$20.
7 bags of candy weigh 350 grams. How much do 10 bags of candy weigh?
Determine the type of proportionality relationship between the two quantities.
As the number of bags of candy increases, so does the weight. This is a direct proportion problem.
Calculate the constant of proportionality, \textbf{k} .
For direct proportion, the constant of proportionality k is the weight of one bag of candy.
Using k=\cfrac{y}{x} where y is the weight of a bag of candy and x is the number of bags of candy, you can calculate the value of k.
k=\cfrac{350}{7}=50
k=50 and so a bag of candy weighs 50g and you can say y=50x.
Calculate the unknown value.
Substituting x=10 into y=50x, you have y=50\times{10}=500.
Write the solution.
The weight of 10 bags of candy is 500g.
8 laps of a race track has a total of 12 \, km. What would the distance be for 20 laps of the race track?
Determine the type of proportionality relationship between the two quantities.
As the number of laps of the track increases, so does the total distance. This is a direct proportion problem.
Calculate the constant of proportionality, \textbf{k} .
For direct proportion, the constant of proportionality k is the distance of one lap of the track.
Using k=\cfrac{y}{x} where y is the distance traveled and x is the number of laps, you can calculate the value of k.
k=\cfrac{12}{8}=1.5
k=1.5 and so one lap of the track is 1.5 \, km and you can say y=1.5x.
Calculate the unknown value.
Substituting x=20 into y=1.5x, you have y=1.5\times{20}=30.
Write the solution.
The distance covered in 20 laps is 30 \, km.
A worker takes 10 days to fit a bathroom. How long would it take 2 workers to fit a bathroom?
Determine the type of proportionality relationship between the two quantities.
As the number of workers increases, the time taken to fit a bathroom decreases. This is an inverse proportion problem.
Calculate the constant of proportionality, \textbf{k} .
For inverse proportion, the constant of proportionality k is the time it takes one person to fit a bathroom.
As one worker takes 10 days to fit a bathroom, you can say that k=10 and so you have the equation y=\cfrac{10}{x} where y is the time taken for x number of workers to complete a bathroom.
Calculate the unknown value.
Substituting x=2 into y=\cfrac{10}{x}, you have y=10\div{2}=5.
Write the solution.
It takes 2 workers 5 days to complete a bathroom.
An oil tank takes 25 hours to be filled by 3 hose pipes. How long does it take 5 hose pipes to fill the same oil tank?
Determine the type of proportionality relationship between the two quantities.
As the number of hose pipes increases, the time taken to fill the oil tank decreases. This is an inverse proportion problem.
Calculate the constant of proportionality, \textbf{k} .
For inverse proportion, the constant of proportionality k is the time it takes one hose to fill the oil tank.
Using k={x}{y} where x is the number of hoses and y is the time taken to fill the oil tank, you can calculate the value of k.
k=3\times{25}=75
k=75 and so one hose would take 75 hours to fill the oil tank and you can say y=\cfrac{75}{x}.
Calculate the unknown value.
Substituting x=5 into y=\cfrac{75}{x}, you have y=75\div{5}=15.
Write the solution.
It takes 5 hoses 15 hours to fill an oil tank.
10 computers can do a task in 15 minutes. How long does it take 3 computers to do the same task?
Determine the type of proportionality relationship between the two quantities.
As the number of computers increases, the time taken to do a task decreases. This is an inverse proportion problem.
Calculate the constant of proportionality, \textbf{k} .
For inverse proportion, the constant of proportionality k is the time it takes one computer to complete a task.
Using k=xy where x is the number of computers and y is the time taken to complete the task, you can calculate the value of k.
k=10\times{15}=150
k=150 and so one computer would take 150 minutes to complete a task and you can say y=\cfrac{150}{x}.
Calculate the unknown value.
Substituting x=3 into y=\cfrac{150}{x}, you have y=150\div{3}=50.
Write the solution.
It takes 3 computers 50 minutes to complete a task.
1. One tennis ball weighs 57 grams. Find the weight of 4 tennis balls.
7 grams
228 grams
14.25 grams
61 grams
k=57 and y=kx where y is the weight of x number of tennis balls.
This means that y=57x.
When x=4, \, y=57\times 4=228 grams.
2. One worker takes 30 hours to build a wall. Find the time it would take 5 workers to build a similar wall.
150 hours
35 hours
6 hours
120 hours
k=30 and y=\cfrac{k}{x} where y is the time taken to build a wall with x number of people.
This means that y=\cfrac{30}{x}.
When x=5, \, y=30\div{5}=6 hours.
3. 4 computer games cost \$18. Find the cost of 5 computer games.
k=\cfrac{y}{x} where y is the cost and x Β is the number of computer games.Β
k=\cfrac{18}{4} = 4.5
This means that 1 Β computer game costs \$4.50
And so y=4.5x
When k=4.5, \, y=4.5\times{3}=\${22.50}
4. 7 workers take 20 weeks to build a house. How long would it take 10 workers to build the same house?
17 weeks
28.6 weeks
14 weeks
140 weeks
k=xy. When x=7, \, y=20 and so k=7 \times 20=140.
This means that it would take 1 person 140 weeks to build the house and so y=\cfrac{140}{x}.
When x=10, \, y=140\div{10}=14 weeks.
5. 5 pens cost 65 cents. Find the cost of 8 pens.
When x=5, \, y=65 and so k=65\div{5}=13.
This means that 1 pen costs \$0.13 and so y=0.13x.
When x=8, \, y=0.13\times{8}=\${1.04}
6. 4 machines take 15 hours to complete a job. Find how long it would take 3 machines to complete the same job.
45 hours
5 hours
60 hours
20 hours
k=xy. When x=4, \, y=15 and so k=4 \times 15=60.
This means that it would take 1 machine 60 hours to complete the job and so y=\cfrac{60}{x}.
When x=3, \, y=60\div{3}=20 hours.
Direct variation describes a relationship between two variables in which one variable increases or decreases proportionally with the other.
The direct variation equation is y=kx where k is the constant of variation.
Inverse variation occurs when one variable increases as the other decreases. The product of the two variables remains constant.
The inverse variation equation is y=\cfrac{k}{x} where k is the constant of variation.
Students typically learn about direct and inverse variation in middle school (Grades 7-8) as they explore basic proportional relationships.
In high school, concepts are covered in more depth in Algebra 1 and Algebra II, where students work with equations and real-world applications.
Advanced courses like Pre-Calculus may revisit these concepts through functions and transformations.
Joint variation is a type of proportional relationship where a variable depends on the product of two or more other variables. Itβs represented by the equation z=kxy.
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