Last month Alex spent $91 on video game rentals, spending $5 for each one-day rental, $8 for each three-day rental,...

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Understanding the Question

Let's understand what we're being asked. Alex spent $91 total on video game rentals with three pricing tiers:

• $5 for 1-day rentals
• $8 for 3-day rentals
• $10 for 7-day rentals

He never rented the same game twice, and he rented fewer than 6 games as 1-day rentals.

What we need to determine: The total number of different games Alex rented.

Let's define:

• x = number of 1-day rentals (with constraint: \(0 \leq \mathrm{x} \leq 5\))
• y = number of 3-day rentals
• z = number of 7-day rentals

This gives us the equation: \(5\mathrm{x} + 8\mathrm{y} + 10\mathrm{z} = 91\)

The question asks for \(\mathrm{x} + \mathrm{y} + \mathrm{z}\) (the total number of games).

For sufficiency, we need to determine if we can find one unique value for \(\mathrm{x} + \mathrm{y} + \mathrm{z}\).

Analyzing Statement 1

Statement 1 tells us that Alex spent $20 on 7-day rentals.

Since each 7-day rental costs $10, we have: \(10\mathrm{z} = 20\), so \(\mathrm{z} = 2\).

Substituting into our equation: \(5\mathrm{x} + 8\mathrm{y} + 10(2) = 91\), which simplifies to \(5\mathrm{x} + 8\mathrm{y} = 71\).

Now, for y to be a whole number (you can't rent a fraction of a game), we need \(71 - 5\mathrm{x}\) to be divisible by 8.

Here's the key insight: Since \(71 = 8 \times 8 + 7\), we need \(5\mathrm{x}\) to leave a remainder of 7 when divided by 8. Let's think about this strategically:

• When \(\mathrm{x} = 3\): \(5 \times 3 = 15 = 8 \times 1 + 7\) ✓ (remainder is 7)

This gives us the remainder we need, so y will be an integer. Let's verify:

• \(5(3) + 8\mathrm{y} = 71\)
• \(15 + 8\mathrm{y} = 71\)
• \(8\mathrm{y} = 56\)
• \(\mathrm{y} = 7\)

Within our constraint (\(0 \leq \mathrm{x} \leq 5\)), only \(\mathrm{x} = 3\) gives us an integer solution.

Therefore, we have exactly one solution: \(\mathrm{x} = 3\), \(\mathrm{y} = 7\), \(\mathrm{z} = 2\), giving us a total of \(3 + 7 + 2 = 12\) games.

[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 tells us that Alex spent $56 on 3-day rentals.

Since each 3-day rental costs $8, we have: \(8\mathrm{y} = 56\), so \(\mathrm{y} = 7\).

Substituting into our equation: \(5\mathrm{x} + 8(7) + 10\mathrm{z} = 91\), which simplifies to \(5\mathrm{x} + 10\mathrm{z} = 35\).

Dividing by 5: \(\mathrm{x} + 2\mathrm{z} = 7\)

Now we need integer solutions where \(0 \leq \mathrm{x} \leq 5\). Let's examine the possibilities:

• If \(\mathrm{z} = 1\): \(\mathrm{x} = 5\) (within constraint) → Total games = \(5 + 7 + 1 = 13\)
• If \(\mathrm{z} = 2\): \(\mathrm{x} = 3\) (within constraint) → Total games = \(3 + 7 + 2 = 12\)
• If \(\mathrm{z} = 3\): \(\mathrm{x} = 1\) (within constraint) → Total games = \(1 + 7 + 3 = 11\)

We have three different possible totals: 11, 12, or 13 games.

Statement 2 is NOT sufficient.

This eliminates choices B and D.

The Answer: A

Since Statement 1 alone gives us a unique answer (12 games) but Statement 2 alone gives us multiple possibilities (11, 12, or 13 games), the answer is A.

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