Loading...
A university professor teaches an introductory class with \(\mathrm{n}\) students. On the first exam, the average (arithmetic mean) of the \(\mathrm{n}\) scores of these students was exactly \(72.5\). After the exam, a new student transferred into the professor's class. The student had taken the same exam in another professor's class. The new average score on the exam, including the \(\mathrm{n}\) scores of the original students, and the new student's score of \(\mathrm{S}\), was exactly \(72.4\).
Select for n and for S values that are jointly consistent. Make only 2 selections, one for each.
62
63
64
65
66
Let's create a simple diagram showing the before and after states:
Before new student:
n students Average = 72.5 Total score = n × 72.5
After new student:
n + 1 students (original n + new student with score S) Average = 72.4 Total score = (n+1) × 72.4
The total score after equals the total score before plus the new student's score:
We need to find values for n (number of original students) and S (new student's score) that work together mathematically. Both must come from our answer choices: {62, 63, 64, 65, 66}.
Let's solve for S in terms of n:
This formula tells us that as n increases by 10, S decreases by 1. We need to find which value of n from our choices produces a value of S that's also in our choices.
Let's systematically check each possible value of n:
If n = 62:
\(\mathrm{S = 72.4 - 0.1(62) = 72.4 - 6.2 = 66.2}\)
Is 66.2 in our choices? No, continue.
If n = 63:
\(\mathrm{S = 72.4 - 0.1(63) = 72.4 - 6.3 = 66.1}\)
Is 66.1 in our choices? No, continue.
If n = 64:
\(\mathrm{S = 72.4 - 0.1(64) = 72.4 - 6.4 = 66.0 = 66}\)
Is 66 in our choices? Yes! ✓
? Stop here - we found our answer.
Let's verify n = 64 and S = 66:
The values that work together are:
These values satisfy our mathematical relationship \(\mathrm{S = 72.4 - 0.1n}\) and produce the correct averages as stated in the problem.