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A theater sold 500 tickets to a concert. The tickets for adults cost $20 each, and the tickets for children cost $12 each. If the total revenue was $8,000, how many adult tickets were sold?
200
250
350
300
250
Assume all 500 tickets were child tickets: 500 \times 12 = \6,000.Theextra$2,000revenuecomesfromreplacingchildticketswithadulttickets.Differenceperticketis20 - 12 = $8,soadulttickets=82000 = 250$.
Total tickets sold = 500, Adult ticket price = \20,Child ticket price=$12,Totalrevenue=$8,000$
x+y=N,Pax+Pcy=R
Assume all 500 tickets were child tickets: 500 \times 12 = \6,000.Theextra$2,000revenuecomesfromreplacingchildticketswithadulttickets.Differenceperticketis20 - 12 = $8,soadulttickets=82000 = 250$.
Solving for y (child tickets) instead of x (adult tickets) and accidentally picking the option for 250 when both counts turn out to be equal.
Formulate the Equations
Let x be the number of adult tickets sold and y be the number of child tickets sold. Based on total tickets and total revenue, we set up two linear equations:
x+y=500and20x+12y=8000
Express One Variable in Terms of the Other
From the total tickets equation, express y in terms of x:
y=500−x
Substitute and Solve for Adult Tickets (x)
Substitute y=500−x into the revenue equation:
20x+12(500−x)=8000
Simplify and Calculate
Expand the terms and isolate x:
20x+6000−12x=8000⟹8x=2000⟹x=250
B is correct because solving the linear equations gives x=250 adult tickets sold.
This problem is functionally identical to Mixture & Alligation problems, where ticket prices represent the concentration or cost rate of components.
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