SAT · Heart of Algebra · Linear inequalities and systems of inequalities
Question sat-q-00046
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The question
A student has a budget of $47. Each notebook costs $3 and a backpack costs $6. If she buys exactly one backpack and x notebooks, what is the maximum value of x she can afford?
- 12
- 41
- 14
- 13
Show answer & worked solution
Answer: 13
Worked solution. Set up the inequality 3x + 6 ≤ 47. Subtract 6: 3x ≤ 41. Divide by 3: x ≤ 13.67. The maximum integer x is 13.
Why each wrong choice is wrong. Off-by-one distractors catch students who solve correctly but round up instead of taking the floor. The fourth distractor is the leftover dollars after the backpack — it ignores the per-notebook division.
Test-day tactic. Whenever a real-world inequality has an integer answer, take the floor — buying 7.4 notebooks doesn't make sense.
About this question type
A linear inequality is solved like a linear equation, except multiplying or dividing both sides by a negative reverses the inequality. Systems of linear inequalities define a feasible region in the plane; the SAT often asks which (x, y) point satisfies all inequalities, or for the maximum value of a linear expression on a feasible region (introductory linear programming, without using that name).
You will see a question shaped like this one on roughly every other official SAT form, typically at a moderate position in the section — solidly within the range that separates a 600-band student from a 700-band student. Treat any miss in this subtopic as a signal to drill the subtopic page before you do another full practice test.
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