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Difficulty: HardAlgebraic Word Problems and Equation Modeling

A software consulting firm calculates the total cost of a project using a fixed administrative fee plus a constant hourly rate for each hour worked. For a project requiring HH hours, the firm originally charged a total of $2,400\$2,400, where the fixed administrative fee was equal to the total hourly charges. Under a new pricing model, the fixed administrative fee is reduced by 25%25\%, while the hourly rate is increased by 20%20\%. Under this new model, a project requiring H+15H + 15 hours costs a total of $2,880\$2,880. What is the value of HH?

  1. A
    25
  2. B
    30
  3. 40Answer
  4. D
    50
  5. E
    60

Answer

The value of HH is 40.
The original cost model breaks into F=$1,200F = \$1,200 and R⋅H=$1,200R \cdot H = \$1,200. Reducing FF by 25%25\% gives $900\$900. Increasing RR by 20%20\% gives a new hourly charge of 1.20R(H+15)=1.20RH+18R1.20 R(H + 15) = 1.20 RH + 18 R. Substituting RH=1,200RH = 1,200 yields 1,440+18R=1,9801,440 + 18 R = 1,980, solving to R=30R = 30. Dividing 1,2001,200 by 3030 gives H=40H = 40.

Step-by-Step Solution

1
Formulate equations for the original pricing structure.
Let FF be the fixed administrative fee and RR be the original hourly rate. The total original cost is given by F+R⋅H=2,400F + R \cdot H = 2,400. Since FF equals the total hourly charges (R⋅HR \cdot H), we have F=R⋅H=1,200F = R \cdot H = 1,200.
Splitting the total cost of $2,400\$2,400 into two equal parts establishes the exact baseline values for FF and R⋅HR \cdot H.
2
Formulate the equation under the new pricing model.
The new fixed fee is 0.75×1,200=9000.75 \times 1,200 = 900. The new hourly rate is 1.20R1.20 R. The new total cost equation for H+15H + 15 hours is 900+1.20R(H+15)=2,880900 + 1.20 R(H + 15) = 2,880.
Applies the specified percentage changes to the fixed fee and hourly rate individually.
3
Solve for the hourly rate RR.
Subtract 900900 from both sides: 1.20R(H+15)=1,980  ⟹  1.20RH+18R=1,9801.20 R(H + 15) = 1,980 \implies 1.20 RH + 18 R = 1,980. Substituting RH=1,200RH = 1,200 gives 1.20(1,200)+18R=1,980  ⟹  1,440+18R=1,980  ⟹  18R=540  ⟹  R=301.20(1,200) + 18 R = 1,980 \implies 1,440 + 18 R = 1,980 \implies 18 R = 540 \implies R = 30.
Expands the algebraic expression and uses the substitution RH=1,200RH = 1,200 to isolate RR.
4
Calculate the value of HH.
Since R⋅H=1,200R \cdot H = 1,200 and R=30R = 30, we find H=1,20030=40H = \frac{1,200}{30} = 40.
Divides the total hourly charge by the hourly rate to obtain the number of hours.

Key Concept

Linear Equation Modeling with Percentage Adjustments
Estimated Time:2m 0s
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Topics

Algebraic Word Problems and Equation Modeling

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