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Algebra / Linear inequalities in one or two variables Difficulty: Medium

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?

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Explanation

Choice A is correct. It’s given that the truck can tow a trailer if the combined weight of the trailer and the boxes it contains is no more than $4,600$ pounds. If the trailer has a weight of $500$ pounds and each box weighs $120$ pounds, the expression $500 plus 120 b$ , where $b$ is the number of boxes, gives the combined weight of the trailer and the boxes. Since the combined weight must be no more than $4,600$ pounds, the possible numbers of boxes the truck can tow are given by the inequality $500 + 120 b \leq 4,600$ . Subtracting $500$ from both sides of this inequality yields $120 b less than or equals 4,100$ . Dividing both sides of this inequality by $120$ yields $b \leq \frac{205}{6}$ , or $b$ is less than or equal to approximately $34.17$ . Since the number of boxes, $b$ , must be a whole number, the maximum number of boxes the truck can tow is the greatest whole number less than $34.17$ , which is $34$ .

Choice B is incorrect. Towing the trailer and $35$ boxes would yield a combined weight of $4,700$ pounds, which is greater than $4,600$ pounds.

Choice C is incorrect. Towing the trailer and $38$ boxes would yield a combined weight of $5,060$ pounds, which is greater than $4,600$ pounds.

Choice D is incorrect. Towing the trailer and $39$ boxes would yield a combined weight of $5,180$ pounds, which is greater than $4,600$ pounds.