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A moving truck can tow a trailer if the combined weight of the trailer and the boxes it contains is...

GMAT Algebra : (Alg) Questions

Source: Practice Test
Algebra
Linear inequalities in 1 or 2 variables
MEDIUM
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Notes
Post a Query

A moving truck can tow a trailer if the combined weight of the trailer and the boxes it contains is no more than 4,600 pounds. What is the maximum number of boxes this truck can tow in a trailer with a weight of 500 pounds if each box weighs 120 pounds?


Solution

  • Total weight limit: \(4,600\) pounds
  • Trailer weight: \(500\) pounds
  • Weight per box: \(120\) pounds
  • Maximum weight available for boxes: \(4,600 - 500 = 4,100\) pounds
  • Maximum number of boxes: \(4,100 \div 120 = 34.166...\)
  • Since we can only have whole boxes, the maximum number of boxes is \(34\)

Answer: 34 boxes

A

34

B

35

C

38

D

39

Solution

1. TRANSLATE the problem information

  • Given information:
    • Maximum combined weight: 4,600 pounds
    • Trailer weight: 500 pounds
    • Each box weight: 120 pounds
    • Need to find: maximum number of boxes
  • Let \(\mathrm{b}\) = number of boxes
  • Combined weight = \(\mathrm{500 + 120b}\)

2. TRANSLATE the constraint into mathematical form

  • "No more than 4,600 pounds" means: \(\mathrm{500 + 120b \leq 4,600}\)
  • This inequality represents our weight limit

3. SIMPLIFY by solving the inequality

  • \(\mathrm{500 + 120b \leq 4,600}\)
  • Subtract 500 from both sides: \(\mathrm{120b \leq 4,100}\)
  • Divide both sides by 120: \(\mathrm{b \leq 4,100/120}\)
  • Calculate: \(\mathrm{b \leq 34.166...}\) (use calculator)

4. APPLY CONSTRAINTS to select the final answer

  • Since we can't have a fractional box, b must be a whole number
  • The maximum whole number ≤ 34.166... is 34
  • Verify: \(\mathrm{500 + 120(34)}\)
    \(\mathrm{= 500 + 4,080}\)
    \(\mathrm{= 4,580 \leq 4,600}\)

Answer: A. 34




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak APPLY CONSTRAINTS reasoning: Students correctly solve \(\mathrm{b \leq 34.166...}\) but round UP to 35 instead of recognizing they must round DOWN for counting problems.

Their reasoning: "34.166... is closer to 34 than to 35, but since we want the maximum, let's try 35."

Check: \(\mathrm{500 + 120(35)}\)
\(\mathrm{= 500 + 4,200}\)
\(\mathrm{= 4,700 \gt 4,600}\) (violates the constraint!)

This may lead them to select Choice B (35).

Second Most Common Error:

Poor TRANSLATE execution: Students set up the inequality incorrectly as \(\mathrm{500 + 120b \geq 4,600}\) (thinking "at least" instead of "no more than") or forget to include the trailer weight in their constraint.

This leads to confusion and incorrect setup, causing them to abandon systematic solution and guess.

The Bottom Line:

This problem tests whether students understand that "maximum" in constraint problems means finding the largest valid solution, and that practical counting scenarios require applying whole number constraints correctly.

Answer Choices Explained
A

34

B

35

C

38

D

39

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