Using Algebraic Manipulation

Oct 13, 2025 by wbrum.org 23 views

Introduction

In this article, we will delve into a mathematical problem that requires critical thinking and analytical skills. Mr. Koger, a school administrator, is tasked with making copies of the school literary magazine to distribute to students and faculty. He has access to two copy machines: an older machine and a new machine. If he used only the older machine, it would take 80 minutes to make 240 copies. However, he also has the option to use the new machine, which is faster. The question is, how long would it take to make 240 copies using the new machine?

Understanding the Problem

To solve this problem, we need to understand the concept of work rate and time. The work rate of a machine is the number of copies it can make per unit of time. In this case, the older machine can make 240 copies in 80 minutes, which means its work rate is 240 copies / 80 minutes = 3 copies per minute. The new machine, on the other hand, is faster, but its work rate is unknown.

Analyzing the Situation

Let's assume that the new machine can make x copies per minute. Since it is faster than the older machine, its work rate is greater than 3 copies per minute. We need to find the value of x.

Formulating the Equation

We know that the new machine can make 240 copies in t minutes. Since its work rate is x copies per minute, we can set up the equation:

240 = xt

We also know that the older machine can make 240 copies in 80 minutes, so we can set up another equation:

240 = 3(80)

Simplifying the second equation, we get:

240 = 240

This equation is true, which means that the older machine can indeed make 240 copies in 80 minutes.

Solving for x

Now, we can solve for x by substituting the value of t into the first equation:

240 = xt

Since the new machine can make 240 copies in t minutes, we can set up the equation:

t = 240/x

Substituting this expression for t into the first equation, we get:

240 = (240/x)x

Simplifying the equation, we get:

240 = 240

This equation is true, which means that the new machine can indeed make 240 copies in some unknown time.

Finding the Value of x

To find the value of x, we need to use the fact that the new machine is faster than the older machine. Since the older machine can make 240 copies in 80 minutes, the new machine can make 240 copies in less than 80 minutes. Let's assume that the new machine can make 240 copies in t minutes, where t < 80.

We can set up the equation:

240 = xt

Substituting the expression for t, we get:

240 = (240/x)x

Simplifying the equation, we get:

240 = 240

This equation is true, which means that the new machine can indeed make 240 copies in some unknown time.

Using Algebraic Manipulation

To find the value of x, we can use algebraic manipulation. We can start by dividing both sides of the equation by 240:

1 = (x/240)x

Simplifying the equation, we get:

1 = x^2/240

Multiplying both sides of the equation by 240, we get:

240 = x^2

Taking the square root of both sides of the equation, we get:

x = ±√240

Since the work rate of the new machine cannot be negative, we take the positive square root:

x = √240

Evaluating the Answer

Now that we have found the value of x, we can evaluate the answer. The new machine can make 240 copies in t minutes, where t = 240/x. Substituting the value of x, we get:

t = 240/√240

Simplifying the expression, we get:

t = 240/√(240 × 1)

t = 240/√240

t = 240/√(4 × 60)

**Q&A: Understanding the Mathematical Dilemma** =====================================================

Q: What is the work rate of the older machine?

A: The work rate of the older machine is 3 copies per minute, as it can make 240 copies in 80 minutes.

Q: How long would it take to make 240 copies using the new machine?

A: To find the time it would take to make 240 copies using the new machine, we need to find the work rate of the new machine. Let's assume that the new machine can make x copies per minute. We can set up the equation:

240 = xt

where t is the time it would take to make 240 copies using the new machine.

Q: How do we find the value of x?

A: To find the value of x, we can use the fact that the new machine is faster than the older machine. Since the older machine can make 240 copies in 80 minutes, the new machine can make 240 copies in less than 80 minutes. Let's assume that the new machine can make 240 copies in t minutes, where t < 80.

We can set up the equation:

240 = xt

Substituting the expression for t, we get:

240 = (240/x)x

Simplifying the equation, we get:

240 = 240

This equation is true, which means that the new machine can indeed make 240 copies in some unknown time.

Q: How do we find the value of x using algebraic manipulation?

A: To find the value of x using algebraic manipulation, we can start by dividing both sides of the equation by 240:

1 = (x/240)x

Simplifying the equation, we get:

1 = x^2/240

Multiplying both sides of the equation by 240, we get:

240 = x^2

Taking the square root of both sides of the equation, we get:

x = ±√240

Since the work rate of the new machine cannot be negative, we take the positive square root:

x = √240

Q: What is the value of x?

A: The value of x is √240, which is approximately 15.49 copies per minute.

Q: How long would it take to make 240 copies using the new machine?

A: Now that we have found the value of x, we can find the time it would take to make 240 copies using the new machine. We can set up the equation:

240 = xt

Substituting the value of x, we get:

240 = (240/15.49)t

Simplifying the equation, we get:

t = 240/15.49

t ≈ 15.49 minutes

Q: What is the difference in time between the older machine and the new machine?

A: The older machine takes 80 minutes to make 240 copies, while the new machine takes approximately 15.49 minutes to make 240 copies. The difference in time is:

80 - 15.49 ≈ 64.51 minutes

Q: What does this tell us about the new machine?

A: This tells us that the new machine is significantly faster than the older machine, taking approximately 64.51 minutes less to make 240 copies. This is a significant improvement in efficiency and productivity.