Last Saturday a certain circus charged $6 per adult ticket and $3 per child ticket. Was the circus' revenue from...

Understanding the Question

We need to determine if the circus' revenue from adult ($6 each) and child ($3 each) tickets exceeded $2,500.

What We Need to Determine: Can we definitively answer YES or NO to whether total revenue > $2,500?

Given Information:

• Adult ticket price: $6
• Child ticket price: $3
• Question: Is \(6\mathrm{A} + 3\mathrm{C} > 2,500\)? (where \(\mathrm{A}\) = adult tickets, \(\mathrm{C}\) = child tickets)

Key Insight: Since adult tickets cost exactly twice as much as child tickets, the mix of tickets directly impacts total revenue. We need to check if either statement's constraint forces a definitive answer about exceeding $2,500.

Analyzing Statement 1

Statement 1 tells us: \(\mathrm{A} > \mathrm{C}/3\) (adult tickets sold > 1/3 of child tickets sold)

This gives us information about the ratio of adult to child tickets, but tells us nothing about the total number of tickets sold.

Let's test whether this ratio constraint alone determines if revenue exceeds $2,500:

Small Scale Example:

• 300 child tickets and 101 adult tickets
• Check ratio: 101 > 300/3 = 100 ✓ (satisfies Statement 1)
• Revenue = 6(101) + 3(300) = $606 + $900 = $1,506
• This is less than $2,500

Large Scale Example:

• 600 child tickets and 201 adult tickets
• Check ratio: 201 > 600/3 = 200 ✓ (satisfies Statement 1)
• Revenue = 6(201) + 3(600) = $1,206 + $1,800 = $3,006
• This is greater than $2,500

Since we found examples where revenue both exceeds and falls short of $2,500 while satisfying Statement 1, we cannot determine a definitive answer.

Statement 1 alone is NOT sufficient.

This eliminates answer choices A and D.

Analyzing Statement 2

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

Statement 2 tells us: \(\mathrm{A} + \mathrm{C} > 850\) (total tickets sold > 850)

This constraint forces a minimum scale for the event. The key question: Does having more than 850 tickets guarantee revenue exceeds $2,500?

Strategic Insight: To find the minimum possible revenue, we should maximize the number of cheaper tickets (child tickets at $3 each).

Minimum Revenue Scenario:

• Assume the smallest total that satisfies Statement 2: exactly 851 tickets
• To minimize revenue, make all 851 tickets child tickets
• Minimum possible revenue = 851 × $3 = $2,553

Since even the absolute worst-case scenario (all child tickets) produces revenue of $2,553, which exceeds $2,500, any mix that includes adult tickets would only increase the revenue further.

Therefore, Statement 2 guarantees that revenue > $2,500.

[STOP – Sufficient!]

Statement 2 alone is sufficient.

This eliminates answer choices C and E, leaving only B.

The Answer: B

Statement 2 alone forces revenue above $2,500 by requiring a minimum number of tickets sold. Even in the worst-case scenario (all child tickets), the revenue exceeds our threshold.

Statement 1's ratio constraint doesn't establish any minimum scale, allowing for scenarios both above and below $2,500.

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