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A juice company fills two bottle sizes: each small bottle contains 300 milliliters of juice, and each large bottle contains...

GMAT Algebra : (Alg) Questions

Source: Prism
Algebra
Linear equations in 2 variables
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A juice company fills two bottle sizes: each small bottle contains 300 milliliters of juice, and each large bottle contains 500 milliliters of juice. The company fills s small bottles and b large bottles using exactly 14,000 milliliters of juice. Which equation best represents this relationship?

  1. \(300\mathrm{s} + 500\mathrm{b} = 14{,}000\)
  2. \(\frac{\mathrm{s}}{300} + \frac{\mathrm{b}}{500} = 14{,}000\)
  3. \(300\mathrm{s} + 500\mathrm{b} = 14\)
  4. \(300\mathrm{s} + 500\mathrm{b} = 140\)
A
\(300\mathrm{s} + 500\mathrm{b} = 14,000\)
B
\(\frac{\mathrm{s}}{300} + \frac{\mathrm{b}}{500} = 14,000\)
C
\(300\mathrm{s} + 500\mathrm{b} = 14\)
D
\(300\mathrm{s} + 500\mathrm{b} = 140\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Small bottles: \(\mathrm{300\, mL}\) each, s bottles made
    • Large bottles: \(\mathrm{500\, mL}\) each, b bottles made
    • Total juice used: exactly \(\mathrm{14{,}000\, mL}\)
  • What this tells us: We need an equation where the total volume from both bottle types equals \(\mathrm{14{,}000\, mL}\)

2. INFER the relationship structure

  • Each small bottle contributes \(\mathrm{300\, mL}\) to the total
  • Each large bottle contributes \(\mathrm{500\, mL}\) to the total
  • Total volume = contributions from all small bottles + contributions from all large bottles

3. Build the equation step by step

  • Volume from small bottles: \(\mathrm{300\, mL}\) per bottle × s bottles = \(\mathrm{300s\, mL}\)
  • Volume from large bottles: \(\mathrm{500\, mL}\) per bottle × b bottles = \(\mathrm{500b\, mL}\)
  • Total equation: \(\mathrm{300s + 500b = 14{,}000}\)

Answer: (A) \(\mathrm{300s + 500b = 14{,}000}\)




Why Students Usually Falter on This Problem


Most Common Error Path:

Weak TRANSLATE skill: Students confuse the direction of the multiplication relationship and write \(\mathrm{\frac{s}{300} + \frac{b}{500} = 14{,}000}\)

They incorrectly think "s bottles divided by \(\mathrm{300\, mL}\) each" instead of "s bottles times \(\mathrm{300\, mL}\) each." This creates an equation that adds fractions representing "bottles per milliliter" rather than total milliliters, which doesn't match the physical situation.

This leads them to select Choice B (\(\mathrm{\frac{s}{300} + \frac{b}{500} = 14{,}000}\))


Second Most Common Error:

Poor TRANSLATE reasoning with units: Students correctly set up \(\mathrm{300s + 500b}\) but use the wrong total value, either converting \(\mathrm{14{,}000\, mL}\) incorrectly or misreading the problem

They might think \(\mathrm{14{,}000\, mL}\) means 14 or 140 in some other unit, leading to equations like \(\mathrm{300s + 500b = 14}\) or \(\mathrm{300s + 500b = 140}\).

This may lead them to select Choice C (\(\mathrm{300s + 500b = 14}\)) or Choice D (\(\mathrm{300s + 500b = 140}\))


The Bottom Line:

This problem tests whether students can correctly translate a rate-based word problem into a linear equation, particularly understanding that "\(\mathrm{rate \times quantity = contribution}\)" and all contributions sum to the total.

Answer Choices Explained
A
\(300\mathrm{s} + 500\mathrm{b} = 14,000\)
B
\(\frac{\mathrm{s}}{300} + \frac{\mathrm{b}}{500} = 14,000\)
C
\(300\mathrm{s} + 500\mathrm{b} = 14\)
D
\(300\mathrm{s} + 500\mathrm{b} = 140\)
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