Question

If the average test score of four students is \(85,\) which of the following scores could a fifth student receive such that the average of all five scores is greater than 84 and less than \(86 ?\) Indicate all such scores. a. 88 b. 86 c. 85 d. 83 e. 80

Short answer

Correct scores are: a. 88, b. 86, c. 85, d. 83.

Step-by-step solution

- Calculate the total score of the first four students

Given the average score of the first four students is 85, use the formula for average to find the total score. The formula for average is \( \text{average} = \frac{\text{total sum}}{\text{number of scores}} \). Solving for the total sum of scores, we get: \[ 85 = \frac{\text{total sum}}{4} \] Multiply both sides by 4 to get: \[ \text{total sum} = 85 \times 4 = 340 \]

- Find the total score for all five students

Let the score of the fifth student be denoted by \( x \). The total score for all five students is the sum of the scores of the first four students and the score of the fifth student. So, the total score for all five students is: \[ 340 + x \]

- Set up the inequality for the average of all five scores

We want the average of all five scores to be greater than 84 and less than 86. Set up the following inequality using the formula for average: \[ 84 < \frac{340 + x}{5} < 86 \]

- Solve the inequality

First, solve the lower bound of the inequality: Multiply both sides by 5: \[ 84 \times 5 < 340 + x \]\[ 420 < 340 + x \] Subtract 340 from both sides to find \( x \): \[ 420 - 340 < x \]\[ 80 < x \]Next, solve the upper bound of the inequality: Multiply both sides by 5: \[ \frac{340 + x}{5} < 86 \]\[ (340 + x) < 86 \times 5 \]\[ 340 + x < 430 \] Subtract 340 from both sides to find \( x \): \[ x < 430 - 340 \]\[ x < 90 \]

- Determine the valid scores from the options given

We found that the score of the fifth student \( x \) must satisfy the inequality: \[ 80 < x < 90 \]Checking the given options: - a. 88: satisfies \( 80 < 88 < 90 \) - b. 86: satisfies \( 80 < 86 < 90 \) - c. 85: satisfies \( 80 < 85 < 90 \) - d. 83: satisfies \( 80 < 83 < 90 \) - e. 80: does not satisfy \( 80 < x < 90 \)

Key concepts

Average Calculation

To solve many GRE math problems, like this one, you'll often need to calculate averages. An average represents the sum of a set of values divided by the number of values. For instance, here we know the average score of four students is 85. We use the formula \( \text{average} = \frac{\text{total sum}}{\text{number of scores}} \) to find the total score. So, rearranging the formula, the total score is calculated as follows:
\[ \text{total sum} = \text{average} \times \text{number of scores} \]
Substituting in our values, we get:
\[ 340 = 85 \times 4 \]
This tells us that the total sum of the scores for the four students is 340. Understanding how to backtrack from the average to the total sum is fundamental.

Inequality Solving

In this problem, we're tasked with finding a score that would keep the average within a certain range. Inequalities help us define that range. Here, we work with the inequality \( 84 < \frac{340 + x}{5} < 86 \). First, solve the lower bound:
1. Multiply both sides by 5:
\[ 420 < 340 + x \]
2. Subtract 340 from both sides:
\[ 80 < x \]
Next, solve the upper bound:
1. Multiply both sides by 5:
\[ 340 + x < 430 \]
2. Subtract 340 from both sides:
\[ x < 90 \]
This tells us that the fifth student's test score should be between 80 and 90. Inequality solving is crucial for understanding the range within which our solutions must fall.

Test Score Analysis

Part of test score analysis involves interpreting given data to find out which specific scores meet certain criteria. In this exercise, we identified that the range for the fifth score must be between 80 and 90. We then checked the given options:

• a. 88: This works because \( 80 < 88 < 90 \).
• b. 86: This works because \( 80 < 86 < 90 \).
• c. 85: This works because \( 80 < 85 < 90 \).
• d. 83: This works because \( 80 < 83 < 90 \).
• e. 80: This does not work because \( 80 \) is not greater than \( 80 \).

The process of validating each option helps ensure that we only pick the correct scores. Test score analysis requires careful checking of the constraints and given options.

Basic Algebra

Basic algebra is at the heart of solving this problem. We used algebra to isolate the variable and then found the range of possible values. The formula for average, solving for the total sum, and setting up inequalities all involve basic algebraic manipulation. While simplifying inequalities, we followed basic steps like multiplying both sides of an equation/inequality and adding/subtracting terms to isolate the variable. Though these might be elementary steps, they are powerful tools in finding accurate solutions. Basic algebra allows you to convert complex word problems into manageable equations you can solve step by step.