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Difficulty: MediumTranslating and Solving Algebraic Word Problems

A commercial printing press charges a business a setup fee of 100100 plus 0.150.15 per brochure printed. To encourage green practices, the press offers a recycling discount equal to 3030 less than 0.050.05 per brochure printed. If the business has a budget of at most 380380 for their brochure order, what is the maximum number of brochures they can print?

  1. A
    1,250
  2. B
    1,550
  3. 2,500Answer
  4. D
    3,100
  5. E
    3,500

Answer

2,500
The correct answer is 2,500 brochures. The total cost is the setup fee of 100plustheprintingcostof100 plus the printing cost of 0.15 per brochure, minus the discount of 30lessthan30 less than 0.05 per brochure. Setting bb as the number of brochures, the discount is 0.05b−300.05b - 30. The total cost equation is 100+0.15b−(0.05b−30)=130+0.10b100 + 0.15b - (0.05b - 30) = 130 + 0.10b. Setting this less than or equal to 380380 yields 130+0.10b≤380⇒0.10b≤250⇒b≤2,500130 + 0.10b \leq 380 \Rightarrow 0.10b \leq 250 \Rightarrow b \leq 2,500.

Step-by-Step Solution

1
Define the variable and write the algebraic expressions for the cost components.
Let bb be the number of brochures printed. The total cost before the discount is 100+0.15b100 + 0.15b. The recycling discount is represented as 0.05b−300.05b - 30.
Establishing clear variables and translating the textual relationships into mathematical expressions is the first step in solving word problems.
2
Set up the inequality representing the budget constraint.
100+0.15b−(0.05b−30)≤380100 + 0.15b - (0.05b - 30) \leq 380
The total cost (original cost minus the discount) must be at most the budget of 380380.
3
Simplify the left side of the inequality.
100+0.15b−0.05b+30≤380⇒130+0.10b≤380100 + 0.15b - 0.05b + 30 \leq 380 \Rightarrow 130 + 0.10b \leq 380
Distributing the subtraction sign across the discount expression and combining like terms simplifies the inequality.
4
Solve for the variable bb.
0.10b≤250⇒b≤25000.10b \leq 250 \Rightarrow b \leq 2500
Isolating bb gives the maximum number of brochures that can be printed.

Key Concept

Translating and Solving Algebraic Word Problems
Estimated Time:1m 30s
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Topics

Translating and Solving Algebraic Word Problems

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