Math  /  Algebra

QuestionA student agrees to throw a party, provided that each guest shares the expenses equally with him. A four-person local band will play for $200\$ 200 and free drinks and snacks. The student estimates that the mean cost of drinks and snacks will be $2\$ 2 per person. a. Suppose that nn people (including the host and band members) attend the party. Let C(n)C(n) be the party's total cost (in dollars). Find an equation of CC. C(n)=200+2nC(n)=200+2 n (Do not factor.) b. Let P(n)P(n) be the equal share of expenses (in dollars) each guest and host contributes. Find an equation of PP. [Hint: Recall that band members get free drinks and snacks.] P(n)=P(n)= \square (Do not factor.)

Studdy Solution

STEP 1

What is this asking? We need to figure out the total cost of the party and how much each person has to chip in, keeping in mind that the band gets free food and drinks! Watch out! Don't forget that the band members are included in the total number of people, but they don't pay for refreshments!

STEP 2

1. Find the total cost.
2. Find each person's share.

STEP 3

The band's cost is a fixed $200\$200, no matter how many people show up.
We can write this as 200200.

STEP 4

Each person (except the band!) consumes $2\$2 worth of food and drinks.

STEP 5

Since there are nn people total, and 44 band members get free refreshments, there are n4n - 4 people paying for food.

STEP 6

So, the total cost of refreshments is $2\$2 per person multiplied by the number of people paying, which is 2(n4)2 \cdot (n - 4).

STEP 7

Let's simplify that: 2(n4)=2n82 \cdot (n - 4) = 2n - 8.

STEP 8

The total cost, C(n)C(n), is the band's fee plus the cost of refreshments: C(n)=200+2n8C(n) = 200 + 2n - 8.

STEP 9

Combining the constants gives us the **total cost**: C(n)=192+2nC(n) = 192 + 2n.

STEP 10

Remember, *everyone* chips in, including the host and the band members, so there are nn payers.

STEP 11

To find each person's share, P(n)P(n), we divide the **total cost** by the number of payers: P(n)=C(n)n P(n) = \frac{C(n)}{n} P(n)=192+2nn P(n) = \frac{192 + 2n}{n}

STEP 12

We can write this as: P(n)=192n+2nn P(n) = \frac{192}{n} + \frac{2n}{n} P(n)=192n+2 P(n) = \frac{192}{n} + 2

STEP 13

a. C(n)=192+2nC(n) = 192 + 2n b. P(n)=192n+2P(n) = \frac{192}{n} + 2

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