Question

Adult tickets to the high school basketball game cost $8 each, and student tickets cost $4. Marcus and his family purchased a total of 12 tickets for $60. If they purchased twice as many adult tickets as student tickets, how many of each type of ticket did they purchase? Explain your answer.

Answer 1

Let a = number of adult tickets and s = number of student tickets.

From the total tickets: a + s = 12.
From the prices: 8a + 4s = 60.
If they bought twice as many adult tickets as student tickets: a = 2s.

Use a + s = 12 and a = 2s => 3s = 12 so s = 4 and a = 8. But that gives cost 8(8) + 4(4) = 64 + 16 = 80, not $60. So those three conditions cannot all be true.

Solve the two consistent equations a + s = 12 and 8a + 4s = 60 (dropping the ratio): multiply the first by 4: 4a + 4s = 48. Subtract from the price equation: 4a = 12, so a = 3 and s = 9. This satisfies the ticket count and total cost, but not the “twice as many adult as student” condition.

Conclusion: There is no possible integer solution that satisfies all three given conditions. If the ratio condition is removed, the actual purchase consistent with the totals is 3 adult tickets and 9 student tickets. If you meant a different ratio, tell me and I’ll solve that version.