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Difficulty: HardAlgebraic Word Problems and Equation Modeling

A boutique marketing agency charges a corporate client a fixed monthly retainer fee of $1,600\$1,600 plus a uniform monthly fee for each active digital ad campaign managed. The client has a total monthly ad management budget of $4,000\$4,000. If the agency were to increase the fee per campaign by $40\$40, the number of ad campaigns the client could afford under the same total budget would decrease by 22. What is the original monthly fee charged per ad campaign?

  1. A
    $160\$160
  2. B
    $180\$180
  3. $200\$200Answer
  4. D
    $240\$240
  5. E
    $250\$250

Answer

The original monthly fee charged per ad campaign is $200\$200.
Subtracting the $1,600\$1,600 retainer from the $4,000\$4,000 total budget leaves $2,400\$2,400 for campaign fees. Letting pp represent the original price per campaign, the original number of campaigns is 2,400p\frac{2,400}{p} and the new number of campaigns at price p+40p+40 is 2,400p+40\frac{2,400}{p+40}. The condition that the client can afford 22 fewer campaigns yields the equation 2,400p−2,400p+40=2\frac{2,400}{p} - \frac{2,400}{p+40} = 2. Simplifying leads to p2+40p−48,000=0p^2 + 40p - 48,000 = 0, which factors as (p−200)(p+240)=0(p - 200)(p + 240) = 0. Since price must be positive, the original fee per campaign is $200\$200.

Step-by-Step Solution

1
Determine the portion of the budget available for ad campaigns.
Variable Budget = Total Budget - Fixed Retainer = $4,000−$1,600=$2,400\$4,000 - \$1,600 = \$2,400.
The fixed retainer fee must be subtracted before calculating how many campaigns can be purchased.
2
Define variables and set up the equation for the number of campaigns.
Let pp be the original fee per campaign. The original number of campaigns is 2,400p\frac{2,400}{p}. With the fee increase, the new number of campaigns is 2,400p+40\frac{2,400}{p + 40}. The difference is given as 22, so 2,400p−2,400p+40=2\frac{2,400}{p} - \frac{2,400}{p + 40} = 2.
Model the real-world reduction in campaigns using algebraic expressions.
3
Simplify the equation and form a standard quadratic equation.
Divide the entire equation by 22: 1,200p−1,200p+40=1\frac{1,200}{p} - \frac{1,200}{p + 40} = 1. Multiply by p(p+40)p(p + 40): 1,200(p+40)−1,200p=p(p+40)  ⟹  48,000=p2+40p  ⟹  p2+40p−48,000=01,200(p + 40) - 1,200p = p(p + 40) \implies 48,000 = p^2 + 40p \implies p^2 + 40p - 48,000 = 0.
Clear fractions to transform the rational equation into quadratic form.
4
Solve the quadratic equation for pp.
(p+240)(p−200)=0(p + 240)(p - 200) = 0. Since price must be positive, p=200p = 200.
Factoring gives two roots (200200 and −240-240); reject the negative root in physical pricing context.

Key Concept

Algebraic Word Problems and Equation Modeling
Estimated Time:2m 0s
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Topics

Algebraic Word Problems and Equation Modeling

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