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Last month Alex spent \(\$91\) on video game rentals, spending \(\$5\) for each one-day rental, \(\$8\) for each three-day rental, and \(\$10\) for each seven-day rental. Alex never rented the same game twice, and he rented fewer than \(6\) games in one-day rentals. How many different games did Alex rent last month?
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Let's understand what we're being asked. Alex spent $91 total on video game rentals with three pricing tiers:
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:
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 \(\mathrm{x} + \mathrm{y} + \mathrm{z}\).
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:
This gives us the remainder we need, so y will be an integer. Let's verify:
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.
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:
We have three different possible totals: 11, 12, or 13 games.
Statement 2 is NOT sufficient.
This eliminates choices B and D.
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."