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At a certain photoprocessing shop, the first standard-size print of a negative costs \(\$4\), and each additional print of the same negative costs \(\$1\). What is the total cost, in dollars, of \(\mathrm{y}\) standard-size prints of each of \(\mathrm{x}\) different negatives?
Let's break down what's happening at this photoprocessing shop in plain English:
So if I have 3 different negatives and I want 5 prints of each, I'm getting 15 total prints, but the pricing isn't simply \(15 \times \text{some price}\) because the first print of each negative is more expensive.
Process Skill: TRANSLATE - Converting the pricing structure into mathematical understanding
Let's think about what happens when I order y prints of just one negative:
Let me check this with a concrete example:
If I want 5 prints of one negative:
Using our formula: \(\$4 + \$1 \times (5 - 1) = \$4 + \$4 = \$8\) ✓
Now I need to apply this same cost structure to all x negatives:
Let me verify with numbers:
If \(\mathrm{x} = 2\) negatives and \(\mathrm{y} = 3\) prints each:
Using our formula: \(\$4(2) + 2(3 - 1) = \$8 + \$4 = \$12\) ✓
Our derived formula is: \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\)
Looking at the answer choices:
Our expression exactly matches choice E: \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\)
Let me double-check by expanding choice E:
\(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1) = 4\mathrm{x} + \mathrm{xy} - \mathrm{x} = 3\mathrm{x} + \mathrm{xy}\)
And checking our logic: x negatives each costing \(\$4\) gives us \(4\mathrm{x}\), plus x negatives each with \((\mathrm{y}-1)\) additional prints at \(\$1\) each gives us \(\mathrm{x}(\mathrm{y}-1)\).
The total cost in dollars of y standard-size prints of each of x different negatives is \(4\mathrm{x} + \mathrm{x}(\mathrm{y} - 1)\).
The answer is E.
Students often misread the problem and think that ALL prints cost \(\$4\) each, or that only the FIRST print overall costs \(\$4\) while all others cost \(\$1\). The key detail is that the FIRST print of EACH NEGATIVE costs \(\$4\), while additional prints of that SAME negative cost \(\$1\). This means if you have 3 different negatives, you'll pay \(\$4\) three times (once for each negative's first print).
Some students interpret this as y total prints distributed among x negatives, rather than y prints PER negative. The problem specifically states 'y prints of EACH of x negatives,' meaning if \(\mathrm{x}=3\) and \(\mathrm{y}=5\), you're getting 15 total prints (5 prints each of 3 different negatives), not 5 prints total shared among 3 negatives.
Students might try to calculate total prints first \((\mathrm{xy})\) and then apply some average price, missing that the pricing is tiered per negative. The correct approach requires calculating the cost per negative first \([\$4 + (\mathrm{y}-1)\times\$1]\), then multiplying by x negatives.
When expanding expressions like \(\mathrm{x}[\$4 + (\mathrm{y}-1)]\), students often make errors such as getting \(4\mathrm{x} + \mathrm{x}(\mathrm{y}+1)\) instead of \(4\mathrm{x} + \mathrm{x}(\mathrm{y}-1)\), or incorrectly distributing to get \(4\mathrm{x} + \mathrm{xy} - 1\) instead of \(4\mathrm{x} + \mathrm{xy} - \mathrm{x}\).
Students might use \((\mathrm{x}-1)\) instead of \((\mathrm{y}-1)\) for additional prints, thinking about additional negatives rather than additional prints per negative. Since we want y prints of each negative, we need \((\mathrm{y}-1)\) additional prints after the first print.
Students might expand their correct expression \(4\mathrm{x} + \mathrm{x}(\mathrm{y}-1)\) to get \(4\mathrm{x} + \mathrm{xy} - \mathrm{x} = 3\mathrm{x} + \mathrm{xy}\), then mistakenly select choice C \((4\mathrm{x} + \mathrm{xy})\) thinking it's close enough, when they should have kept the original form to match choice E exactly.
When checking their work with small numbers, students might find that multiple answer choices give the same result for their test case (like \(\mathrm{x}=1, \mathrm{y}=2\)) and then guess among the 'working' options rather than testing with different values or checking their logical setup more carefully.
Step 1: Choose convenient values
Let's use \(\mathrm{x} = 2\) negatives and \(\mathrm{y} = 3\) prints of each negative. These are small, manageable numbers that will make our calculations clear.
Step 2: Calculate cost for one negative
For each negative with 3 prints:
• First print: \(\$4\)
• Additional prints: \((3 - 1) = 2\) prints at \(\$1\) each = \(\$2\)
• Total per negative: \(\$4 + \$2 = \$6\)
Step 3: Calculate total cost for all negatives
Total cost for 2 negatives: \(2 \times \$6 = \$12\)
Step 4: Test answer choices
Using \(\mathrm{x} = 2\) and \(\mathrm{y} = 3\):
Answer: E