Solving Algebraic Equations: Finding the Cost of Pens and Pencils

In real-life scenarios, algebraic equations can be used to solve practical problems. This article provides a clear and detailed solution to a simple arithmetic problem involving the cost of pens and pencils.

Introduction to the Problem

Let's consider a common problem: The cost of four pens and three pencils is Rs. 150. Given that the cost of one pencil is Rs. 10, how can we find the cost of a pen?

Step-by-Step Solution

The problem can be broken down into a simple algebraic equation. We start by defining the variables:

Defining the Variables

P will represent the cost of one pen.

Formulating the Equation

Given that the total cost of 4 pens and 3 pencils is Rs. 150, and the cost of one pencil is Rs. 10, we can form the following equation:

4P 3 times; 10 150

Simplifying the Equation

Let's simplify the equation step by step:

4P 30 150

4P 150 - 30

4P 120

P 120 / 4

P 30

Conclusion

The cost of one pen is Rs. 30.

Detailed Explanation of the Solution

1. Identify the Variables and Given Information:
- Cost of 1 pencil Rs. 10
- Total cost of 4 pens and 3 pencils Rs. 150
- We need to find the cost of 1 pen.

2. Form the Equation:
- Let the cost of 1 pen be P.
- The equation becomes: 4P 3 times; 10 150

3. Apply the Distributive Property:
- 4P 30 150

4. Solve for P:
- Isolate P by subtracting 30 from both sides of the equation: 4P 120
- Divide both sides by 4: P 120 / 4
- Thus, P 30

Therefore, the cost of one pen is Rs. 30.

Summary and Conclusion

The problem of finding the cost of a pen when the total cost of pens and pencils is given can be easily solved using basic algebraic techniques. By defining the variable and formulating the equation, we can systematically solve for the unknown value, demonstrating the practical application of algebra in real-life situations.

Hints for Recap:

Determining the cost of an unknown item based on given total costs and known values. Using substitution to simplify and solve equations. Verifying the solution by plugging the value back into the original equation.

For more such practical applications and problem-solving techniques, stay tuned to this blog!