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A moving company charges a fixed booking fee of F dollars, plus 18 cents per pound of cargo. A customer...

GMAT Algebra : (Alg) Questions

Source: Prism
Algebra
Linear inequalities in 1 or 2 variables
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A moving company charges a fixed booking fee of \(\mathrm{F}\) dollars, plus \(\mathrm{18}\) cents per pound of cargo. A customer wants the total charge to be at least \(\mathrm{2.5}\) times and at most \(\mathrm{3.5}\) times the booking fee. Which inequality represents all possible cargo weights \(\mathrm{w}\), in pounds, where \(\mathrm{w \geq 0}\), that meet this goal?

  1. \(\frac{1.5}{18}\mathrm{F} \leq \mathrm{w} \leq \frac{2.5}{18}\mathrm{F}\)
  2. \((1.5 \times 0.18)\mathrm{F} \leq \mathrm{w} \leq (2.5 \times 0.18)\mathrm{F}\)
  3. \(1.5\mathrm{F} \leq \mathrm{w} \leq 2.5\mathrm{F}\)
  4. \(\frac{1.5}{0.18}\mathrm{F} \leq \mathrm{w} \leq \frac{2.5}{0.18}\mathrm{F}\)
  5. \(\frac{2.5}{0.18}\mathrm{F} \leq \mathrm{w} \leq \frac{3.5}{0.18}\mathrm{F}\)
A
\(\frac{1.5}{18}\mathrm{F} \leq \mathrm{w} \leq \frac{2.5}{18}\mathrm{F}\)
B
\((1.5 \times 0.18)\mathrm{F} \leq \mathrm{w} \leq (2.5 \times 0.18)\mathrm{F}\)
C
\(1.5\mathrm{F} \leq \mathrm{w} \leq 2.5\mathrm{F}\)
D
\(\frac{1.5}{0.18}\mathrm{F} \leq \mathrm{w} \leq \frac{2.5}{0.18}\mathrm{F}\)
E
\(\frac{2.5}{0.18}\mathrm{F} \leq \mathrm{w} \leq \frac{3.5}{0.18}\mathrm{F}\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Fixed booking fee: \(\mathrm{F}\) dollars
    • Variable charge: 18 cents per pound of cargo
    • Customer wants total charge between \(\mathrm{2.5F}\) and \(\mathrm{3.5F}\) (inclusive)
    • Need to find possible weights \(\mathrm{w}\)
  • Key insight: Convert 18 cents to \(\$0.18\) for consistent units

2. INFER the approach

  • Set up pricing model equation: \(\mathrm{Total\ charge = F + 0.18w}\)
  • Create compound inequality: \(\mathrm{2.5F \leq Total\ charge \leq 3.5F}\)
  • Substitute and solve for \(\mathrm{w}\)

3. TRANSLATE the constraints into mathematical form

  • Customer's requirement: \(\mathrm{2.5F \leq Total\ charge \leq 3.5F}\)
  • Substitute pricing model: \(\mathrm{2.5F \leq F + 0.18w \leq 3.5F}\)

4. SIMPLIFY to isolate w

  • Subtract \(\mathrm{F}\) from all parts:

\(\mathrm{2.5F - F \leq 0.18w \leq 3.5F - F}\)

  • This gives us:

\(\mathrm{1.5F \leq 0.18w \leq 2.5F}\)

  • Divide all parts by 0.18:

\(\mathrm{(1.5/0.18)F \leq w \leq (2.5/0.18)F}\)

Answer: D




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Using 18 instead of 0.18 for the per-pound charge

Students often forget to convert cents to dollars, keeping the per-pound rate as 18 rather than 0.18. This leads them to set up: \(\mathrm{2.5F \leq F + 18w \leq 3.5F}\), which simplifies to \(\mathrm{1.5F \leq 18w \leq 2.5F}\), and finally \(\mathrm{(1.5/18)F \leq w \leq (2.5/18)F}\).

This may lead them to select Choice A \(\mathrm{((1.5/18)F \leq w \leq (2.5/18)F)}\)


Second Most Common Error:

Poor SIMPLIFY execution: Algebraic manipulation errors when setting up or solving the compound inequality

Students might incorrectly set up the initial inequality as \(\mathrm{F \leq Total\ charge \leq 2.5F}\) (misreading "at least 2.5 times" as "at most 2.5 times"), or make sign errors during the subtraction step. These errors lead to different coefficient values in the final answer.

This causes them to get stuck and guess among the remaining choices.


The Bottom Line:

This problem requires careful attention to unit conversion and systematic algebraic manipulation of compound inequalities. The key insight is recognizing that "18 cents" must become "\(\$0.18\)" for the mathematics to work correctly with the dollar-denominated booking fee.

Answer Choices Explained
A
\(\frac{1.5}{18}\mathrm{F} \leq \mathrm{w} \leq \frac{2.5}{18}\mathrm{F}\)
B
\((1.5 \times 0.18)\mathrm{F} \leq \mathrm{w} \leq (2.5 \times 0.18)\mathrm{F}\)
C
\(1.5\mathrm{F} \leq \mathrm{w} \leq 2.5\mathrm{F}\)
D
\(\frac{1.5}{0.18}\mathrm{F} \leq \mathrm{w} \leq \frac{2.5}{0.18}\mathrm{F}\)
E
\(\frac{2.5}{0.18}\mathrm{F} \leq \mathrm{w} \leq \frac{3.5}{0.18}\mathrm{F}\)
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