A university professor teaches an introductory class with n students. On the first exam, the average (arithmetic mean) of the...

Phase 1: Owning the Dataset

Visualization

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

Key Relationships

The total score after equals the total score before plus the new student's score:

• Total before + S = Total after
• \(\mathrm{n \times 72.5 + S = (n+1) \times 72.4}\)

Phase 2: Understanding the Question

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}.

Simplifying the Relationship

Let's solve for S in terms of n:

• \(\mathrm{n \times 72.5 + S = (n+1) \times 72.4}\)
• \(\mathrm{n \times 72.5 + S = n \times 72.4 + 72.4}\)
• \(\mathrm{S = n \times 72.4 - n \times 72.5 + 72.4}\)
• \(\mathrm{S = n(72.4 - 72.5) + 72.4}\)
• \(\mathrm{S = -0.1n + 72.4}\)
• \(\mathrm{S = 72.4 - 0.1n}\)

Key Insight

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.

Phase 3: Finding the Answer

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.

Verification

Let's verify n = 64 and S = 66:

• Original total: \(\mathrm{64 \times 72.5 = 4,640}\)
• New total: \(\mathrm{4,640 + 66 = 4,706}\)
• New average: \(\mathrm{4,706 ÷ 65 = 72.4}\) ✓

Phase 4: Solution

The values that work together are:

• n = 64 (number of original students)
• S = 66 (new student's score)

These values satisfy our mathematical relationship \(\mathrm{S = 72.4 - 0.1n}\) and produce the correct averages as stated in the problem.