A student has $40 to spend on notebooks costing $4 each and pens costing $2 each, and wants at least 12 items total. If n is the number of notebooks and p is the number of pens, which system models the situation?

Question

Accepted Answer

A. 4n + 2p ≤ 40 and n + p ≥ 12. Total spending 4n + 2p can be at most $40 (≤), and total items n + p must be at least 12 (≥). Choice B flips both inequality directions; choice D swaps the notebook and pen prices.

Answer

B. 4n + 2p ≥ 40 and n + p ≤ 12 — You flipped both inequalities: $40 is a maximum you can spend (≤), and 12 items is a minimum (≥), not the other way around.

Answer

C. 4n + 2p ≤ 40 and n + p ≤ 12 — You got the spending constraint right, but 'at least 12 items' means n + p ≥ 12, not ≤ 12.

Answer

D. 2n + 4p ≤ 40 and n + p ≥ 12 — You swapped the prices — notebooks cost $4 each (4n) and pens cost $2 each (2p), so the cost expression is 4n + 2p.

Algebra practice question

Answer: A. 4n + 2p ≤ 40 and n + p ≥ 12

Why the other answers are wrong

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