Amy's grade was 90th percentile of the 80 grades for her class. Of the 100 grades from another class, 19...

1. Translate the problem requirements: Understand that "90th percentile" means \(90\%\) of grades are below Amy's grade, and we need to find what percentile Amy's grade represents when combining both classes (total 180 grades).
2. Calculate Amy's position in her original class: Determine exactly how many students scored below Amy in the 80-student class.
3. Determine Amy's position in the second class: Use the given information about how many grades were higher and lower than Amy's in the 100-student class.
4. Combine the data from both classes: Count total grades below Amy's across both classes and calculate her percentile in the combined 180-student population.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what "percentile" means in everyday language. When we say Amy scored at the 90th percentile, we mean that \(90\%\) of all students scored below her grade, and only \(10\%\) scored above her.

Now, what are we trying to find? We want to know what percentile Amy's grade represents when we combine both classes together. This means we need to figure out what percentage of the combined 180 students (80 + 100) scored below Amy's grade.

Process Skill: TRANSLATE - Converting the percentile language into concrete counting of students above and below Amy's grade

2. Calculate Amy's position in her original class

In Amy's original class of 80 students, she scored at the 90th percentile. This means \(90\%\) of the 80 students scored below her.

Let's calculate: \(90\%\) of \(80 = 0.90 \times 80 = 72\) students scored below Amy

Since there are 80 students total and 72 scored below Amy, this means \(80 - 72 = 8\) students scored above Amy in her class.

So in her original class:
• 72 students scored below Amy
• 8 students scored above Amy
• Amy herself makes it 81 total (but we're told no one had the exact same grade)

3. Determine Amy's position in the second class

Now let's look at the second class of 100 students. We're told that:
• 19 students scored higher than Amy
• The rest scored lower than Amy
• No student had the exact same grade as Amy

Since 19 students scored higher and there are 100 students total, the number who scored lower than Amy is: \(100 - 19 = 81\) students scored below Amy in the second class.

4. Combine the data from both classes

Now we can count the total number of students who scored below Amy across both classes:

From her original class: 72 students below Amy
From the second class: 81 students below Amy
Total students below Amy: \(72 + 81 = 153\) students

Total students in both classes combined: \(80 + 100 = 180\) students

To find Amy's percentile in the combined classes, we calculate what percentage of students scored below her:
Percentile = \(\frac{\text{Number of students below Amy}}{\text{Total number of students}} \times 100\)
Percentile = \(\frac{153}{180} \times 100\)

Let's simplify this fraction: \(\frac{153}{180} = \frac{153 \div 9}{180 \div 9} = \frac{17}{20} = 0.85\)

Therefore: \(0.85 \times 100 = 85\%\)

Process Skill: SIMPLIFY - Breaking down the fraction to make the calculation more manageable

4. Final Answer

Amy's grade was at the 85th percentile of the grades of the two classes combined.

This makes sense because when we combined the classes, Amy's relative position changed. Even though she was at the 90th percentile in her original class, the addition of the second class (where more students scored below her than above her) brought her percentile down to the 85th percentile overall.

The answer is D. 85th

Common Faltering Points

Errors while devising the approach

1. Misunderstanding percentile definition: Students may confuse percentile with percentage score. They might think "90th percentile" means Amy scored \(90\%\) on her test, rather than understanding that \(90\%\) of students scored below her grade. This fundamental misunderstanding would lead to a completely wrong approach.

2. Incorrectly interpreting "the rest were lower": When the problem states "19 were higher than Amy's, and the rest were lower," students might forget that Amy herself is not included in the second class. They may mistakenly think there are only 99 students to account for instead of the full 100, leading to incorrect counting.

3. Confusion about what to calculate: Students may get confused about whether they need to find Amy's rank, her percentile in just one class, or her combined percentile. The problem asks for her percentile in the "two classes combined," but students might focus only on one class or try to average the percentiles from both classes.

Errors while executing the approach

1. Arithmetic errors in percentage calculations: When calculating \(90\%\) of 80 students, students might make basic multiplication errors (getting 70 instead of 72) or miscalculate the number of students above Amy (getting 7 instead of 8).

2. Incorrect addition when combining classes: Students may make errors when adding the students who scored below Amy from both classes. For example, they might accidentally add \(72 + 19 = 91\) (mixing up "below Amy" with "above Amy" from the second class) instead of the correct \(72 + 81 = 153\).

3. Fraction simplification errors: When calculating \(\frac{153}{180}\), students might make errors in finding the GCD or in the division process, potentially getting \(\frac{152}{180}\) or \(\frac{154}{180}\), which would lead to percentiles close to but not exactly \(85\%\).

Errors while selecting the answer

1. Rounding errors: If students get a decimal result like \(84.7\%\) or \(85.3\%\) due to calculation errors, they might incorrectly round to the nearest answer choice, potentially selecting 80th or 92nd instead of 85th.

2. Selecting intermediate calculations: Students might accidentally select 81st percentile if they mistakenly use 81 (the number of students below Amy in the second class) as their final answer, or select 90th if they stick with Amy's original percentile without considering the combined classes.