Mathematics High School

Student Council is selling tickets to the Winter
Dance. Tickets cost $5 per person or $8 per
couple. To cover expenses, at least $1200
worth of tickets must be sold. No more than 500
students can fit in the gym where the dance is
being held.
List two combinations of tickets that could be sold, so
that Student Council makes a profit, yet stays under the
room capacity.

Answers

Answer 1

Answer:

Two possible combinations are;

1) 200 single and 30 couples

2) 150 single and 57 couples

Step-by-step explanation:

The given parameter in the question are;

The cost of ticket per person = $5

The cost of tickets per couple = $8

The minimum worth of tickets to be sold = $1,200

The maximum number of students that can fit in the gym = 500

Let 'x' represent the number of single persons that attend the dance and let 'y' represent the number of couple that attend, we have;

5·x + 8·y ≥ 1,200

x + y ≤ 500

Making 'y' the subject of both equations gives;

y ≥ 150 - 5·x/8

y ≤ 500 - x

From the graph of the inequalities created with Microsoft Excel, we have;

When x = 200, and y = 30, we get;

5 × 200 + 8 × 30 = 1240 > 1200

200 + 30 × 2 = 460 < 500

We also have;

x = 150, and y = 57, we get

Equating both values of 'y', we get;

5 × 150 + 8 × 57 = 1206 > 1200

150 + 57 × 2 = 264 < 500

Therefore, from the acceptable region in the graph, we have the following two possible combinations

1) 200 single and 30 couples

2) 150 single and 57 couples

Student Council Is Selling Tickets To The WinterDance. Tickets Cost $5 Per Person Or $8 Percouple. To

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Answers

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__

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Answers

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Step-by-step explanation:

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Calculation of the equation:

Since the equation is [tex](43\times 8) - (234\div 6)[/tex]

So,

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