Problem 36 - Elementary Algebra Within Reach

Question

Course Grade You were 6 points shy of a B in your mathematics course. Your point total for the course was 394. How many points were possible in the course? (Assume that you needed \(80 \%\) of the course total for a B.)

Step-by-Step Solution

Answer

The total possible points in the course were 500.

Step 1: Calculate Points Required for a B-grade

The number of points required to get a B grade would be the student's course total of 394 points plus the additional 6 points they were shy of a B. Therefore, the required points for a B grade is \(394 + 6 = 400\) points.

Step 2: Calculate Total Possible Points

Since 400 points represents the 80% of the total possible points for getting a B grade, we can set up the equation: \[400 = 0.80 * x\] where \(x\) is the total possible points in the course. Solving this equation for \(x\), we find that \(x=500\). Therefore, there were 500 total possible points in the course.

Step 3: Final Result

So, the total number of possible points in the course was 500 points.

Key Concepts

Linear Equations

Linear equations are fundamental in algebra and represent a relationship between two variables where there is a constant rate of change. They can be as simple as something like \( y = 2x + 3 \) or as complex as a system of several variables. Each variable has a power of 1. The basic structure is usually \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.

When solving problems using linear equations, we often look for the value that makes the equation true. The process of finding this value involves various techniques such as graphing, substitution, or using algebraic manipulation. In our textbook exercise, we construct a linear equation based on the given conditions to reveal the total possible points in the course. By understanding the properties of linear equations, you can tackle a vast array of problems.

Percentage Problems

Percentage problems are a staple in algebra, often involving comparisons of ratios to 100. They appear in a variety of contexts, from real-world financial situations to academic scenarios. The key to solving percentage problems lies in understanding the percent formula: \( \text{Part} = \text{Percent} \times \text{Whole} \).

When solving percentage problems, such as figuring out discounts, increases, or, in our textbook case, the percentage needed for a grade, it is crucial to understand what represents the \(\text{Part}\), the \(\text{Percent}\), and the \(\text{Whole}\), and how to translate words into a mathematical expression. For instance, in the grade calculation, knowing that 80% (or 0.80 as a decimal) is needed for a B-grade helps us establish the relationship between the points earned (part) and the total possible points (whole).

Equation Setup and Solving

Equation setup and solving is the process of translating a word problem into an algebraic expression and then finding the unknown value that solves the equation. This skill hinges on identifying the key pieces of information that relate to one another and determining the correct mathematical operations to use.

In the context of our textbook problem, the first step is to recognize that the course grade is tied to a certain percentage. Once we have that information, we can set up an equation, \(400 = 0.80 \times x\), with \(x\) representing the variable we're solving for— the total possible points. We use algebraic skills to isolate and solve for \(x\), bringing us to the final solution. Mastering this process is crucial as it not only helps in algebra but also enhances logical thinking and problem-solving skills in a variety of disciplines.