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Algebra / Systems of two linear equations in two variables Difficulty: Medium

A company that provides whale-watching tours takes groups of $21$ people at a time. The company’s revenue is $80$ dollars per adult and $60$ dollars per child. If the company’s revenue for one group consisting of adults and children was $1,440$ dollars, how many people in the group were children?

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Explanation

Choice C is correct. Let $x$ represent the number of children in a whale-watching tour group. Let $y$ represent the number of adults in this group. Because it's given that $21$ people are in a group and the group consists of adults and children, it must be true that $x + y = 21$ . Since the company's revenue is $60$ dollars per child, the total revenue from $x$ children in this group was $60 x$ dollars. Since the company's revenue is $80$ dollars per adult, the total revenue from $y$ adults in this group was $80 y$ dollars. Because it's given that the total revenue for this group was $1,440$ dollars, it must be true that $60 x + 80 y = 1,440$ . The equations $x + y = 21$ and $60 x + 80 y = 1,440$ form a linear system of equations that can be solved to find the value of $x$ , which represents the number of children in the group, using the elimination method. Multiplying both sides of the equation $x + y = 21$ by $80$ yields $80 x + 80 y = 1,680$ . Subtracting $60 x + 80 y = 1,440$ from $80 x + 80 y = 1,680$ yields $(80 x + 80 y) - (60 x + 80 y) = 1,680 - 1,440$ , which is equivalent to $80 x - 60 x + 80 y - 80 y = 240$ , or $20 x equals 240$ . Dividing both sides of this equation by $20$ yields $x equals 12$ . Therefore, $12$ people in the group were children.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect. This is the number of adults in the group, not the number of children in the group.

Choice D is incorrect and may result from conceptual or calculation errors.