Linear Algebra · Question of the Day

Linear Algebra Question of the Day

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Friday, October 23, 2026

A theater sells three types of tickets: adult ( $a$ ), student ( $s$ ), and child ( $c$ ). On Friday, they sold twice as many adult tickets as student tickets, and the number of child tickets was 15 more than half the number of adult tickets. If the total number of tickets sold was 245, which system of equations correctly represents this situation?

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Question of the Day

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$a = 2s$ , $c = \frac{1}{2}a + 15$ , $a + s + c = 245$ (correct answer)
$s = 2a$ , $c = \frac{1}{2}a + 15$ , $a + s + c = 245$
$a = 2s$ , $c = \frac{1}{2}s + 15$ , $a + s + c = 245$
$a = 2s$ , $a = \frac{1}{2}c + 15$ , $a + s + c = 245$

Explanation: The correct answer is A. 'Twice as many adult tickets as student tickets' means $a = 2s$ . 'Child tickets was 15 more than half the number of adult tickets' means $c = \frac{1}{2}a + 15$ . The total constraint gives $a + s + c = 245$ . Choice B reverses the adult-student relationship. Choice C incorrectly relates child tickets to student tickets instead of adult tickets. Choice D inverts the relationship between adult and child tickets.