Problem 53 - Algebra for College Students

Question

Use an algebraic approach to solve each problem. Aura took three biology exams and has an average score of 88 . Her second exam score was 10 points better than her first, and her third exam score was 4 points better than her second exam. What were her three exam scores?

Step-by-Step Solution

Answer

First score: 80, second score: 90, third score: 94.

Step 1: Understand the Problem

Aura has three biology exam scores with an average score of 88. We need to find each of her exam scores.

Step 2: Define Variables

Let \( x \) be Aura's first exam score. Her second exam score is \( x + 10 \) (10 points better than her first), and her third exam score is \( x + 10 + 4 = x + 14 \) (4 points better than her second).

Step 3: Set Up the Equation Using the Average

The average of the three scores is given by the equation: \( \frac{x + (x + 10) + (x + 14)}{3} = 88 \).

Step 4: Simplify the Equation

Combine the terms in the numerator: \( \frac{3x + 24}{3} = 88 \).

Step 5: Solve the Equation for \( x \)

Multiply both sides by 3 to eliminate the fraction: \( 3x + 24 = 264 \). Next, subtract 24 from both sides to isolate the term with \( x \): \( 3x = 240 \). Finally, divide both sides by 3 to solve for \( x \): \( x = 80 \).

Step 6: Calculate Each Exam Score

Using \( x = 80 \), the first score is 80, the second score is \( 80 + 10 = 90 \), and the third score is \( 90 + 4 = 94 \).

Key Concepts

System of Equations

In algebra, solving problems often involves using a system of equations. A system of equations consists of multiple equations that are solved together, as they share common variables.
For instance, in the problem with Aura's biology exams, you can consider all conditions given as parts of a system of equations. Each condition represents one piece of that system, all interconnecting and sharing the same unknown, which in this case, we express as variable "x".
Here, one equation calculates the average score, while others are implicit through descriptions of the scores' relationships:

• First score: Let it be "x".
• Second score: "x + 10", which is 10 points more than the first score.
• Third score: "x + 14", which is 4 points more than the second score.

By setting up these relationships, you effectively create a unified system that can be solved using algebraic manipulation.

Average Calculation

An average gives us a single value representing the overall tendency of a set of numbers. To calculate an average, sum all the numbers and then divide by the total count of numbers.
In Aura's scenario, we had to calculate the average of her three exam scores. The problem statement provided that the average was 88. We express this information algebraically with the equation:\[\frac{x + (x + 10) + (x + 14)}{3} = 88\]This equation sums all scores and divides by 3, as there are three exams, to give the known average score of 88.
Calculating averages this way is crucial as it links different parts of a problem by creating a clear mathematical statement of balance.

Variable Manipulation

Variable manipulation involves altering and simplifying equations to find the values of unknown variables.
To find Aura's exam scores, we had to isolate and solve for the variable "x". Initially, we started with the equation:\[\frac{3x + 24}{3} = 88\]The aim was to manipulate the equation by eliminating processes that complicate the isolation of "x". First, multiply throughout by 3 to remove the denominator:\[3x + 24 = 264\]Next, subtract 24 from both sides to further simplify:\[3x = 240\]Finally, divide both sides by 3 to solve for "x":\[x = 80\]This approach shows how understanding and applying operations like addition, subtraction, multiplication, or division systematically help deduce unknown values from equations.