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245 LearnersLast updated on November 18, 2025

Have you ever bought candies or toys and noticed that if you take more, the total cost goes up? That’s called direct variation. It means when one thing changes, the other changes too. Can you think of other examples, like how the more apples you buy, the more money you spend? Let’s learn how direct variation helps us find discounts and the total cost.
Direct variation exists between two variables when one quantity depends directly on the other. In other words, if one quantity increases, the other increases too, and if one decreases, the other also decreases. This means the two variables are directly proportional, with one variable being a constant multiple of the other.
There are two main types of proportionalities: direct variation and inverse variation. Proportionality means that two quantities are connected through a constant. In direct variation, the ratio of the two quantities remains the same, while in inverse variation, the product of the two quantities stays constant.
In this topic, we will explore the definition of direct variation, its properties, and go through examples to understand how it works in real-life situations.
Learn what the constant of variation (k) is and how it simply affects a direct variation graph.
A direct variation equation describes the relationship between two variables, in which the value of one changes as the other changes. For example, in the formula \(b = ka\), the value of b is directly proportional to the value of a, and k is the constant of variation. The constant k does not affect the proportional relationship but helps define the specific ratio between the two variables.
Direct variation can be expressed in various mathematical equations.
For example,
Let’s take area of a circle
A = \(πr^2\)
Here,
A represents the area of the circle.
r is the radius of the circle.
π is the constant value, either 3.14 or 22/7.
According to this direct variation formula, the area of a circle varies with the radius. If the radius increases, the area also increases; if the radius decreases, the area decreases. This relationship can also be shown on a direct variation graph, which typically passes through the origin and shows how one variable changes directly with the other.
Direct variation shows how one value increases as the other increases, while inverse variation shows how one value increases as the other decreases.
| Direct Variation | Inverse Variation |
| One quantity increases or decreases along with the other quantity. | One quantity increases when the other decreases, and vice versa. |
| The ratio of the two quantities remains constant (constant of proportionality). | The product of the two quantities remains constant (constant of proportionality). |
| Represented as\( y 𝛼 x\) | Represented as \( y 𝛼 \frac{1}{x}\) |
| Formula:\( y=kx\) | Formula: \(y=\frac{k}{x}\) |
| The graph is a straight line passing through the origin. | The graph is a rectangular hyperbola. |
The graph of two quantities in direct variation forms a straight line, indicating that direct variation is represented by a linear equation in two variables. The equation is written as \(y = kx\), where the ratio of change\(\frac{Δy}{Δx} \)is equal to k. This ratio represents the slope of the line. The direct variation graph can be drawn based on this relationship.
Cross multiplication is a method used to solve direct variation equations. When two ratios are set equal to each other, you can multiply the numerator of one ratio by the denominator of the other and vice versa. This helps you quickly find the missing value in a direct variation equation.
For example, if \(y_1x_1=y_2x_2\), then:
This method is beneficial for solving direct variation problems efficiently.
Learning direct variation helps children solve many real-life problems efficiently. Here, we will look at a few tips and tricks that help you understand the concept better:
Students often confuse direct variation with other relationships or misapply the constant of proportionality. In this section, we will learn a few common mistakes and the ways to avoid them.
Direct variation is seen in many daily situations where two quantities change at the same rate, making it easy to understand and predict how one value affects the other.
Let's assume that Sara’s savings from photography is directly proportional to her working hours. What would her savings be for 20 hours if she saved $200 for 10 hours of work?
Sara will save $400 after 20 hours of her work.
Assume y = savings and x = number of hours worked. Since her savings is directly proportional to her working hours:
\(y = kx\)
Substituting the values, x = 20 hours and y = $200:
\(200 = k × 20\)
\(k = 200/10\)
= 20
We will now use k = 20 to find her savings for 20 hours of work.
\(y = 20 × 20 = 400\)
Therefore, Sara will save $400 after 20 hours of her work.
Assume that y and x are directly proportional. When y is 60, x equals 20. Calculate the value of x when the value of y becomes 100.
The value of x when y = 100 is 33.33 (approx).
Here, we use the formula \( y = kx\)
\(60 = k × 20\)
\(k = 60 / 20 = 3\)
Now, we find the value of x when y = 100:
Substituting k = 3 and y = 100,
\(100 = 3 × x\)
\(x = 100/3 ≈ 33.33\)
Therefore, the value of x when y = 100 is approximately 33.33.
Leona bakes 50 cookies using 5 cups of sugar. Calculate the number of cups of sugar she would require for 75 cookies, given that the number of cups of sugar is directly proportional to the number of cookies she baked.
Leona would require 7.5 cups of sugar to bake 75 cookies.
Assume, y = amount of sugar; x = number of cookies
\(y = kx \) (Since the amount of sugar \(\propto \) the number of cookies).
Now, we find k using x = 50 cookies and y = 5 cups of sugar:
\(5 = k × 50\)
\( k = 5/50 = 0.1\)
To find the number of cups of sugar for 75 cookies, multiply the value of k by 75.
So, \(y = 0.1 × 75 = 7.5\)
Therefore, Leona would require 7.5 cups of sugar to bake 75 cookies.
Suppose the value of y changes proportionally with x. Given that y = 15 when x = 5, calculate the value of y when x becomes 10.
When x = 10, the value of y will be 30.
From the question, we understand that y \(\propto\) x
So we use the formula, \(y = kx\)
Substitute the given values:
\(15 = k × 5\)
\(k = 15/5 = 3\)
When x = 10:
\(y = 3 × 10\) (substituting k = 3)
\(y = 30\)
Therefore, when x = 10, the value of y will be 30.
Let's say a and b are directly proportional. Given that a = 36 and b = 6, express their relationship in the form of a direct variation equation.
We can express their relationship as: \(a = 6b\).
Since a and b are directly proportional,
\(a = kb\)
Where k is the proportionality constant.
Substitute a = 36 and b = 6
\(36 = k × 6\)
\(k = 36 / 6 = 6\)
Therefore, we can express it as \(a = 6b\).
Dr. Sarita Tiwari is a passionate educator specializing in Commercial Math, Vedic Math, and Abacus, with a mission to make numbers magical for young learners. With 8+ years of teaching experience and a Ph.D. in Business Economics, she blends academic rigo
: She believes math is like music—once you understand the rhythm, everything just flows!






