Last week a certain comedian had an audience of 120 people for each of the afternoon performances and 195 people...

Understanding the Question

We need to find the average audience size across all performances last week.

Given Information

• Afternoon performances: 120 people each
• Evening performances: 195 people each
• The comedian gave ONLY afternoon and evening performances

What We Need to Determine

The average (arithmetic mean) number of people per performance.

Key Insight

This is a weighted average problem. The overall average depends entirely on the MIX of afternoon versus evening performances, not their absolute numbers. More evening shows (195 people) pull the average higher. More afternoon shows (120 people) pull the average lower.

To determine sufficiency, we need either:

• The exact number of each type of performance, OR
• The ratio between afternoon and evening performances

Analyzing Statement 1

Statement 1: The comedian gave 3 more evening performances than afternoon performances.

This tells us the DIFFERENCE between the number of performances, but not their ratio. Let's test whether this gives us a unique answer:

Test Case 1: 1 afternoon and 4 evening performances

• Total audience: \(1(120) + 4(195) = 120 + 780 = 900\)
• Total performances: 5
• Average: \(900 ÷ 5 = 180\) people

Test Case 2: 10 afternoon and 13 evening performances

• Total audience: \(10(120) + 13(195) = 1,200 + 2,535 = 3,735\)
• Total performances: 23
• Average: \(3,735 ÷ 23 ≈ 162\) people

The key observation: Different numbers of performances yield different averages (180 vs 162). This happens because the ratio changes with scale—it's \(1:4\) in the first case but \(10:13\) in the second.

Conclusion: Statement 1 is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

Now we forget Statement 1 and analyze Statement 2 independently.

Statement 2: The comedian gave twice as many evening performances as afternoon performances.

This gives us a fixed RATIO of \(2:1\) (evening to afternoon). Here's why this is sufficient:

For every 3 performances total:

• 1 is an afternoon show (120 people)
• 2 are evening shows (195 people each)

This means:

• \(1/3\) of all performances have audiences of 120
• \(2/3\) of all performances have audiences of 195

The weighted average is:

• \((1/3) × 120 + (2/3) × 195 = 40 + 130 = 170\) people

Critical insight: When the ratio is fixed at \(2:1\), the weighted average is always 170, regardless of scale. Whether it's 1 afternoon and 2 evening performances or 100 afternoon and 200 evening performances, the proportions—and thus the average—remain constant.

[STOP - Sufficient!]

Conclusion: Statement 2 is sufficient.

This eliminates choices C and E.

The Answer: B

Statement 2 alone provides the ratio between performance types, which uniquely determines the weighted average. Statement 1 only provides the difference, which leads to different averages at different scales.

Answer Choice B: "Statement 2 alone is sufficient, but Statement 1 alone is not sufficient."