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Difficulty: HardSolving Linear Equations

A school club sold 20 rolls of wrapping paper and 25 boxes of greeting cards for a fundraiser. The price of a roll of wrapping paper was 34\frac{3}{4} of the price of a box of greeting cards. If the club raised a total of $800 from these sales, what was the price of a single roll of wrapping paper?

  1. A
    $5.00
  2. B
    $20.00
  3. $15.00Answer
  4. D
    $20.65
  5. E
    $26.67

Answer

The price of a single roll of wrapping paper was $15.00.
The correct answer is 15.00.Byrepresentingthepriceofarollofwrappingpaperas15.00. By representing the price of a roll of wrapping paper as W andthepriceofaboxofgreetingcardsas and the price of a box of greeting cards as G ,theproblemstatesthat, the problem states that W = \frac{3}{4}G ,whichmeans, which means G = \frac{4}{3}W .Thetotalrevenueequationis. The total revenue equation is 20W + 25G = 800 .Substituting. Substituting G gives gives 20W + 25\left(\frac{4}{3}W\right) = 800 ,whichsimplifiesto, which simplifies to 20W + \frac{100}{3}W = 800 .Multiplyingtheentireequationby3toclearthefractionresultsin. Multiplying the entire equation by 3 to clear the fraction results in 60W + 100W = 2400 .Combiningliketermsgives. Combining like terms gives 160W = 2400 ,whichyields, which yields W = 15$.

Step-by-Step Solution

1
Define variables for the unknowns and translate the price relationship into an algebraic equation.
Let WW be the price of a roll of wrapping paper and GG be the price of a box of greeting cards. The relationship is given by W=34GW = \frac{3}{4}G, which can be rearranged to express GG in terms of WW: G=43WG = \frac{4}{3}W.
Expressing one variable in terms of another allows us to set up a single-variable linear equation.
2
Write the linear equation representing the total revenue from the fundraiser sales.
The total revenue from selling 20 rolls of wrapping paper and 25 boxes of greeting cards is 20W+25G=80020W + 25G = 800. Substituting G=43WG = \frac{4}{3}W gives: 20W+25(43W)=80020W + 25\left(\frac{4}{3}W\right) = 800, which simplifies to 20W+1003W=80020W + \frac{100}{3}W = 800.
This sets up the equation that we need to solve to find the value of WW.
3
Clear the fraction by multiplying all terms by 3 and solve for WW.
Multiplying the entire equation by 3 yields: 3(20W)+3(1003W)=3(800)  ⟹  60W+100W=24003(20W) + 3\left(\frac{100}{3}W\right) = 3(800) \implies 60W + 100W = 2400. Combining like terms gives 160W=2400160W = 2400. Dividing by 160 yields W=15W = 15.
Clearing the denominator simplifies the equation to a standard linear form that can be solved directly.

Key Concept

Solving linear equations derived from real-world contexts, particularly those involving fractional relationships and multi-step isolation.

Alternative Method

Instead of expressing GG in terms of WW first, solve for GG directly by substituting W=34GW = \frac{3}{4}G into the revenue equation. This gives 20(34G)+25G=800  ⟹  15G+25G=800  ⟹  40G=800  ⟹  G=2020\left(\frac{3}{4}G\right) + 25G = 800 \implies 15G + 25G = 800 \implies 40G = 800 \implies G = 20. Then, calculate W=34(20)=15W = \frac{3}{4}(20) = 15.
Estimated Time:2m 30s
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Topics

Solving Linear Equations

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