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

A small business owner budgets $dollar sign 2,200$ to purchase candles. The owner must purchase a minimum of $200$ candles to maintain the discounted pricing. If the owner pays $dollar sign 4.90$ per candle to purchase small candles and $dollar sign 11.60$ per candle to purchase large candles, what is the maximum number of large candles the owner can purchase to stay within the budget and maintain the discounted pricing?

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Explanation

The correct answer is $182$ . Let $s$ represent the number of small candles the owner can purchase, and let $l$ represent the number of large candles the owner can purchase. It’s given that the owner pays $dollar sign 4.90$ per candle to purchase small candles and $dollar sign 11.60$ per candle to purchase large candles. Therefore, the owner pays $4.90 s$ dollars for $s$ small candles and $11.60 script l$ dollars for $l$ large candles, which means the owner pays a total of $4.90 s plus 11.60 script l$ dollars to purchase candles. It’s given that the owner budgets $dollar sign 2,200$ to purchase candles. Therefore, $4.90 s + 11.60 l \leq 2,200$ . It’s also given that the owner must purchase a minimum of $200$ candles. Therefore, $s + l \geq 200$ . The inequalities $4.90 s + 11.60 l \leq 2,200$ and $s + l \geq 200$ can be combined into one compound inequality by rewriting the second inequality so that its left-hand side is equivalent to the left-hand side of the first inequality. Subtracting $l$ from both sides of the inequality $s + l \geq 200$ yields $s \geq 200 - l$ . Multiplying both sides of this inequality by $4.90$ yields $4.90 s \geq 4.90 (200 - l)$ , or $4.90 s \geq 980 - 4.90 l$ . Adding $11.60 script l$ to both sides of this inequality yields $4.90 s + 11.60 l \geq 980 - 4.90 l + 11.60 l$ , or $4.90 s + 11.60 l \geq 980 + 6.70 l$ . This inequality can be combined with the inequality $4.90 s + 11.60 l \leq 2,200$ , which yields the compound inequality $980 + 6.70 l \leq 4.90 s + 11.60 l \leq 2,200$ . It follows that $980 + 6.70 l \leq 2,200$ . Subtracting $980$ from both sides of this inequality yields $6.70 script l less than or equals 2,200$ . Dividing both sides of this inequality by $6.70$ yields approximately $script l less than or equals 182.09$ . Since the number of large candles the owner purchases must be a whole number, the maximum number of large candles the owner can purchase is the largest whole number less than $182.09$ , which is $182$ .