example of direct variation problem solving

• Direct Variation

When two variables change in proportion it is called as direct variation. In direct variation one variable is constant times of other. If one variable increases other will increase, if one decrease other will also decease. This means that the variables change in a same ratio which is called as constant of variation.

Direct variation is the simplest type of variation and in practical life we can find many situations which can be co-related with direct variation.

If two variables A and B are so related that when A increases ( or decreases ) in a given ratio, B also increases ( or, decreases ) in the same ratio, then  A is said to vary directly as B  ( or, A is said to vary as B ).

This is symbolically written as, A ∝ B (read as, ‘A varies as B’ ). 

Suppose a train moving at a uniform speed travels d km. in t minutes. Now, consider the following table:

Like in a math examination if for one problem solving we can score 10 numbers, so five problems solving we can get 50 numbers. This can be explained with a direct variation equation. If T denotes total numbers scored, N denotes numbers of problem solved and K denotes numbers can be scored for solving a problem, then the direct variation equation for this situation will be T = KN. 

As the numbers for a problem can be scored is fixed, it is a constant = K = \(\frac{T}{N}\) = 10

For solving 5 problems total numbers scored T = KN = 10 x 5 = 50.

From the above example we can understand that the ratio of two variables is a constant K and T, N are the variables which changes in proportion with value of constant.

Direct variation can be by a linear equation Y = KX where K is a constant. When the value of constant is higher, the change of variable Y is significantly for small change of X. But when the value of K is very small, Y changes very less with change of X. For this case K is equivalent to the ratio of change of two variables. So \(\frac{σY}{σX}\) = K when K is very small.

Now we will solve some problems on direct variation:

1.  If P varies directly as Q and the value of P is 60 and Q is 40, what is the equation that describes this direct variation of P and Q?

As P varies directly with Q, ratio of P and Q is constant for any value of P and Q.

So constant K = \(\frac{P}{Q}\) = \(\frac{60}{40}\) = \(\frac{3}{2}\)

So the equation that describes the direct variation of P and Q is P = \(\frac{3}{2}\)Q.

2.  If a car runs at a constant speed and takes 3 hrs to run a distance of 180 km, what time it will take to run 100 km?

If T is the time taken to cover the distance and S is the distance and V is the speed of the car, the direct variation equation is S= VT where V is constant.

For the case given in the problem,

180 = V × 3 or V = \(\frac{180}{3}\) = 60

So speed of the car is 60kmph and it is constant.

For 100 km distance 

S = VT or 100 = 60 × T

T = \(\frac{100}{60}\) = \(\frac{5}{3}\) hrs = 1 hr 40 mins.

So it will take 1 hr 40 mins time.

3.  In X is in direct variation with square of Y and when X is 4, Y is 3. What is the value of X when Y is 6?

From the given problem direct variation equation can be expressed as

For the given case

4 = K × 3 2

or, K = \(\frac{4}{9}\)

So when Y is 6,

X = \(\frac{4}{9}\) × 6 = \(\frac{8}{3}\)

So the value of X is \(\frac{8}{3}\).

●   Variation

• What is Variation?
• Inverse Variation
• Joint Variation
• Theorem of Joint Variation
• Worked out Examples on Variation
• Problems on Variation

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Direct Variation: Definition, Formula, Equation, Examples, FAQs

What is direct variation in math, what is a direct variation equation (direct variation formula), difference between direct variation and inverse variation, solved examples of direct variation, practice problems on direct variation, frequently asked questions about direct variation.

Direct variation is a type of proportionality in which one quantity directly varies with respect to a change in another quantity, by the same factor. 

This means that an increase in one quantity results in a proportionate increase in the other quantity. Similarly, a decrease in one quantity results in a proportionate decrease in the other quantity. There will be a proportionate increase or decrease.

Note that the ratio between two quantities that are in direct proportion or direct variation always remains the same. 

In direct variation, the variable that represents the cause of the relationship is called the independent variable, generally denoted by x. The other variable depends on the value of the independent variable; it is called the dependent variable, generally denoted by y. 

( Note: It’s important to note that the roles of independent and dependent variables vary depending on the specific context and the relationship between the defined variables. It is always better to first understand the relationship between two quantities in comparison.)

Direct variation real-life example:

Suppose you are working on an hourly pay basis. In this case, the number of hours you work is directly proportional to your earnings. The more you work, the more you earn. You can earn double by working twice as many hours.  

Here, $x =$ Number of hours you work and $y =$ The amount of money you earn

Direct Variation Definition

Direct variation or direct proportionality is a mathematical relationship between two variables where one variable varies in direct proportion with respect to the other variable.

Direct Variation Symbol

Suppose that a variable y is directly proportional to x. In other words, y varies directly as x.

We can write this mathematically as

$y \propto x$

Symbol “$\propto$” stands for “is proportional to.”

The relationship between two variables x and y, which are in direct variation, can be represented by an equation

$k = \frac{y}{x}$ is the constant of variation representing the constant ratio between the two variables.

Here, k is a non-zero real constant. The value of k can be both positive or negative.

The above equation can be decoded as “y varies directly with x” or “y varies directly as x.” The equation can be used to derive different formulas according to the requirements of a solution. 

Different forms of the direct variation formula are as follows:

• To solve for the constant of variation, k, we can shuffle the formula.

$k = \frac{y}{x}$ and $x \neq 0$

• To solve for one of the variables (x or y), we can shuffle the formula.

$x = \frac{y}{k}$

Direct Variation on a Graph

Note that the graph of the direct variation linear equation $y = kx$ is a straight line passing through the origin.

• If k is positive, then the line will rise from left to right, passing through the origin.
• If k is negative, then the line will fall from left to right, passing through the origin.

The slope of the line represents the constant of variation in a direct variation. It indicates how much the dependent variable changes for each unit increase in the independent variable.

So, to understand how to graph direct variation, we will identify the plot points satisfying the equation $y = kx$.

Cross Multiplication Method to Solve Direct Variation Problems

We can easily solve direct variation problems using the cross multiplication method. Consider two variables, x and y, in direct variation. 

Let y = B for x = A.

Let y = D for x = C.

Since x and y are in direct variation, we can write the proportion as

$\frac{A}{B} = \frac{C}{D}$ 

By cross multiplication, we get

Thus, $A \times D = B \times C$

We can find any missing value using this method.

In a direct variation, the variables change in proportion to each other, while in an inverse variation, the variables change in inverse proportion to each other.

See this comparison table that shows the difference between Direct and Inverse variations:

Facts on Direct Variation

• The constant of variation (k) is the fixed ratio that does not change even if the values of x and y change.
• The graph line passes through the origin (0,0) because the direct variation equation always includes the term y = 0 when x = 0.
• The constant of variation (k) gets bigger with a rise in the steepness of the slope of the line.
• A horizontal graph line indicates that the variable y is constant and does not change with x; hence, it does not have direct variation.
• A vertical graph line indicates that the variable x is constant and does not change with y; hence, it does not have direct variation.

Okay, so we now have a clear idea of the concepts relating to direct variation, including direct variation function, definition, graph, differences, and much more. Let us now look at some examples and practice problems to boost our understanding of the concept.

1. The cost of printing 100 pages is $50. What will be the cost of printing 150 pages?

Cost $\propto$ Number of pages

Cost $= k \times$ Number of pages

$k = \frac{y}{x} = \frac{cost}{Number \;of \;pages}$

$k = \frac{50}{100} = 0.5$

Now, the cost of 150 pages can be calculated by a direct variation equation.

Cost $= 0.5 \times $Number of pages

Cost $= 0.5 \times 150$

Cost $= 75$

Therefore, $\$75$ is the cost of printing 150 pages.

Another method:

$\$50 \times 150 = 100 \times$ ?

?$ = \frac{\$7500}{100}$

2. If the number and cost of notebooks have a direct variation, what will be the cost of 15 notebooks if 5 notebooks cost $\$10$ ?

As the cost varies directly with the number of notebooks, we can solve it by setting up the proportions.

5 notebooks cost $\$10$.

Let the cost of 15 notebooks to be x.

The proportion can be written as

$\frac{5}{10} = \frac{15}{x}$

Cross multiply.

$5 \times x = 10 \times 15$

Solve for x

Therefore, $\$30$ is the cost of 15 notebooks.

3. If a car covers 240 miles in 4 hours time. How many miles will it travel in 6 hours?

Solution:  

We know that the distance covered is directly proportional to the time taken. So, we can solve this question accordingly.

240 miles$\times 6 =$ ?$\times 4$ hours

?$= 360$ miles

Therefore, the car will cover 360 miles in 6 hours.

4. A company produces 100 units of a product per day. How many units will it produce in 5 days?

The number of units produced are directly proportional to the number of days.

Suppose that the company produces y units in 5 days.

$y = 100 \times 5$

Therefore, 500 units can be produced in 5 days.

5. Find the constant of direct variation where the cost of 4 dozen oranges is $\$24$ . Also, find the cost of 8 dozen oranges.

Assume the cost of 8 dozen oranges to be y. 

As we can see that the cost of oranges varies directly with the number of dozens. So, we can set up a proportion as

$\frac{4}{24} = 8y$

Isolate y on one side by cross-multiplying both sides

$4y = 24 \times 8$

Simplify it

Therefore, 8 dozen oranges will cost $48.

Now, let us find the constant of direct variation by using the equation y = kx

where 

y = the cost,

x = the number of dozens, and

k = the constant of variation

$k = \frac{1}{6}$

Therefore, $\frac{1}{6}$ or 0.1667 (approx) is the constant of direct variation.

Attend this quiz & Test your knowledge.

What will be the cost of 10 donuts if 4 donuts cost $\$2$?

If it takes 2 hours to mow a 1 acre lawn, how many hours will it take to mow a lawn that is $\frac{1}{4}$ an acre, if a car can go 30 miles with one gallon of gas, then how far can it go with 6 gallons of gas, 5 pounds of apples cost $\$10$. what will be the cost of 8.5 pounds of apples, which equation represents a direct variation.

Are the terms direct variation and direct proportion the same?

Yes, the terms direct variation and direct proportion are used interchangeably, but mean the same thing.

How can you identify direct variation?

In direct variation, the ratio of two quantities is always constant. As one variable increases, the other variable also increases. Also, if one variable decreases, the other variable also decreases.

Can the constant of proportionality be negative?

Yes, it can be positive or negative. It can never be 0. When we study the positive rate of change, we only consider positive values of k.

Are direct variations always linear?

We mostly study direct variations that are linear, but they do not have to be linear. We can say that linear equations are a specific type of direct variation, but not all direct variations follow a linear relationship. In direct variation, the ratio between the two variables remains constant, regardless of the type of relationship they exhibit.

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Chapter 6: Proportions and Modeling Using Variation

Solve direct variation problems.

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula e = 0.16 s tells us her earnings, e , come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other.

The graph below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[/latex] is used for direct variation. The value k  is a nonzero constant greater than zero and is called the constant of variation . In this case, k  = 0.16 and n  = 1.

A General Note: Direct Variation

If x and y  are related by an equation of the form

then we say that the relationship is direct variation and y   varies directly with the n th power of x . In direct variation relationships, there is a nonzero constant ratio [latex]k=\frac{y}{{x}^{n}}[/latex], where k  is called the constant of variation , which help defines the relationship between the variables.

How To: Given a description of a direct variation problem, solve for an unknown.

• Identify the input, x , and the output, y .
• Determine the constant of variation. You may need to divide y  by the specified power of x  to determine the constant of variation.
• Use the constant of variation to write an equation for the relationship.
• Substitute known values into the equation to find the unknown.

Example 1: Solving a Direct Variation Problem

The quantity y  varies directly with the cube of x . If y  = 25 when x  = 2, find y  when x  is 6.

The general formula for direct variation with a cube is [latex]y=k{x}^{3}[/latex]. The constant can be found by dividing y  by the cube of x .

Now use the constant to write an equation that represents this relationship.

Substitute x = 6 and solve for y .

Analysis of the Solution

The graph of this equation is a simple cubic, as shown below.

Do the graphs of all direct variation equations look like Example 1?

No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0, 0).

The quantity y  varies directly with the square of x . If y  = 24 when x  = 3, find y  when x  is 4.

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Direct Variation

Related Topics: More Lessons for Grade 9 Math Worksheets

Videos, worksheets, solutions, and activities to help Algebra students learn about direct variation or direct proportion. What is direct variation and how to solve direct variation problems?

What is Direct Variation? We often use the term direct variation to describe a form of dependence of one variable on another. An equation that makes a line and crosses the origin is a form of direct variation, where the magnitude of x increases or decreases directly as y increases or decreases. Direct variation and inverse variation are used often in science when modeling activity, such as speed or velocity.

Some real life examples of direct variation are: The number of hours you work and the amount of your paycheck The amount of weight on a spring and the distance the spring will stretch The speed of a car and the distance traveled in a certain amount of time

How to solve direct variation problems? In general, if two quantities vary directly, if one goes up the other goes up or down proportionally. Graphically, we have a line that passed through the origin. The slope is k, which is called the constant of proportionality.

• Y varies directly with x. y = 54 when x = 9. Determine the direct variation equation and then determine y when x = 3.5.
• Hooke’s Law states that the displacement, d that a spring is stretched by a hanging object caries directly as the mass, m of the object. If the distance is 10 cm when the mass is 3 kg, what is the distance when the mass is 5 kg?
• Y varies directly with with the square of x. y = 32 when x = 4. Determine the direct variation equation and then determine y when x = 6.

What is the Direct Variation or Direct Proportionality Formula? Ever heard of two things being directly proportional? Well, a good example is speed and distance. The bigger your speed, the farther you’ll go over a given time period. So as one variable goes up, the other goes up too, and that’s the idea of direct proportionality. But you can express direct proportionality using equations, and that’s an important thing to do in algebra. See how to do that in the tutorial.

Algebra Word Problem: Variation Application Solving a Variation problem Example: Art’s wages are directly proportional to the number of hours he works per week. If Art works 36 hours in a week, he earns $540. What are his wages if he works 40 hours in a week?

Direct Variation Models Examples:

• The current standard for low-flow showerheads is 2.5 gallons per minute. Calculate how long it would take to fill a 30 gallon bathtub using such a showerhead to supply the water.
• Amen is using a hose to fill his swimming pool for the first time. He starts the hose at 10 P.M. and leaves it running all night. At 6 A.M. he measures the depth and calculates that the pool is four-sevenths full. At what time will his new pool be full?

Direct Variation Word Problems Determine the constant of variation and find the other missing value using the given information. Solve a direct variation equation y = kx, where k is the constant of variation. A direct variation graph is a straight line graph that passes through the origin.

Example: The distance required to stop a car varies directly as the square of its speed. If 200 feet are required to stop a car traveling 60 miles per hour, how many feet are required to stop a car traveling 100 miles per hour?

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Direct Variation Lesson

• Demonstrate an understanding of linear equations in two variables
• Learn how to solve a direct variation problem
• Learn how to solve a direct variation as a power problem
• Learn how to solve a word problem that involves direct variation

How to Solve a Direct Variation Problem

Solving a direct variation problem.

• Write the variation equation: y = kx or k = y/x
• Substitute in for the given values and find the value of k
• Rewrite the variation equation: y = kx with the known value of k
• Substitute the remaining values and find the unknown

Direct Variation as a Power

Direct variation word problems, skills check:.

Solve each direct variation problem.

The distance a body falls from rest varies directly as the square of the time it falls (disregarding air resistance). If a sky diver falls 64 feet in 2 seconds, how far will he fall in 8 seconds?

Please choose the best answer.

The annual simple interest earned on a savings account varies directly with the rate of interest. If the annual simple interest earned is $48 when the interest rate is 5%, find the annual simple interest earned when the interest rate is 4.2%.

The area of a circle varies directly with the square of its radius. A circle with a radius of 7 inches has an area of 153.94 in 2 (approx). What is the area of a circle with a radius of 2.9 inches?

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Direct Variation

Direct variation is a type of proportionality wherein one quantity directly varies with respect to a change in another quantity. This implies that if there is an increase in one quantity then the other quantity will experience a proportionate increase. Similarly, if one quantity decreases then the other quantity also decreases.  Direct variation is a linear relationship hence, the graph will be a straight line.

Further, if two quantities are in direct variation then one will be a constant multiple of the other. In this article, we will elaborate on direct variation, its definition, formula, graph and associated examples.

What is Direct Variation?

Direct variation exists between any two variables when one quantity is directly dependent on the other i.e. if one quantity increases with respect to the other quantity and vice versa. It is the relationship between two variables where one of the variables is a constant multiple of the other. Since the two variables are directly related to each other it is also termed as directly proportional. 

Direct variation and inverse variation are two types of proportionalities. Proportionality refers to a relationship where two quantities are multiplicatively connected by a constant. In a direct variation, the ratio of the two quantities remains the same whereas in an inverse variation the product of the two quantities remains constant. Here we shall check in detail the definition and examples of direct variation.

Direct Variation Definition

Two quantities as said to follow a direct variation if both increase or decrease by the same factor. Thus, an increase in one quantity leads to an increase in the other while a decrease in one quantity leads to a decrease in the other. In other words, if the ratio of the first quantity to the second quantity is a constant term, then the quantities are said to be directly proportional to each other. This constant value is known as the coefficient or constant of proportionality .

Direct Variation Example

The following two quick examples are helpful for an easy understanding of this concept of direct variation.

Example I: The formula for the circumference of a circle is given by C = 2πr or C = πd. Here, r is the radius and d is the diameter . This is an example of a direct variation. Thus, the circumference of a circle and its corresponding diameter are in direct variation with π being the constant of proportionality.

Example II:  The quantity of Iron boxes made is directly proportional to the number of iron blocks. The number of iron blocks needed for 40 boxes is 160. How many iron blocks are needed for a box?

In the given problem, the number of iron blocks needed for 40 boxes is referred to as y = 160, and the number of boxes is referred to as x = 40. The number of iron blocks needed for a box is k. Here we use the direct variation formula of y = kx.

160 = k × 40 k = 160/40 k = 4

Thus 4 iron blocks are needed for a box.

What Is Direct Variation Formula?

Direct variation formula helps relate two quantities, having a mathematical relationship such that one of the variables is a constant multiple of the other. When two quantities are directly proportional to each other or are in direct variation they are represented using the symbol "\(\propto\)". Suppose there are two quantities x and y that are in direct variation then they are expressed as follows:

y \(\propto\) x

When the proportionality sign is removed then the direct variation formula is given as follows.

Direct Variation Formula: y = kx

Here k is the constant of proportionality. If x is not equal to zero then the value of the constant of proportionality can be given as k = y/x. Thus, the ratio of these two variables is always a constant number. Another way of expressing the direct variation equation is x = y / k. This means that x is directly proportional to y with the constant of proportionality equalling 1 / k.

The formula for the direction variation for a set of two quantities that are linearly dependent is as follows. 

Let us understand the formula of direct variation with the help of a simple example. Example: Let us assume that y varies directly with x, and y = 30 when x = 6. What is the value of y when x = 100?

The given quantities are y 1   = 30, x 1  = 6, x 2  = 100, y 2  = ? Using direct variation formula we have the following expression.

y 1  / x 1  = y 2  / x 2 30/6 = y 2  / 1005 = y 2  / 100 y 2  = 500

Therefore the value of y when x = 100 is 500.

Direct Variation Graph

The graph of two quantities in a direct variation will result in a straight line. Thus, direct variation represents a linear equation in two variables . The linear equation is given by y = kx. The ratio of change \(\frac{\Delta y}{\Delta x}\) is also equal to k. This change represents the slope of the line . The direct variation graph is given as follows:

Difference Between Direct Variation and Inverse Variation

The difference between direct variation and inverse variation provides the relationship between two mathematical quantities. The two quantities are said to be in direct variation if one quantity is proportionally increasing with respect to another quantity, and the two quantities are said to be in inverse variation if one quantity is increasing and the other quantity is simultaneously decreasing. The relationship between the two quantities, y, and x is as follows.

Inverse and direct variation are both types of proportionalities. The difference between inverse and direct variation is given in the table below:

Related Articles:

• Direct Proportion
• Ratio and Proportion
• Percent Proportion

Important Notes on Direct Variation

• A direct variation is a proportionality relationship in which two quantities follow a direct relationship. This implies that an increase (or decrease) in one quantity leads to a corresponding increase (or decrease) in another quantity.
• The direct variation equation is a linear equation in two variables and is given by y = kx where k is the constant proportionality.
• The direct variation graph in a coordinate plane is a straight line.
• The ratio of two quantities in a direct variation is a constant.

Examples on Direct Variation

• Example 2: If x = 10 and y = 20 follow a direct variation then find the constant of proportionality Solution: As x and y are in a direct variation thus y = kx or k = y / x. k = 20 / 10 = 2 Answer: k = 2
• Example 3: Let x and y be in direct variation, x = 6 and y = 21. Then find the direct variation equation. Solution: As x and y are in a direct variation thus y = kx or k = y / x. k = 21 / 6 = 7 / 2 y = \(\frac{7}{2}x\) Answer: The direct variation equation is y = \(\frac{7}{2}x\)

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Practice Questions on Direct Variation

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FAQs on Direct Variation

When one quantity directly varies with respect to another quantity it is known as direct variation . This implies that if one quantity increases or decreases the other quantity also increases or decreases proportionately.

What is the Direct Variation Equation?

The direct variation equation is given as y = kx where y and x are the two varying quantities and k is the constant of proportionality.

What is a Real-Life Direct Variation Example?

One example of direct variation is the speed of a car and the distance covered by it. If the speed increases the distance traveled within a certain time will also increase. Similarly, if the speed of the car decreases the distance covered within that interval of time will also decrease.

How to Solve Direct Variation?

To solve questions on direct variation the formula used is y = kx. If the constant of proportionality needs to be determined then y is divided by x to get the result. Suppose k is given and either x or y need to be determined then these values can be substituted in the aforementioned equation to find the unknown value.

How to Tell if an Equation is a Direct Variation?

If an equation is not in the form of y = kx then it does not follow a direct variation. For example y = 3x + 5 is not of the form y = kx hence, it doesn't represent a direct variation. On the other hand, y = 1.2x is of the form y = kx and denotes a direct variation.

How to Graph a Direct Variation Equation?

To graph a direct variation equation, y = kx, the steps are as follows:

• Substitute x with numerical values.
• Find the corresponding values of y using the direct variation formula.
• Using these test points, plot a graph. The graph will be a straight line passing through the origin.

Is y = 2x a Direct Variation?

Yes. As y = 2x is of the form y = kx thus, it represents a direct variation. The value of y will increase or decrease with an increase or decrease in the value of x.Yes. 

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Chapter 2: Linear Equations

2.7 Variation Word Problems

Direct variation problems.

There are many mathematical relations that occur in life. For instance, a flat commission salaried salesperson earns a percentage of their sales, where the more they sell equates to the wage they earn. An example of this would be an employee whose wage is 5% of the sales they make. This is a direct or a linear variation, which, in an equation, would look like:

[latex]\text{Wage }(x)=5\%\text{ Commission }(k)\text{ of Sales Completed }(y)[/latex]

[latex]x=ky[/latex]

A historical example of direct variation can be found in the changing measurement of pi, which has been symbolized using the Greek letter π since the mid 18th century. Variations of historical π calculations are Babylonian [latex]\left(\dfrac{25}{8}\right),[/latex] Egyptian [latex]\left(\dfrac{16}{9}\right)^2,[/latex] and Indian [latex]\left(\dfrac{339}{108}\text{ and }10^{\frac{1}{2}}\right).[/latex] In the 5th century, Chinese mathematician Zu Chongzhi calculated the value of π to seven decimal places (3.1415926), representing the most accurate value of π for over 1000 years.

Pi is found by taking any circle and dividing the circumference of the circle by the diameter, which will always give the same value: 3.14159265358979323846264338327950288419716… (42 decimal places). Using an infinite-series exact equation has allowed computers to calculate π to 10 13 decimals.

[latex]\begin{array}{c} \text{Circumference }(c)=\pi \text{ times the diameter }(d) \\ \\ \text{or} \\ \\ c=\pi d \end{array}[/latex]

All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are:

• [latex]x[/latex] varies directly as [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex]
• [latex]x[/latex] varies directly proportional to [latex]y[/latex]
• [latex]x[/latex] is proportional to [latex]y[/latex]
• [latex]x[/latex] varies directly as the square of [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex] squared
• [latex]x[/latex] is proportional to the square of [latex]y[/latex]
• [latex]x[/latex] varies directly as the cube of [latex]y[/latex]
• [latex]x[/latex] varies as [latex]y[/latex] cubed
• [latex]x[/latex] is proportional to the cube of [latex]y[/latex]
• [latex]x[/latex] varies directly as the square root of [latex]y[/latex]
• [latex]x[/latex] varies as the root of [latex]y[/latex]
• [latex]x[/latex] is proportional to the square root of [latex]y[/latex]

Example 2.7.1

Find the variation equation described as follows:

The surface area of a square surface [latex](A)[/latex] is directly proportional to the square of either side [latex](x).[/latex]

[latex]\begin{array}{c} \text{Area }(A) =\text{ constant }(k)\text{ times side}^2\text{ } (x^2) \\ \\ \text{or} \\ \\ A=kx^2 \end{array}[/latex]

Example 2.7.2

When looking at two buildings at the same time, the length of the buildings’ shadows [latex](s)[/latex] varies directly as their height [latex](h).[/latex] If a 5-story building has a 20 m long shadow, how many stories high would a building that has a 32 m long shadow be?

The equation that describes this variation is:

[latex]h=kx[/latex]

Breaking the data up into the first and second parts gives:

[latex]\begin{array}{ll} \begin{array}{rrl} \\ &&\textbf{1st Data} \\ s&=&20\text{ m} \\ h&=&5\text{ stories} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ h&=&kx \\ 5\text{ stories}&=&k\text{ (20 m)} \\ k&=&5\text{ stories/20 m}\\ k&=&0.25\text{ story/m} \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ s&=&\text{32 m} \\ h&=&\text{find 2nd} \\ k&=&0.25\text{ story/m} \\ \\ &&\text{Find }h\text{:} \\ h&=&kx \\ h&=&(0.25\text{ story/m})(32\text{ m}) \\ h&=&8\text{ stories} \end{array} \end{array}[/latex]

Inverse Variation Problems

Inverse variation problems are reciprocal relationships. In these types of problems, the product of two or more variables is equal to a constant. An example of this comes from the relationship of the pressure [latex](P)[/latex] and the volume [latex](V)[/latex] of a gas, called Boyle’s Law (1662). This law is written as:

[latex]\begin{array}{c} \text{Pressure }(P)\text{ times Volume }(V)=\text{ constant} \\ \\ \text{ or } \\ \\ PV=k \end{array}[/latex]

Written as an inverse variation problem, it can be said that the pressure of an ideal gas varies as the inverse of the volume or varies inversely as the volume. Expressed this way, the equation can be written as:

[latex]P=\dfrac{k}{V}[/latex]

Another example is the historically famous inverse square laws. Examples of this are the force of gravity [latex](F_{\text{g}}),[/latex] electrostatic force [latex](F_{\text{el}}),[/latex] and the intensity of light [latex](I).[/latex] In all of these measures of force and light intensity, as you move away from the source, the intensity or strength decreases as the square of the distance.

In equation form, these look like:

[latex]F_{\text{g}}=\dfrac{k}{d^2}\hspace{0.25in} F_{\text{el}}=\dfrac{k}{d^2}\hspace{0.25in} I=\dfrac{k}{d^2}[/latex]

These equations would be verbalized as:

• The force of gravity [latex](F_{\text{g}})[/latex] varies inversely as the square of the distance.
• Electrostatic force [latex](F_{\text{el}})[/latex] varies inversely as the square of the distance.
• The intensity of a light source [latex](I)[/latex] varies inversely as the square of the distance.

All inverse variation relationship are verbalized in written problems as inverse variations or as inversely proportional. Examples of inverse variation or inversely proportional equations are:

• [latex]x[/latex] varies inversely as [latex]y[/latex]
• [latex]x[/latex] varies as the inverse of [latex]y[/latex]
• [latex]x[/latex] varies inversely proportional to [latex]y[/latex]
• [latex]x[/latex] is inversely proportional to [latex]y[/latex]
• [latex]x[/latex] varies inversely as the square of [latex]y[/latex]
• [latex]x[/latex] varies inversely as [latex]y[/latex] squared
• [latex]x[/latex] is inversely proportional to the square of [latex]y[/latex]
• [latex]x[/latex] varies inversely as the cube of [latex]y[/latex]
• [latex]x[/latex] varies inversely as [latex]y[/latex] cubed
• [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex]
• [latex]x[/latex] varies inversely as the square root of [latex]y[/latex]
• [latex]x[/latex] varies as the inverse root of [latex]y[/latex]
• [latex]x[/latex] is inversely proportional to the square root of [latex]y[/latex]

Example 2.7.3

The force experienced by a magnetic field [latex](F_{\text{b}})[/latex] is inversely proportional to the square of the distance from the source [latex](d_{\text{s}}).[/latex]

[latex]F_{\text{b}} = \dfrac{k}{{d_{\text{s}}}^2}[/latex]

Example 2.7.4

The time [latex](t)[/latex] it takes to travel from North Vancouver to Hope varies inversely as the speed [latex](v)[/latex] at which one travels. If it takes 1.5 hours to travel this distance at an average speed of 120 km/h, find the constant [latex]k[/latex] and the amount of time it would take to drive back if you were only able to travel at 60 km/h due to an engine problem.

[latex]t=\dfrac{k}{v}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} &&\textbf{1st Data} \\ v&=&120\text{ km/h} \\ t&=&1.5\text{ h} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ k&=&tv \\ k&=&(1.5\text{ h})(120\text{ km/h}) \\ k&=&180\text{ km} \end{array} & \hspace{0.5in} \begin{array}{rrl} \\ \\ \\ &&\textbf{2nd Data} \\ v&=&60\text{ km/h} \\ t&=&\text{find 2nd} \\ k&=&180\text{ km} \\ \\ &&\text{Find }t\text{:} \\ t&=&\dfrac{k}{v} \\ \\ t&=&\dfrac{180\text{ km}}{60\text{ km/h}} \\ \\ t&=&3\text{ h} \end{array} \end{array}[/latex]

Joint or Combined Variation Problems

In real life, variation problems are not restricted to single variables. Instead, functions are generally a combination of multiple factors. For instance, the physics equation quantifying the gravitational force of attraction between two bodies is:

[latex]F_{\text{g}}=\dfrac{Gm_1m_2}{d^2}[/latex]

• [latex]F_{\text{g}}[/latex] stands for the gravitational force of attraction
• [latex]G[/latex] is Newton’s constant, which would be represented by [latex]k[/latex] in a standard variation problem
• [latex]m_1[/latex] and [latex]m_2[/latex] are the masses of the two bodies
• [latex]d^2[/latex] is the distance between the centres of both bodies

To write this out as a variation problem, first state that the force of gravitational attraction [latex](F_{\text{g}})[/latex] between two bodies is directly proportional to the product of the two masses [latex](m_1, m_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating the two masses. From this information, the necessary equation can be derived. All joint variation relationships are verbalized in written problems as a combination of direct and inverse variation relationships, and care must be taken to correctly identify which variables are related in what relationship.

Example 2.7.5

The force of electrical attraction [latex](F_{\text{el}})[/latex] between two statically charged bodies is directly proportional to the product of the charges on each of the two objects [latex](q_1, q_2)[/latex] and inversely proportional to the square of the distance [latex](d)[/latex] separating these two charged bodies.

[latex]F_{\text{el}}=\dfrac{kq_1q_2}{d^2}[/latex]

Solving these combined or joint variation problems is the same as solving simpler variation problems.

First, decide what equation the variation represents. Second, break up the data into the first data given—which is used to find [latex]k[/latex]—and then the second data, which is used to solve the problem given. Consider the following joint variation problem.

Example 2.7.6

[latex]y[/latex] varies jointly with [latex]m[/latex] and [latex]n[/latex] and inversely with the square of [latex]d[/latex]. If [latex]y = 12[/latex] when [latex]m = 3[/latex], [latex]n = 8[/latex], and [latex]d = 2,[/latex] find the constant [latex]k[/latex], then use [latex]k[/latex] to find [latex]y[/latex] when [latex]m=-3[/latex], [latex]n = 18[/latex], and [latex]d = 3[/latex].

[latex]y=\dfrac{kmn}{d^2}[/latex]

[latex]\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ && \textbf{1st Data} \\ y&=&12 \\ m&=&3 \\ n&=&8 \\ d&=&2 \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ 12&=&\dfrac{k(3)(8)}{(2)^2} \\ \\ k&=&\dfrac{12(2)^2}{(3)(8)} \\ \\ k&=& 2 \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ y&=&\text{find 2nd} \\ m&=&-3 \\ n&=&18 \\ d&=&3 \\ k&=&2 \\ \\ &&\text{Find }y\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ y&=&\dfrac{(2)(-3)(18)}{(3)^2} \\ \\ y&=&12 \end{array} \end{array}[/latex]

For questions 1 to 12, write the formula defining the variation, including the constant of variation [latex](k).[/latex]

• [latex]x[/latex] is jointly proportional to [latex]y[/latex] and [latex]z[/latex]
• [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex]
• [latex]x[/latex] is jointly proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
• [latex]x[/latex] is inversely proportional to [latex]y[/latex] to the sixth power
• [latex]x[/latex] is jointly proportional with the cube of [latex]y[/latex] and inversely to the square root of [latex]z[/latex]
• [latex]x[/latex] is inversely proportional with the square of [latex]y[/latex] and the square root of [latex]z[/latex]
• [latex]x[/latex] varies jointly as [latex]z[/latex] and [latex]y[/latex] and is inversely proportional to the cube of [latex]p[/latex]
• [latex]x[/latex] is inversely proportional to the cube of [latex]y[/latex] and square of [latex]z[/latex]

For questions 13 to 22, find the formula defining the variation and the constant of variation [latex](k).[/latex]

• If [latex]A[/latex] varies directly as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=15[/latex] and [latex]B=5.[/latex]
• If [latex]P[/latex] is jointly proportional to [latex]Q[/latex] and [latex]R,[/latex] find [latex]k[/latex] when [latex]P=12, Q=8[/latex] and [latex]R=3.[/latex]
• If [latex]A[/latex] varies inversely as [latex]B,[/latex] find [latex]k[/latex] when [latex]A=7[/latex] and [latex]B=4.[/latex]
• If [latex]A[/latex] varies directly as the square of [latex]B,[/latex] find [latex]k[/latex] when [latex]A=6[/latex] and [latex]B=3.[/latex]
• If [latex]C[/latex] varies jointly as [latex]A[/latex] and [latex]B,[/latex] find [latex]k[/latex] when [latex]C=24, A=3,[/latex] and [latex]B=2.[/latex]
• If [latex]Y[/latex] is inversely proportional to the cube of [latex]X,[/latex] find [latex]k[/latex] when [latex]Y=54[/latex] and [latex]X=3.[/latex]
• If [latex]X[/latex] is directly proportional to [latex]Y,[/latex] find [latex]k[/latex] when [latex]X=12[/latex] and [latex]Y=8.[/latex]
• If [latex]A[/latex] is jointly proportional with the square of [latex]B[/latex] and the square root of [latex]C,[/latex] find [latex]k[/latex] when [latex]A=25, B=5[/latex] and [latex]C=9.[/latex]
• If [latex]y[/latex] varies jointly with [latex]m[/latex] and the square of [latex]n[/latex] and inversely with [latex]d,[/latex] find [latex]k[/latex] when [latex]y=10, m=4, n=5,[/latex] and [latex]d=6.[/latex]
• If [latex]P[/latex] varies directly as [latex]T[/latex] and inversely as [latex]V,[/latex] find [latex]k[/latex] when [latex]P=10, T=250,[/latex] and [latex]V=400.[/latex]

For questions 23 to 37, solve each variation word problem.

• The electrical current [latex]I[/latex] (in amperes, A) varies directly as the voltage [latex](V)[/latex] in a simple circuit. If the current is 5 A when the source voltage is 15 V, what is the current when the source voltage is 25 V?
• The current [latex]I[/latex] in an electrical conductor varies inversely as the resistance [latex]R[/latex] (in ohms, Ω) of the conductor. If the current is 12 A when the resistance is 240 Ω, what is the current when the resistance is 540 Ω?
• Hooke’s law states that the distance [latex](d_s)[/latex] that a spring is stretched supporting a suspended object varies directly as the mass of the object [latex](m).[/latex] If the distance stretched is 18 cm when the suspended mass is 3 kg, what is the distance when the suspended mass is 5 kg?
• The volume [latex](V)[/latex] of an ideal gas at a constant temperature varies inversely as the pressure [latex](P)[/latex] exerted on it. If the volume of a gas is 200 cm 3 under a pressure of 32 kg/cm 2 , what will be its volume under a pressure of 40 kg/cm 2 ?
• The number of aluminum cans [latex](c)[/latex] used each year varies directly as the number of people [latex](p)[/latex] using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in a city that has a population of 1,000,000?
• The time [latex](t)[/latex] required to do a masonry job varies inversely as the number of bricklayers [latex](b).[/latex] If it takes 5 hours for 7 bricklayers to build a park wall, how much time should it take 10 bricklayers to complete the same job?
• The wavelength of a radio signal (λ) varies inversely as its frequency [latex](f).[/latex] A wave with a frequency of 1200 kilohertz has a length of 250 metres. What is the wavelength of a radio signal having a frequency of 60 kilohertz?
• The number of kilograms of water [latex](w)[/latex] in a human body is proportional to the mass of the body [latex](m).[/latex] If a 96 kg person contains 64 kg of water, how many kilograms of water are in a 60 kg person?
• The time [latex](t)[/latex] required to drive a fixed distance [latex](d)[/latex] varies inversely as the speed [latex](v).[/latex] If it takes 5 hours at a speed of 80 km/h to drive a fixed distance, what speed is required to do the same trip in 4.2 hours?
• The volume [latex](V)[/latex] of a cone varies jointly as its height [latex](h)[/latex] and the square of its radius [latex](r).[/latex] If a cone with a height of 8 centimetres and a radius of 2 centimetres has a volume of 33.5 cm 3 , what is the volume of a cone with a height of 6 centimetres and a radius of 4 centimetres?
• The centripetal force [latex](F_{\text{c}})[/latex] acting on an object varies as the square of the speed [latex](v)[/latex] and inversely to the radius [latex](r)[/latex] of its path. If the centripetal force is 100 N when the object is travelling at 10 m/s in a path or radius of 0.5 m, what is the centripetal force when the object’s speed increases to 25 m/s and the path is now 1.0 m?
• The maximum load [latex](L_{\text{max}})[/latex] that a cylindrical column with a circular cross section can hold varies directly as the fourth power of the diameter [latex](d)[/latex] and inversely as the square of the height [latex](h).[/latex] If an 8.0 m column that is 2.0 m in diameter will support 64 tonnes, how many tonnes can be supported by a column 12.0 m high and 3.0 m in diameter?
• The volume [latex](V)[/latex] of gas varies directly as the temperature [latex](T)[/latex] and inversely as the pressure [latex](P).[/latex] If the volume is 225 cc when the temperature is 300 K and the pressure is 100 N/cm 2 , what is the volume when the temperature drops to 270 K and the pressure is 150 N/cm 2 ?
• The electrical resistance [latex](R)[/latex] of a wire varies directly as its length [latex](l)[/latex] and inversely as the square of its diameter [latex](d).[/latex] A wire with a length of 5.0 m and a diameter of 0.25 cm has a resistance of 20 Ω. Find the electrical resistance in a 10.0 m long wire having twice the diameter.
• The volume of wood in a tree [latex](V)[/latex] varies directly as the height [latex](h)[/latex] and the diameter [latex](d).[/latex] If the volume of a tree is 377 m 3 when the height is 30 m and the diameter is 2.0 m, what is the height of a tree having a volume of 225 m 3 and a diameter of 1.75 m?

Answer Key 2.7

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1.8: Variation - Constructing and Solving Equations

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Solving Problems involving Direct, Inverse, and Joint variation

Certain relationships occur so frequently in applied situations that they are given special names. Variation equations show how one quantity changes in relation to other quantities. The relationship between the quantities can be described as direct , inverse , or joint variation .

Direct Variation

Many real-world problems encountered in the sciences involve two types of functional relationships. The first type of functional relationship can be explored using the fact that the distance \(s\) in feet an object falls from rest, without regard to air resistance, can be approximated using the following formula:

\(s=16t^{2}\)

Here \(t\) represents the time in seconds the object has been falling. For example, after \(2\) seconds the object will have fallen \(s = 16 ( 2 ) ^ { 2 } = 16 \cdot 4 = 64\) feet.

In this example, we can see that the distance varies over time as the product of a constant \(16\) and the square of the time \(t\). This relationship is described as direct variation  and \(16\) is called the constant of variation or the  constant of proportionality . 

Definition: Direct Variation (\(y=kx\))

Direct variation is a relationship where quantities behave in a like manner. If one increases, so does the other. If one decreases, so does the other.

For two quantities \(x\) and \(y\), this relationship is described as "\(y\) varies directly as \(x\)" or "\(y\) is directly proportional to \(x\)" .

The equation that describes this relationship is \(y=kx\) , where \(k\) is a non-zero constant called the constant of variation or the proportionality constant .

• Translate the given English statement containing the words varies  or proportional , into a model equation.
• Substitute a given set of values into the equation and solve for \(k\), the constant of variation.
• Rewrite the equation obtained in step 1 as a formula with a value for \(k\) found in step 2 defined. Make note of the units used for each variable in the formula.
• Use the equation from step 3, and another set of values (with one value missing) to solve for the unknown quantity.

Example \(\PageIndex{1}\): Direct Variation

An object’s weight on Earth varies directly to its weight on the Moon. If a man weighs \(180\) pounds on Earth, then he will weigh \(30\) pounds on the Moon. Set up an algebraic equation that expresses the weight on Earth in terms of the weight on the Moon and use it to determine the weight of a woman on the Moon if she weighs \(120\) pounds on Earth.

Step 1. Translate “the weight on Earth varies directly to the weight on the Moon.”    \(E = kM \) 

Step 2. Find \(k\) using "If a man weighs \(180\) pounds on Earth, then he will weigh \(30\) pounds on the Moon."    \(E=180\) pounds, \(M=30\) pounds

\(\begin{array} { cr } E = kM & \text{Model equation} \\180=k \cdot 30\\ \frac { 180 } { 30 } = k  \\ { 6 = k } \end{array}\)

Step 3. The formula is \(E = 6M \), where \(E\) is the weight on Earth in pounds and \(M\) is the weight on the moon in pounds.

Step 4. Answer the question: "determine the weight of a woman on the Moon if she weighs \(120\) pounds on Earth."    \(E=120\) pounds, find M

\(\begin{array} { cll } E = 6M & \text{Formula:} & \text{ \(E\) pounds on Earth}\\  && \text{ \(M\) pounds on the Moon}\\{ 120 = 6 M } \\ { \frac { 120 } { 6 } = M } \\ { 20 = M } \end{array}\)

The woman weighs \(20\) pounds on the Moon.

Indirect Variation

The second functional relationship can be explored using the model that relates the intensity of light \(I\) to the square of the distance from its source \(d\).

\(I = \frac { k } { d ^ { 2 } }\)

Here \(k\) represents some constant. A foot-candle is a measurement of the intensity of light. One foot-candle is defined to be equal to the amount of illumination produced by a standard candle measured one foot away. For example, a \(125\)-Watt fluorescent growing light is advertised to produce \(525\) foot-candles of illumination. This means that at a distance \(d=1\) foot, \(I=525\) foot-candles and we have:

\(\begin{array} { l } { 525 = \frac { k } { ( 1 ) ^ { 2 } } } \\ { 525 = k } \end{array}\)

Using \(k=525\) we can construct a formula which gives the light intensity produced by the bulb:

\(I = \frac { 525 } { d ^ { 2 } }\)

Here \(d\) represents the distance the growing light is from the plants. In the chart above, we can see that the amount of illumination fades quickly as the distance from the plants increases.

This type of relationship is described as an inverse variation . We say that I is inversely proportional  to the square of the distance \(d\), where \(525\) is the constant of proportionality.

Definition: Indirect Variation (\(y=\frac{k}{x}\))

Indirect variation is a relationship between quantities where if one increases, the other decreases.

For two quantities \(x\) and \(y\), this relationship is described as "\(y\) varies indirectly as \(x\)" or "\(y\) is inversely proportional to \(x\)" .

The equation that describes this relationship is \(y=\dfrac{k}{x}\) , where \(k\) is a non-zero constant called the constant of variation or the proportionality constant .

Example \(\PageIndex{2}\): Indirect Variation

The weight of an object varies inversely as the square of its distance from the center of Earth. If an object weighs \(100\) pounds on the surface of Earth (approximately \(4,000\) miles from the center), how much will it weigh at \(1,000\) miles above Earth’s surface?

Step 1. Translate  “\(w\) varies inversely as the square of \(d\)”    \(w = \frac { k } { d ^ { 2 } }\)

Step 2. Find \(k\) using "An object weighs \(100\) pounds on the surface of Earth, approximately \(4,000\) miles from the center".     \(w = 100\) when \(d = 4,000\)

\(\begin{aligned} \color{Cerulean}{( 4,000 ) ^ { 2 }}\color{black}{ \cdot} 100 & =\color{Cerulean}{ ( 4,000 ) ^ { 2 }}\color{black}{ \cdot} \frac { k } { ( 4,000 ) ^ { 2 } } \\ 1,600,000,000 &= k \\ 1.6 \times 10 ^ { 9 } &= k \end{aligned}\)

Step 3. The formula is \(w = \frac { 1.6 \times 10 ^ { 9 } } { d ^ { 2 } }\), where \(w\) is the weight of the object in pounds and \(d\) is the distance of the object from the center of the Earth in miles.

Step 4. Answer the question: "how much will it weigh at \(1,000\) miles above Earth’s surface?"     Since the object is \(1,000\) miles above the surface, the distance of the object from the center of Earth is  \(d = 4,000 + 1,000 = 5,000 \:\:\text{miles}\)

\(\begin{aligned} y & = \frac { 1.6 \times 10 ^ { 9 } } { ( \color{OliveGreen}{5,000}\color{black}{ )} ^ { 2 } } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 25,000,000 } \\ & = \frac { 1.6 \times 10 ^ { 9 } } { 2.5 \times 10 ^ { 9 } } \\ & = 0.64 \times 10 ^ { 2 } \\ & = 64 \end{aligned}\)

The object will weigh \(64\) pounds at a distance \(1,000\) miles above the surface of Earth.

Joint Variation

Lastly, we define relationships between multiple variables.

Definition: Joint Variation and Combined Variation

Joint variation is a relationship in which one quantity is proportional to the product of two or more quantities.

Combined variation exists when combinations of direct and/or inverse variation occurs

Example \(\PageIndex{3}\): Joint Variation

Step 1. If we let \(A\) represent the area of an ellipse, then we can use the statement “area varies jointly as \(a\) and \(b\)” to write

Step 2. To find the constant of variation \(k\), use the fact that the area is \(300π\) when \(a=10\) and \(b=30\).

\(\begin{array} { c } { 300 \pi = k ( \color{OliveGreen}{10}\color{black}{ )} (\color{OliveGreen}{ 30}\color{black}{ )} } \\ { 300 \pi = 300 k } \\ { \pi = k } \end{array}\)

Step 3. Therefore, the formula for the area of an ellipse is

\(A=πab\)

The constant of proportionality is \(π\) and the formula for the area of an ellipse is \(A=abπ\).

Given that \(y\) varies directly as the square of \(x\) and inversely with \(z\), where \(y=2\) when \(x=3\) and \(z=27\), find \(y\) when \(x=2\) and \(z=16\).

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Variation Word Problems

Equations Word Problems More Prob's

It's one thing to be able to take the words for a variation equation (such as " y varies directly as the square of x and inversely as the cube root of z ") and turn this into an equation that you can solve or use. It's another thing to extract the words from a word problem. But, because the lingo for variation equations is so specific, it's not really that hard. Just look for the keywords, and you're nearly home and dry.

The only other keywords (or "key-phrases", really) you might need to know are "is proportional to" which, in the strictly-mathematical sense, means "varies directly as"; and "is inversely proportional to" which means "varies inversely with".

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Suppose that y is inversely proportional to x , and that y  = 0.4 when x  = 2.5 . Find y when x  = 4 .

Translating the above from the English into algebra, I see the key-phrase "inversely proportional to", which means "varies indirectly as". In practical terms, it means that the variable part that does the varying is going to be in the denominator. So I get the formula:

Plugging in the data point they gave me, I can solve for the value of k :

0.4 = k /(2.5)

(0.4)(2.5) = k = 1

Now that I have found the value of the variation constant, I can plug in the x -value they gave me, and find the value of y when x  = 4 :

Then my answer is:

Most word problems, of course, are not nearly as simple as the above example (or the ones on the previous page). Instead, you have to figure out which values go where, what the equation is, and how to interpret it. Fortunately, the keywords and key-phrases should generally be fairly clear, telling you exactly what format to use.

According to Hooke's Law, the force needed to stretch a spring is proportional to the amount the spring is stretched. If fifty pounds of force stretches a spring five inches, how much will the spring be stretched by a force of 120 pounds?

"Is proportional to" means "varies directly with", so the formula for Hooke's Law is:

...where " F " is the force and " d " is the distance that the spring is stretched.

Note: In physics, "weight" is a force. These Hooke's Law word problems, among other types, are often stated in terms of weight, and the weight they list is the force they mean.

First I have to solve for the value of k . They've given me the data point ( d , F) = (5, 50) , so I'll plug this in to the formula:

Now I know that the formula for this particular spring is:

(Hooke's Law doesn't change, but each spring is different, so each spring will have its own " k ".)

Once I know the formula, I can answer their question: "How much will the spring be stretched by a force of 120 pounds?" I'll plug the value they've given me for the force into the equation I've found:

Note that they did not ask "What is the value of ' d '?". They asked me for a distance. I need to be sure to answer the question that they actually asked. That final answer is the distance that the spring is stretched, including the units (which are "inches", in this case):

Note: If you give the above answer as being only " 12 ", the grader will be perfectly correct to count your answer as being at least partly wrong. The answer is not a number, but is a number of units.

Kepler's third law of planetary motion states that the square of the time required for a planet to make one revolution about the sun varies directly as the cube of the average distance of the planet from the sun. If you assume that Mars is 1.5 times as far from the sun as is the earth, find the approximate length of a Martian year.

This one is a bit different from the previous exercise. Normally, I've given given a relation in terms of variation with a plain old variable, like y . In this case, though, the variation relationship is between the square of the time and the cube of the distance. This means that the left-hand side of my equation will have a squared variable!

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My variation formula is:

t 2 = k d 3

If I take " d  = 1 " to mean "the distance is one AU", an AU being an "astronomical unit" (the distance of earth from the sun), then the distance for Mars is 1.5 AU. Also, I will take " t  = 1 " to stand for "one earth year". Then, in terms of the planet Earth, I get:

(1) 2 = k  (1) 3

Then the formula, in terms of Earth, is:

Now I'll plug in the information for Mars (in comparison to earth):

t 2 = (1.5) 3

This is one of those times when a calculator's decimal approximation is probably going to be a little more useful in answering the question. I'll show the exact answer in my working, but I'll use a sensible approximation in my final answer. The decimal expansion starts as:

In other words, the Martian year is approximately the length of:

1.837 earth years

By the way, you can make the above answer more intuitive by finding the number of (Earth) days, approximately, represented by that " 0.837 " part of the answer above. Since the average Earth year, technically, has about 365.25 days, then the 0.837 of an Earth year is:

0.837 × 365.25 = 305.71425

Letting the "average" month have 365.25 ÷ 12 = 30.4375 days, then the above number of (Earth) days is:

305.71425 ÷ 30.4375 = 10.044

In other words, the Martian year is almost exactly one Earth year and ten Earth months long.

If you were writing for an audience (like a fellow student, as you'll be required to do in some class projects or essay questions), this "one year and ten months" form would probably be the best way to go.

The weight of a body varies inversely as the square of its distance from the center of the earth. If the radius of the earth is 4000 miles, how much would a 200 -pound man weigh 1000 miles above the surface of the earth?

Remembering that "weight" is a force, I'll let the weight be designated by " F ". The distance of a body from the center of the earth is " d   ". Then my variation formula is the following:

(200)(16,000,000) = k

3,200,000,000 = k

(Hey; there's nothing that says that the value of the variation constant k has to be small!)

The distance is always measured from the center of the earth. If the guy is in orbit a thousand miles up (from the surface of the planet), then his distance (from the center of the planet) is the 4000 miles from the center to the surface plus the 1000 miles from the surface to his ship. That is, d = 5000 . I'll plug this in to my equation, and solve for the value of the force (which is here called "weight") F :

Remembering my units, my answer is that, up in his spacecraft, the guy weighs:

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Inverse and Direct Variation Word Problems - Examples - Expii

Inverse and direct variation word problems - examples, explanations (4).

(Videos) Variation and Proportionality Word Problems

by mathman1024

This video by mathman1024 goes over a word problem based on variation.

The problem says, Johnny earns $594 when he works twenty-two hours. How much would he earn if he were to work thirty-seven hours?

We know that this is a direct variation problem because it deals with hours worked and pay. If you work more hours, you get paid more.

Once we know the variation we're using we can define our variables . P=amount of paycheck h=number of hours

So we can now define our paycheck to be: P=kh

The goal for these problems is to find the value of k (the constant of proportionality) and use it to solve for the desired value. To find k, take the specific example the problem has given. The given problem states Johnny earns $594 when he works twenty-two hours. So we can plug those into our equation. P=kh(594)=k(22)59422=22k2227=k

So what does this mean in terms of this problem? Well, it means that you make $27 per hour. Thus, k is now a known quantity, and our equation becomes: P=27h

Now we can answer the question they ask, How much would he earn if he were to work thirty-seven hours? P=27hP=27(37)P=999

We state, When he works 37 hourshe earns 999 dollars.

Related Lessons

Direct proportionality in science.

Directly proportional variables are used in the sciences. They come up most in the math-y sciences like chemistry and physics (i.e., the sciences that use variables).

That's where we're going to get our examples from today.

If you need a refresher on direct proportionality in general, click here .

There's also a page on inverse proportionality here .

By Masterdeis CC BY-SA 3.0 , from Wikimedia Commons

Chemistry: Gas Laws

Chemical gas laws are a great example of proportionality. There are several of these, but they all involve the same 3 variables:

• P (pressure)
• T (temperature).

Let's look at Charles' Law , which relates the volume and temperature of a gas.

Charles' Law V1T1=V2T2

V1= original volumeV2= new volumeT1= original temperatureT2= new temperature

In this case, volume and temperature are directly proportional . If volume goes up, temperature does too. The ratio between temperature and volume is constant.

Let's practice.

In a jar, Math Gas has a volume of 20 liters (L) and a temperature of 100 Kelvin (K).

What is the ratio of volume to temperature for Math Gas?

Proportionality: Word Problems

Real quick, before we jump into problems. Here are some links in case you need refreshers on any of the topics involved:

Direct proportionality: general explanation here and science examples here

Inverse proportionality: general explanation here

If you're ready to test out some word problems, let's get started!

The best approach to word problems is to translate it from words to numbers.

Here's a hint for translating problems directly proportional variables :

Image source: by Hannah Bonville

Let's try this method out.

Anya eats 12 apples over 4 days. Assuming that she eats the same number of apples each day, what is the ratio of apples to days?

(Video) Algebraic Inverse Proportion

by HEGARTYMATHS

Thies video by HEGARTYMATHS goes over some word problems based on inverse proportionality.

Word problems involving proportionality will usually give you one set of points and ask you to find part of a second set given only some of the information. Using the first set of points you solve for the constant k. Then, with k and part of the second set, you can find the final part of the second set.

Let's work through the first problem together.

y is inversely proportional to x. If y=10 when x=2, find:(i) A Formula for y in terms of x(ii) y when x=5(iii) x when y=8

The first step is to write our basic formula. We know that basic inverse formula has the form y=k⋅1x, or y=kx.

We know that y=10 when x=2. Let's plug those in then solve for k. y=kx10=k220=k

Our constant, k, is 20. We're going to use k=20 and x=5 to solve for y in the second part. y=kxy=205y=4

When x=5, y=4.

We use the constant k=20 to solve the second part as well. y=kx8=20x8x=20x=208x=52 or 2.5

Here's a graphic with another example!

Image source: By Caroline Kulczycky

Test Yourself: Amy's phone's battery life is inversely proportional to the number of new features on it. Her old phone had 7 new features and a battery life of 14 hours. Her new phone has 16 new features. How long is her battery life now?

7.245 hours

6.75 hours

6.125 hours

5.585 hours