Solving Direct Variation Problems A Comprehensive Guide

In the realm of mathematics, direct variation represents a fundamental concept illustrating the relationship between two variables. When we say that y varies directly with x, we are essentially stating that there exists a constant ratio between y and x. This relationship can be expressed mathematically as y = kx, where k is the constant of variation. Understanding direct variation is crucial for solving various problems in algebra, physics, and other fields. This article will delve into three specific problems that exemplify the application of direct variation, offering step-by-step solutions and insights into the underlying principles.

The first problem involves finding the value of y when x is given, given that y varies directly with x and their values are known for one instance. The second problem extends this concept by exploring the direct variation between p and q, requiring us to find q when p is known. Finally, the third problem transitions to a real-world scenario, examining the direct variation between the cost of petrol and the number of liters purchased. By dissecting these problems, we aim to provide a comprehensive understanding of direct variation and its practical applications. This exploration will not only enhance problem-solving skills but also provide a solid foundation for more advanced mathematical concepts. Understanding direct variation is not just about solving equations; it's about grasping the proportional relationships that govern many real-world phenomena. So, let's embark on this mathematical journey and unravel the intricacies of direct variation through these illustrative examples.

The first problem presents a classic scenario of direct variation: "The value of y varies directly with x, and y = 12 when x = 10. Find y when x = 25.4." To tackle this, we first need to establish the direct variation equation. We know that y varies directly with x, so we can write the equation as y = kx, where k is the constant of variation. The initial condition given is that y = 12 when x = 10. This allows us to determine the value of k. Substituting these values into the equation, we get 12 = k(10). Solving for k, we divide both sides by 10, yielding k = 12/10 = 1.2. This constant of variation, k = 1.2, signifies that for every unit increase in x, y increases by 1.2 units.

Now that we have found the constant of variation, we can write the specific direct variation equation for this problem as y = 1.2x. This equation represents the relationship between x and y for all values. The problem asks us to find the value of y when x = 25.4. To do this, we simply substitute x = 25.4 into our equation: y = 1.2 * 25.4. Performing this multiplication, we find that y = 30.48. Therefore, when x = 25.4, y equals 30.48. This result demonstrates the direct proportionality between x and y; as x increases, y increases proportionally, governed by the constant of variation. Understanding this direct relationship is key to solving similar problems and grasping the broader concept of direct variation in mathematics and real-world applications. The ability to calculate and interpret these relationships is a fundamental skill in various scientific and engineering disciplines, where proportional changes are frequently encountered.

The second problem delves further into direct variation, presenting the scenario: "If p varies directly as q and if p = 21 when q = 7, then find q when p = 15.5." Similar to the first problem, our initial step is to establish the direct variation equation. Since p varies directly as q, we express this relationship as p = kq, where k represents the constant of variation. We are given that p = 21 when q = 7. Substituting these values into the equation, we get 21 = k(7). To solve for k, we divide both sides of the equation by 7, which gives us k = 21/7 = 3. This constant of variation, k = 3, indicates that p is three times the value of q. This constant is crucial as it defines the proportionality between the two variables.

Having determined the constant of variation, we can now formulate the specific direct variation equation for this problem: p = 3q. The problem requires us to find the value of q when p = 15.5. To do this, we substitute p = 15.5 into our equation: 15.5 = 3q. To solve for q, we divide both sides of the equation by 3, yielding q = 15.5 / 3. Performing this division, we find that q ≈ 5.17 (rounded to two decimal places). Thus, when p = 15.5, q is approximately 5.17. This result reinforces the concept of direct proportionality; as p changes, q changes in direct proportion, governed by the constant of variation. This problem further illustrates the practical application of direct variation in solving for unknown variables, a skill essential in various mathematical and scientific contexts. The ability to manipulate these equations and solve for unknowns is a cornerstone of algebraic thinking and problem-solving.

The third problem shifts our focus to a real-world application of direct variation: "If the cost of petrol is ₹ 68.75 per litre, write a direct variation equation to represent the relationship between the cost and the quantity of petrol." This problem highlights how direct variation is not just a theoretical concept but also a practical tool for modeling everyday scenarios. In this case, we are dealing with the relationship between the cost of petrol and the number of liters purchased. The core idea is that the total cost varies directly with the number of liters bought, given a constant price per liter.

To formulate the direct variation equation, we need to identify the variables and the constant of variation. Let C represent the total cost of petrol in rupees, and let L represent the number of liters purchased. The problem states that the cost per liter is ₹ 68.75. This price per liter serves as our constant of variation, as it represents the constant ratio between the cost and the quantity of petrol. Therefore, we can express the direct variation equation as C = 68.75L. This equation signifies that the total cost C is directly proportional to the number of liters L, with the constant of proportionality being the price per liter, ₹ 68.75. This equation is a simple yet powerful representation of a real-world relationship. For instance, if you buy 2 liters of petrol, the cost would be C = 68.75 * 2 = ₹ 137.50. Similarly, for 10 liters, the cost would be C = 68.75 * 10 = ₹ 687.50. This problem vividly demonstrates how direct variation can be used to model and predict costs in everyday transactions, making it a valuable concept for both mathematical understanding and practical application.

In summary, this article has explored the concept of direct variation through three distinct problems, each illustrating a different facet of this mathematical principle. The first problem demonstrated how to find an unknown variable when a direct variation relationship exists between two variables and some initial values are given. By calculating the constant of variation, we were able to determine the value of y for a specific value of x. The second problem expanded on this, requiring us to find q when p was known, further solidifying our understanding of how to manipulate direct variation equations to solve for unknowns. Finally, the third problem showcased the real-world applicability of direct variation by modeling the relationship between the cost of petrol and the quantity purchased. This example underscored how direct variation is not just an abstract mathematical concept but a practical tool for understanding and predicting everyday phenomena.

Through these problems, we have seen that direct variation is characterized by a constant ratio between two variables, expressed mathematically as y = kx. The constant of variation, k, plays a crucial role in defining the relationship and allowing us to solve for unknown values. The ability to identify direct variation relationships, calculate the constant of variation, and formulate the appropriate equations is a fundamental skill in mathematics and its applications. Whether it's determining the cost of goods, understanding physical relationships, or modeling various phenomena, direct variation provides a powerful framework for analyzing and solving problems. The concepts and techniques discussed in this article lay a solid foundation for further exploration of mathematical concepts and their practical applications in various fields.