Question

Mark is in charge of the fundraiser for his church. He is selling pre-sale tickets for $10 and at-the-door tickets for $13. The venue has the capacity to hold 275 people. The graph represents the number of raffle tickets he needs to sell to offset the upfront costs and raise at least $3,100:
What is the minimum number of at-the-door tickets he needs to sell to make his goal?
159
158
117
116

Answer 1

Answer:

117

Step-by-step explanation:

At-the-door-tickets cost $13 each.

Pre-sale tickets cost $10 each.

The venue has room for 275 people, so the total number of tickets sold is 275. If all tickets were pre-sale tickets, then he'd make 275 * $10 = $2750. That is less than the amount of money he needs to raise which is $3100.

Since the at-the-door tickets cost $13 each, if he sold all 275 tickets at the door, he'd make 275 * $13 = $3575 which is more than he needs to make.

The question here is what is the minimum number of the more expensive tickets (at $13 each) that he needs to sell and still raise a minimum of $3100.

The intersection of the lines on the graph is at point (158.333, 116.667). The horizontal axis is Pre-sale Tickets, and the vertical axis is At-The-Door Tickets. The y-coordinate representing at-the-door tickets is 116.667. He cannot sell 116.667 tickets. He must sell a whole number of tickets, so he must sell either 116 or 117 tickets.

Let's look at 116 at-the-door tickets:

If he sells 116 at-the-door tickets, then he can sell 275 - 116 = 159 pre-sale tickets.

The money he raises is:

116 * $13 + 159 * $10 = $3098 which is less than $3100

Let's look at 117 at-the-door tickets:

If he sells 117 at-the-door tickets, then he can sell 275 - 117 = 158 pre-sale tickets.

The money he raises is:

117 * $13 + 158 * $10 = $3101 which is more than $3100 and works

A minimum of 117 at-the-door tickets works.

Answer: The minimum number of at-the-door tickets is 117.

Answer 2

Answer:

the answer is 117 i just took the test and got it right