
Have you ever noticed that the more kilograms of apples you buy, the higher the total cost becomes? Or have you seen that as a car travels for a longer time at a steady speed, the distance it covers increases accordingly? These are not just random; they are perfect examples of a mathematical concept called direct variation.
For many Class 7 students, understanding how two different things change together can feel a bit tricky at first. You might wonder exactly how much one value will grow when the other changes. This article breaks down the definition, explains the essential equation, and provides clear examples to help you master this topic with ease.
Direct variation is a mathematical relationship between two variables where their ratio remains constant. When we talk about this concept, we usually look at two quantities, let’s call them x and y.
If y is said to vary directly with x, it means that as x increases, y increases at the same rate. Conversely, if x decreases, y will also decrease. The most important thing to remember is that they change in total harmony. If you were to divide y by x at any point in a direct variation relationship, you would always get the same number. This unchanging number is known as the constant of variation or the constant of proportionality.
Proportional Growth: Both variables move in the same direction.
The Constant Ratio: The value of y/x never changes.
Origin Connection: If you were to plot a direct variation relationship on a graph, the line would always be straight and pass through the origin (0,0).
To solve math problems, we need a reliable way to write this relationship down. The formula is straightforward:
y = kx
In this equation:
y is the dependent variable.
x is the independent variable.
k is the constant of variation (a non-zero number).
To find the constant k, you can rearrange the formula to k = y/x. Once you know the value of k, you can predict any future value of y just by knowing x.
The constant k tells you the "rate" of the change. For instance, if you are paid £10 per hour, the total money you earn (y) varies directly with the hours you work (x). Here, k is 10. Whether you work 2 hours or 100 hours, that "10" stays the same, acting as the bridge between your time and your pay.
Question: A part-time worker earns 139.20 for 3 days of work. If their pay varies directly with the number of days worked, how much will they earn in the month of June (30 days)?
Solution:
Identify the variables: Let y be the total pay and x be the number of days.
Find the constant of variation (k): k = y / x
k = 139.20 / 3 = 46.40 (This is the daily rate).
Write the equation: y = 46.40x
Solve for 30 days:
$y = 46.40 * 30
Answer: The worker will earn 1,392 in June.
Question: The distance a car travels varies directly with the amount of petrol used. If a car covers 180 km using 3 litres of petrol, how far can it travel on 10 litres?
Solution:
Identify the variables: Let d be distance and p be petrol.
Find the constant (k): k = d / p
k = 180 / 3 = 60 km per litre.
Write the equation: d = 60p
Solve for 10 litres:
d = 60 * 10
Answer: The car can travel 600 km.
Read More - Constants in Maths - Definition, Formula, Examples
Question: At a certain time of day, the length of a building's shadow varies directly with its height. If a 5-story building casts a 20 m shadow, how many stories high is a building that casts a 32 m shadow?
Solution:
Identify the variables: Let h be the height (stories) and s be the shadow length.
Find the constant (k): k = h / s
k = 5 / 20 = 0.25
Write the equation: h = 0.25s
Solve for a 32 m shadow:
$h = 0.25 * 32
Answer: The building is 8 stories high.
Question: A brownie recipe states that 2 cups of flour are needed to produce 8 brownies. If the number of brownies varies directly with the amount of flour, how many cups are needed for 64 brownies?
Solution:
Identify the variables: Let b be the number of brownies and c be the cups of flour.
Find the constant (k): k = b / c
k = 8 / 2 = 4 brownies per cup.
Write the equation: b = 4c
Solve for 64 brownies:
64 = 4c
c = 64 / 4
Answer: 16 cups of flour are needed.
Question: An energy-efficient washing machine saves 10 gallons of water for every 1 load of laundry. How many gallons are saved after 20 loads?
Solution:
Identify the variables: Let w be water saved and l be the number of loads.
Find the constant (k): The problem gives the rate directly: k = 10 gallons per load.
Write the equation: w = 10l
Solve for 20 loads:
w = 10 * 20
Answer: 200 gallons of water will be saved.
Read More - Brackets in Maths: Types, Rules & Examples
Not every relationship where two things increase is a DV. To be sure, you must check if the ratio is constant.
|
Value of X |
Value of Y |
Ratio (y/x) |
|
2 |
10 |
10/2 = 5 |
|
4 |
20 |
20/4 = 5 |
|
6 |
30 |
30/6 = 5 |
In the table above, because y/x always equals 5, we can say y varies directly with x, and the equation for this data is y = 5x.
When you encounter a problem in your exams, follow these simple steps to find the answer:
Step 1: Write down the basic formula (y = kx).
Step 2: Substitute the known values of x and y into the equation to find k.
Step 3: Rewrite the equation using the value of k you just found.
Step 4: Use this new equation to find the "missing" value requested in the problem.
It is easy to get confused between different types of variation. While DV means both variables go up together, inverse variation is the opposite. In inverse variation, as one value goes up, the other goes down (like the speed of a car and the time it takes to reach a destination). However, for your Class 7 studies, focusing on the "same direction" movement of direct proportion is the priority.
CuriousJr’s Mental Maths Course helps children develop strong number sense and improve their calculation skills. With an easy step-by-step method, the course trains students to think faster and solve problems with confidence. The learning process is simple, clear, and enjoyable.
The online classes focus on basic operations like addition, subtraction, multiplication, and division. Equal importance is given to both speed and accuracy. Fun activities, interactive exercises, and regular practice sessions help improve problem-solving ability and build a strong foundation in maths.
With a friendly and structured teaching approach, CuriousJr supports children in overcoming fear or hesitation towards maths. This course is perfect for school students who want to improve their everyday math skills and feel more confident while working with numbers.
