The Cost of 3 Pens and 2 Erasers

Oct 13, 2025 by orovajewels.com 22 views

Introduction

Mathematics is a fascinating subject that involves solving problems and puzzles using logical reasoning and mathematical concepts. In this article, we will explore a mathematical puzzle that involves the cost of pens and erasers. We will use algebraic equations to solve the problem and find the cost of 3 pens and 2 erasers.

The Problem

The cost of 5 pens and 3 erasers is 90p. The cost of 2 pens and 1 eraser is 35p. We need to find the cost of 3 pens and 2 erasers.

Let's Define the Variables

Let's define the variables:

P = cost of 1 pen
E = cost of 1 eraser

Equations

We can write two equations based on the given information:

5P + 3E = 90 (cost of 5 pens and 3 erasers is 90p)
2P + E = 35 (cost of 2 pens and 1 eraser is 35p)

Solving the Equations

We can solve the equations using the method of substitution or elimination. Let's use the elimination method.

First, we can multiply equation (2) by 3 to get:

6P + 3E = 105

Now, we can subtract equation (1) from this new equation to get:

(6P + 3E) - (5P + 3E) = 105 - 90 P = 15

Now that we have the value of P, we can substitute it into equation (2) to get:

2(15) + E = 35 30 + E = 35 E = 5

The Cost of 3 Pens and 2 Erasers

Now that we have the values of P and E, we can find the cost of 3 pens and 2 erasers.

Cost of 3 pens = 3P = 3(15) = 45p Cost of 2 erasers = 2E = 2(5) = 10p

Total cost = 45 + 10 = 55p

Conclusion

In this article, we solved a mathematical puzzle involving the cost of pens and erasers. We used algebraic equations to find the cost of 3 pens and 2 erasers. The cost of 3 pens and 2 erasers is 55p.

Mathematical Concepts Used

Algebraic equations
Substitution method
Elimination method
Variables
Constants

Real-World Applications

This problem can be applied to real-world scenarios, such as:

Calculating the cost of office supplies
Determining the cost of materials for a project
Solving problems involving multiple variables

Tips and Tricks

When solving equations, make sure to check your work by plugging in the values of the variables.
Use the substitution method or elimination method to solve equations.
Make sure to define the variables clearly and use them consistently throughout the problem.

Frequently Asked Questions

Q: What is the cost of 1 pen? A: The cost of 1 pen is 15p.

Q: What is the cost of 1 eraser? A: The cost of 1 eraser is 5p.

Q: How did you solve the equations? A: We used the elimination method to solve the equations.

Introduction

In our previous article, we solved a mathematical puzzle involving the cost of pens and erasers. We used algebraic equations to find the cost of 3 pens and 2 erasers. In this article, we will answer some frequently asked questions related to the problem.

Q&A

Q: What is the cost of 1 pen?

A: The cost of 1 pen is 15p.

Q: What is the cost of 1 eraser?

A: The cost of 1 eraser is 5p.

Q: How did you solve the equations?

A: We used the elimination method to solve the equations. We multiplied equation (2) by 3 to get:

6P + 3E = 105

Then, we subtracted equation (1) from this new equation to get:

(6P + 3E) - (5P + 3E) = 105 - 90 P = 15

Q: What is the total cost of 3 pens and 2 erasers?

A: The total cost of 3 pens and 2 erasers is 55p.

Q: Can you explain the concept of variables in the context of this problem?

A: Yes, in this problem, we defined two variables:

P = cost of 1 pen
E = cost of 1 eraser

We used these variables to write two equations:

5P + 3E = 90
2P + E = 35

Q: How do you determine which method to use to solve equations?

A: There are two main methods to solve equations: substitution and elimination. The substitution method involves substituting one equation into another, while the elimination method involves adding or subtracting equations to eliminate one of the variables. In this problem, we used the elimination method.

Q: Can you explain the concept of constants in the context of this problem?

A: Yes, in this problem, we used constants to represent the cost of pens and erasers. The constants are the numbers that do not change in the equations. In this case, the constants are 90 and 35.

Q: How do you check your work when solving equations?

A: When solving equations, it's essential to check your work by plugging in the values of the variables. This ensures that the solution is correct and that the equations are balanced.

Q: Can you provide more examples of real-world applications of this problem?

A: Yes, this problem can be applied to various real-world scenarios, such as:

Calculating the cost of office supplies
Determining the cost of materials for a project
Solving problems involving multiple variables

Q: What are some tips and tricks for solving equations?

A: Here are some tips and tricks for solving equations:

Make sure to define the variables clearly and use them consistently throughout the problem.
Use the substitution method or elimination method to solve equations.
Check your work by plugging in the values of the variables.
Use algebraic equations to represent the problem.