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Lorenzo purchased a box of cereal and some strawberries at the grocery store. Lorenzo paid $2 for the box of...

GMAT Algebra : (Alg) Questions

Source: Practice Test
Algebra
Linear equations in 1 variable
EASY
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Notes
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Lorenzo purchased a box of cereal and some strawberries at the grocery store. Lorenzo paid \(\$2\) for the box of cereal and \(\$1.90\) per pound for the strawberries. If Lorenzo paid a total of \(\$9.60\) for the box of cereal and the strawberries, which of the following equations can be used to find \(\mathrm{p}\), the number of pounds of strawberries Lorenzo purchased? (Assume there is no sales tax.)

A
\(1.90\mathrm{p} + 2 = 9.60\)
B
\(1.90\mathrm{p} - 2 = 9.60\)
C
\(1.90 + 2\mathrm{p} = 9.60\)
D
\(1.90 - 2\mathrm{p} = 9.60\)
Solution

1. TRANSLATE the problem information

  • Given information:
    • Cereal costs \(\$2\) (fixed amount)
    • Strawberries cost \(\$1.90\) per pound
    • Total spent is \(\$9.60\)
    • \(\mathrm{p}\) = number of pounds of strawberries purchased
  • What this tells us: We need an equation that represents total cost as the sum of cereal cost plus strawberry cost.

2. TRANSLATE each cost component into mathematical expressions

  • Cereal cost: \(\$2\) (this stays as 2)
  • Strawberry cost: \(\$1.90\) per pound × \(\mathrm{p}\) pounds = \(1.90\mathrm{p}\)
  • Total cost: \(\$9.60\)

3. INFER the equation structure

  • Since total cost = cereal cost + strawberry cost
  • We can write: \(1.90\mathrm{p} + 2 = 9.60\)
  • This matches the structure where variable costs plus fixed costs equal total cost

Answer: A. \(1.90\mathrm{p} + 2 = 9.60\)




Why Students Usually Falter on This Problem


Most Common Error Path:

Weak TRANSLATE skill: Students confuse which operation to use between the costs, thinking they should subtract the cereal cost from the total instead of adding it to the strawberry cost.

Their reasoning: "I need to find how much was spent on strawberries, so I subtract the cereal cost from the total." This leads them to write \(1.90\mathrm{p} - 2 = 9.60\), thinking the strawberry cost minus cereal cost equals total cost.

This may lead them to select Choice B (\(1.90\mathrm{p} - 2 = 9.60\)).


Second Most Common Error:

Poor TRANSLATE reasoning: Students mix up which value represents the rate and which represents the quantity, incorrectly thinking the \(\$2\) is multiplied by \(\mathrm{p}\) instead of \(\$1.90\).

Their reasoning: "The \(\$2\) must be the per-pound cost and \(\$1.90\) must be something else." This fundamental misreading of the problem structure leads them to set up \(1.90 + 2\mathrm{p} = 9.60\).

This may lead them to select Choice C (\(1.90 + 2\mathrm{p} = 9.60\)).


The Bottom Line:

This problem requires careful translation of real-world costs into algebraic expressions. Students must distinguish between fixed costs (cereal = \(\$2\)) and variable costs (strawberries = \(\$1.90\) × pounds) and recognize that these costs are added together, not subtracted from each other.

Answer Choices Explained
A
\(1.90\mathrm{p} + 2 = 9.60\)
B
\(1.90\mathrm{p} - 2 = 9.60\)
C
\(1.90 + 2\mathrm{p} = 9.60\)
D
\(1.90 - 2\mathrm{p} = 9.60\)
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