Solving the Petrol Price Puzzle: A Mathematical Approach for Students
In today's world, understanding the dynamics of price changes and making accurate calculations based on given information is crucial for a wide range of real-life applications. Take, for instance, the scenario where we need to find the old price of petrol per liter after a price hike. This problem requires students to analyze and solve equations, applying their basic algebraic skills. In this article, we will explore the process of solving such a problem step-by-step, using the example of a man buying petrol at both the old and new prices and calculating the old price based on the total cost.
Understanding the Problem
The problem at hand is as follows:
Last week, a man bought 20 liters of petrol at the old price. This week, he bought 10 liters of petrol at the new price. The total cost for both weeks is $9.20. It is known that the cost of petrol rises by 2 cents per liter.Our goal is to find the old price of 1 liter of petrol. To solve this, let's define the variables and set up the necessary equations.
Solving the Problem
Defining the Variables
Let's denote the old price of petrol per litre as x cents.
Setting Up the Equations
Knowing the new price is 2 cents more than the old price, we can express the new price as (x 2).
The cost of petrol last week, when 20 liters were bought at the old price, is:
[2]The cost of petrol this week, when 10 liters were bought at the new price, is:
[10(x 2)]The total cost for both weeks is given as $9.20, which is equivalent to 920 cents:
[2 10(x 2) 920]Simplifying the Equation
Expanding and simplifying the equation:
begin{equation*}2 1 20 920end{equation*} begin{equation*}3 20 920end{equation*}Subtracting 20 from both sides:
begin{equation*}3 900end{equation*}Dividing both sides by 30:
begin{equation*}x 30end{equation*}Therefore, the old price for 1 liter of petrol was 30 cents.
Exploring Alternative Approaches
Another way to approach this problem involves using algebraic manipulation and substitution. Let x be the last week's petrol cost per litre and y be the current petrol cost per litre. We know that:
begin{equation*}y x 0.02end{equation*} begin{equation*}10y 2 920end{equation*}Substituting the value of y into the second equation:
begin{equation*}10(x 0.02) 2 920end{equation*}Expanding and simplifying:
begin{equation*}1 0.20 2 920end{equation*} begin{equation*}3 0.20 920end{equation*}Subtracting 0.20 from both sides:
begin{equation*}3 919.80end{equation*}Dividing both sides by 30:
begin{equation*}x 30.66end{equation*}Rounding to the nearest cent, the old price for 1 liter of petrol is approximately 30 cents.
Conclusion
This problem demonstrates the practical application of basic algebraic techniques in solving real-world financial problems. Whether you use the traditional method or the alternative approach, both lead to the same conclusion: the old price for 1 liter of petrol was 30 cents.