In a class of 21, how many scored above the class average (arithmetic mean) on the final exam? The class...

Understanding the Question

We have a class of 21 students and need to find the exact number of students who scored above the class average on the final exam.

This is a value question - we need a specific count, not a range or possibility. For sufficiency, we must be able to determine precisely how many students scored above the average.

Analyzing Statement 1

Statement 1: The class average is 78.

Knowing the average tells us the mean score, but nothing about how individual scores are distributed. Let's test whether different score distributions can produce the same average but different counts above that average:

Test Case 1: All 21 students scored exactly 78

• Average: 78 ✓
• Number above average: 0 students

Test Case 2: 20 students scored 77, and 1 student scored 98

• Average: \((20 \times 77 + 98) \div 21 = 1638 \div 21 = 78\) ✓
• Number above average: 1 student

Test Case 3: 10 students scored 68, and 11 students scored 87

• Average: \((10 \times 68 + 11 \times 87) \div 21 = (680 + 957) \div 21 = 1637 \div 21 \approx 78\) ✓
• Number above average: 11 students

Since we found multiple valid distributions with different answers (0, 1, or 11 students above average), Statement 1 is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices A and D.

Analyzing Statement 2

Important: We now analyze Statement 2 independently, forgetting Statement 1 completely.

Statement 2: The class median is 78.

With 21 students (odd number), the median is the 11th score when arranged from lowest to highest. This tells us:

• Exactly 10 students scored \(\leq 78\)
• The 11th student scored exactly 78
• Exactly 10 students scored \(\geq 78\)

However, this doesn't tell us:

• What the average is
• How many scored strictly above 78 (vs. equal to 78)

Let's test different scenarios:

Test Case 1: Scores are 70, 70, ..., 70 (ten times), 78, 78, ..., 78 (eleven times)

• Median: 78 ✓
• But we don't know the average, so we can't determine how many are above it

Test Case 2: Scores are 70, 70, ..., 70 (ten times), 78 (once), 85, 85, ..., 85 (ten times)

• Median: 78 ✓
• Again, without knowing the average, we can't answer the question

Statement 2 alone is NOT sufficient because we don't even know what the average is.

[STOP - Not Sufficient!] This eliminates choice B.

Combining Statements

Now we use both statements together:

• Average = 78
• Median = 78

Even with both constraints, can we determine a unique answer? Let's construct different valid distributions:

Configuration 1: All 21 students scored exactly 78

• Average: 78 ✓
• Median: 78 ✓
• Students above average (78): 0 students

Configuration 2: Let's carefully construct a symmetric distribution

• 10 students scored 75.1
• 1 student scored 78
• 10 students scored 80.9

Let's verify:

• Average: \((10 \times 75.1 + 78 + 10 \times 80.9) \div 21 = (751 + 78 + 809) \div 21 = 1638 \div 21 = 78\) ✓
• Median: When ordered, the 11th score is 78 ✓
• Students above average (78): 10 students

Since we can construct at least two different distributions that satisfy both conditions but yield different counts (0 vs. 10), the statements together are still NOT sufficient.

[STOP - Not Sufficient!]

The Answer: E

Even knowing that both the average and median equal 78, we cannot uniquely determine how many students scored above the class average. The answer could be 0, 10, or potentially other values.

Answer Choice E: Statements together are not sufficient.