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At a fair, the number of adult tickets is (x) and child tickets is (y). If (x+y=12) and (50x+30y=500), what are (x) and (y)?

Answer and explanation

Correct answer: ( (7,5) )

The direct answer is option C: \((x,y)=(7,5)\). The first equation gives \(x+y=12\). Divide the second equation by 10: \(5x+3y=50\). From the first equation, \(y=12-x\). Substitute: \(5x+3(12-x)=50\), so \(5x+36-3x=50\), \(2x=14\), and \(x=7\). Then \(y=12-7=5\). Option A, \((6,6)\), has the correct total but gives ticket value 480, not 500. Option B, \((8,4)\), totals 12 but gives 520. Option C gives total 12 and \(50(7)+30(5)=350+150=500\), so it is correct. Option D, \((5,7)\), gives 480. A useful check is to test both the count equation and the money equation, not just one.

Related tags

Linear EquationsWord ProblemTicketsMediumClass 10

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