In a photo contest, prize money of $30, $20, and $10 was awarded to the first-, second-, and third-place photos,...

Understanding the Question

Let's understand what we need to find: How many of Julia's photos were awarded second place?

Given Information

• Prize money: \(\$30\) (first place), \(\$20\) (second place), \(\$10\) (third place)
• Julia received a total of \(\$110\) in prize money
• Each photo could only be entered in one category

What We Need to Determine

We need a specific number - the exact count of Julia's second-place photos. This is a value question, so "sufficient" means we can determine one unique value.

Key Insight

The relatively small total of \(\$110\) creates natural constraints. Since \(\$110\) is only 11 times the smallest prize, there are limited ways to combine the prizes to reach this exact amount. Think of this as "spending" a \(\$110\) budget on prizes of \(\$30\), \(\$20\), and \(\$10\).

Analyzing Statement 1

Statement 1: "Twice as many of Julia's photos were awarded third place as were awarded first place."

What Statement 1 Tells Us

This creates a specific spending pattern: for every first-place photo (\(\$30\)), Julia must have exactly two third-place photos (\(2 \times \$10 = \$20\)). Together, that's \(\$50\) per "unit" of (1 first, 2 third).

Testing the Constraint

With \(\$110\) total and \(\$50\) per unit, let's see what's possible:

• Could she have 2 units? That would be \(2 \times \$50 = \$100\), leaving \(\$10\) unaccounted for
• Since \(\$10\) is the smallest prize, we'd need exactly 1 more third-place photo
• But that would give us 5 third-place photos total (\(2 \times 2 + 1\))
• For the constraint to hold, we'd need 2.5 first-place photos - impossible!

Therefore, she can only have 1 unit:

• 1 first-place photo: \(\$30\)
• 2 third-place photos: \(\$20\)
• Remaining budget: \(\$110 - \$50 = \$60\)
• This \(\$60\) must be from second-place photos: \(\$60 \div \$20 = 3\) photos

Conclusion

Statement 1 gives us a unique answer: Julia has exactly 3 second-place photos.

[STOP - Statement 1 is SUFFICIENT!]

This eliminates choices B, C, and E.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: "A total of 6 of Julia's photos received awards."

What Statement 2 Provides

We know Julia has exactly 6 winning photos that together earned \(\$110\). The average per photo is \(\$110 \div 6 \approx \$18.33\), suggesting we need a mix of prizes since no single prize value equals this average.

Testing Different Scenarios

Let's find different ways to distribute \(\$110\) among exactly 6 photos:

Scenario 1: Heavy on second place

• 5 second-place photos: \(5 \times \$20 = \$100\)
• 1 third-place photo: \(1 \times \$10 = \$10\)
• Total: \(\$110\) with 6 photos ✓
• Second-place count: 5

Scenario 2: More balanced distribution

• 1 first-place photo: \(1 \times \$30 = \$30\)
• 3 second-place photos: \(3 \times \$20 = \$60\)
• 2 third-place photos: \(2 \times \$10 = \$20\)
• Total: \(\$110\) with 6 photos ✓
• Second-place count: 3

Conclusion

We found at least two valid distributions giving different numbers of second-place photos (5 vs 3). Since we cannot determine a unique value, Statement 2 is NOT sufficient.

This eliminates choices B and D.

The Answer: A

Statement 1 alone allows us to determine that Julia has exactly 3 second-place photos, while Statement 2 alone permits multiple possibilities.

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