Algebra practice question

Answer: A. 40

Let a be adult tickets and c be child tickets. Then a + c = 130 and 8a + 3c = 840. Substituting a = 130 - c gives 8(130 - c) + 3c = 840, or 1040 - 5c = 840, so 5c = 200 and c = 40 (with a = 90; check: 8(90) + 3(40) = 720 + 120 = 840). Choice B assumes the tickets split evenly, 130 ÷ 2 = 65, ignoring the price information. Choice C is the number of adult tickets, the other variable. Choice D comes from computing 840 ÷ 8 = 105, treating every ticket as an adult ticket.

Why the other answers are wrong

B. 65

You split the 130 tickets evenly (130 ÷ 2 = 65), which ignores the $840 total and the different prices; solving 8a + 3c = 840 with a + c = 130 gives c = 40.

C. 90

90 is the number of adult tickets — the other variable. The child tickets are the remaining 130 - 90 = 40.

D. 105

You computed 840 ÷ 8 = 105, treating every ticket as an $8 adult ticket; the child tickets cost $3, so you need both equations.

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