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A logistics company ships packages using two different services. Service A charges $3.50 per kilogram, while Service B charges $8.20...

GMAT Algebra : (Alg) Questions

Source: Prism
Algebra
Linear equations in 2 variables
EASY
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A logistics company ships packages using two different services. Service A charges \(\$3.50\) per kilogram, while Service B charges \(\$8.20\) per kilogram. For a particular shipment totaling \(15\) kilograms, the company wants to split the entire load between both services so that the average shipping cost is exactly \(\$5.40\) per kilogram. If \(\mathrm{x}\) represents the weight in kilograms sent via Service A and \(\mathrm{y}\) represents the weight sent via Service B, where \(\mathrm{x + y = 15}\), which equation correctly represents this situation?

A
\(3.50\mathrm{x} + 8.20\mathrm{y} = 5.40(15)\)
B
\(8.20\mathrm{x} + 3.50\mathrm{y} = 5.40(15)\)
C
\(3.50\mathrm{x} + 8.20\mathrm{y} = 15\)
D
\(5.40\mathrm{x} + 5.40\mathrm{y} = 15\)
E
\(3.50\mathrm{x} + 8.20\mathrm{y} = 5.40\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Service A charges \(\$3.50\) per kilogram (\(\mathrm{x}\) kilograms sent this way)
    • Service B charges \(\$8.20\) per kilogram (\(\mathrm{y}\) kilograms sent this way)
    • Total shipment: 15 kilograms (so \(\mathrm{x + y = 15}\))
    • Target average cost: \(\$5.40\) per kilogram

2. TRANSLATE costs into mathematical expressions

  • Total cost from Service A: \(3.50\mathrm{x}\) dollars
  • Total cost from Service B: \(8.20\mathrm{y}\) dollars
  • Combined total cost: \(3.50\mathrm{x} + 8.20\mathrm{y}\) dollars

3. INFER what the total cost should equal

  • The company wants the average cost to be exactly \(\$5.40\) per kilogram
  • For 15 kilograms at \(\$5.40\) average: Target total cost = \(5.40 \times 15 = 81\) dollars
  • Therefore: \(3.50\mathrm{x} + 8.20\mathrm{y}\) must equal \(5.40(15)\)

4. Identify the correct equation

The equation representing this situation is: \(3.50\mathrm{x} + 8.20\mathrm{y} = 5.40(15)\)

Answer: A





Why Students Usually Falter on This Problem


Most Common Error Path:

Weak TRANSLATE skill: Students mix up which service gets which variable, thinking x goes with the higher rate since it comes first alphabetically, or y goes with Service B because it's the "second" service.

This leads them to write \(8.20\mathrm{x} + 3.50\mathrm{y} = 5.40(15)\), causing them to select Choice B (\(8.20\mathrm{x} + 3.50\mathrm{y} = 5.40(15)\)).


Second Most Common Error:

Poor INFER reasoning: Students correctly translate the individual costs but don't recognize that the equation should equal the target total cost. Instead, they think the cost equation should equal either the total weight (15) or just the average rate (5.40).

This confusion may lead them to select Choice C (\(3.50\mathrm{x} + 8.20\mathrm{y} = 15\)) or Choice E (\(3.50\mathrm{x} + 8.20\mathrm{y} = 5.40\)).


The Bottom Line:

This problem requires careful attention to which variable represents which service's weight, and understanding that a weighted average creates a target total cost that the individual service costs must sum to achieve.

Answer Choices Explained
A
\(3.50\mathrm{x} + 8.20\mathrm{y} = 5.40(15)\)
B
\(8.20\mathrm{x} + 3.50\mathrm{y} = 5.40(15)\)
C
\(3.50\mathrm{x} + 8.20\mathrm{y} = 15\)
D
\(5.40\mathrm{x} + 5.40\mathrm{y} = 15\)
E
\(3.50\mathrm{x} + 8.20\mathrm{y} = 5.40\)
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