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The senior class at Hartford High School is planning their prom. They’ve narrowed the location down to two choices. The Regal Hotel charges, C , $500 for the ballroom rental fee and $75 per student, s , for the meal. The Greenhill Country Club charges, C , $1,000 for the ballroom rental fee and $65 per student, s , for the meal. Create a pair of linear equations that represent this real-world problem

option:

A). Regal Hotel: C=75s+500; Greenhill Country Club: C=65s+1,000

B). Regal Hotel: C=75s+1,000; Greenhill Country Club: C=65s+500

C). Regal Hotel: C=1,000s+75; Greenhill Country Club: C=500s+65

D). Regal Hotel: C=500s+75; Greenhill Country Club: C=1,000s+65

9 answers

To represent the costs for hosting the prom at each location with linear equations, we need to consider both the fixed costs (ballroom rental fee) and the variable costs (meal cost per student).

  1. For the Regal Hotel:

    • Fixed cost (ballroom rental fee) = $500
    • Variable cost (meal cost per student) = $75 per student
    • Therefore, the equation for the Regal Hotel is: \[ C = 75s + 500 \]

  2. For the Greenhill Country Club:

    • Fixed cost (ballroom rental fee) = $1,000
    • Variable cost (meal cost per student) = $65 per student
    • Therefore, the equation for the Greenhill Country Club is: \[ C = 65s + 1000 \]

Thus, the correct pair of linear equations representing the total cost (C) as a function of the number of students (s) is:

Regal Hotel: \( C = 75s + 500 \)

Greenhill Country Club: \( C = 65s + 1000 \)

The correct option that matches this is:

A) Regal Hotel: \( C = 75s + 500 \); Greenhill Country Club: \( C = 65s + 1,000 \)

The Jackson Jammers are giving away bags and caps at their next game. The bags, b , cost them $10 each, and the caps, c , cost them $5 each. They plan to give away 2,000 total items and have a budget of $12,500. Write the pair of linear equations that model this situation

options:

A). b+c=12,500; 5b+10c=2,000

B). b+c=2,000; 10b+5c=12,500

C). b+c=12,500; 10b+5c=2,000

D). b+c=2,000; 5b+10c=12,500

To model the situation involving the Jackson Jammers giving away bags and caps, we need to establish two equations based on the information provided:

  1. Total number of items: The total number of bags (\( b \)) and caps (\( c \)) given away is 2,000. \[ b + c = 2000 \]

  2. Budget equation: The cost of the bags is $10 each and the cost of the caps is $5 each, with a total budget of $12,500. The expense can be modeled as: \[ 10b + 5c = 12500 \]

Now we have our two equations:

  1. \( b + c = 2000 \)
  2. \( 10b + 5c = 12500 \)

Looking at the given options, we identify the correct pair of linear equations:

B). \( b + c = 2000 \); \( 10b + 5c = 12,500 \)

This option correctly represents the total number of items given away and the budget constraints.

Solve this real-world problem using a system of equations. The O’Donnells are having a birthday party for their son, and they plan to have a clown bring balloons to the party. Balloon Express charges $2 per balloon and $10 for delivery. Balloon Mania charges $1.50 per balloon and $20 for delivery. If the O’Donnells plan to order 30 balloons, which company should they choose, and how much will they save?

options:

A). Balloon Mania; $5

B). Balloon Mania; $25 

C). Balloon Express; $25

D). Balloon Express; $5

To determine which balloon company the O'Donnells should choose and how much they will save, we can set up a system of equations for the total cost of each company based on their prices.

1. Calculate the total cost for Balloon Express:

  • Cost per balloon = $2
  • Delivery fee = $10
  • Number of balloons = 30

The total cost \(C_E\) for Balloon Express is given by: \[ C_E = (2 \times 30) + 10 \] \[ C_E = 60 + 10 = 70 \]

2. Calculate the total cost for Balloon Mania:

  • Cost per balloon = $1.50
  • Delivery fee = $20
  • Number of balloons = 30

The total cost \(C_M\) for Balloon Mania is given by: \[ C_M = (1.50 \times 30) + 20 \] \[ C_M = 45 + 20 = 65 \]

3. Comparison and Savings:

To find out how much the O'Donnells will save by choosing one company over the other, we can subtract the total costs: \[ \text{Savings} = C_E - C_M \] \[ \text{Savings} = 70 - 65 = 5 \]

Thus, the O'Donnells should choose Balloon Mania, and they will save $5.

The correct answer is: A). Balloon Mania; $5.

The Spanish Club at Santa Fe Middle School wants to sell mini quiches to raise money for a field trip. The culinary department at their school will make the quiches for $30 plus $0.50 per quiche for the ingredients. The Spanish Club wants to sell each quiche for $2. How many quiches will they need to sell in order to break even?

option:

20 quiches
20 quiches

75 quiches
75 quiches

12 quiches
12 quiches

45 quiches
45 quiches

To determine how many mini quiches the Spanish Club needs to sell in order to break even, we first need to set up equations for their costs and their revenue.

Costs:

  1. Fixed cost for the culinary department: $30
  2. Variable cost per quiche: $0.50

Let \( q \) be the number of quiches they sell. The total cost \( C \) for making \( q \) quiches can be expressed as: \[ C = 30 + 0.50q \]

Revenue:

The selling price for each quiche is $2. The total revenue \( R \) from selling \( q \) quiches is: \[ R = 2q \]

Break-Even Point:

To find the break-even point, we set the total cost equal to the total revenue: \[ 30 + 0.50q = 2q \]

Now we can solve for \( q \):

  1. Start by rearranging the equation: \[ 30 = 2q - 0.50q \] \[ 30 = 1.5q \]

  2. Now, divide both sides by 1.5: \[ q = \frac{30}{1.5} = 20 \]

Conclusion:

The Spanish Club will need to sell 20 quiches in order to break even.

Thus, the correct answer is: 20 quiches.

The Baynard High School Boosters are selling hot dogs and pretzels at the concession stand. They have 300 hot dogs and pretzels in stock and want to earn an income of $800. They plan to sell the hot dogs for $4 each and the pretzels for $2 each. How many hot dogs and how many pretzels do the boosters need to sell to reach their goal of $800?

option:

100 hot dogs and 200 pretzels
100 hot dogs and 200 pretzels

250 hot dogs and 50 pretzels
250 hot dogs and 50 pretzels

50 hot dogs and 250 pretzels
50 hot dogs and 250 pretzels

200 hot dogs and 100 pretzels

Let's denote the number of hot dogs sold as \( h \) and the number of pretzels sold as \( p \).

We have the following equations based on the information provided:

  1. Total items: The total number of hot dogs and pretzels in stock is 300. \[ h + p = 300 \]

  2. Total income: They want to earn an income of $800, where hot dogs are sold for $4 each and pretzels for $2 each. \[ 4h + 2p = 800 \]

Now, we can solve the system of equations.

  1. From the first equation, we can express \( p \) in terms of \( h \): \[ p = 300 - h \]

  2. Substitute \( p \) into the second equation: \[ 4h + 2(300 - h) = 800 \] \[ 4h + 600 - 2h = 800 \] \[ 2h + 600 = 800 \] \[ 2h = 800 - 600 \] \[ 2h = 200 \] \[ h = 100 \]

  3. Now use the value of \( h \) to find \( p \): \[ p = 300 - h = 300 - 100 = 200 \]

Conclusion:

The Baynard High School Boosters need to sell 100 hot dogs and 200 pretzels to reach their goal of $800.

Thus, the correct answer is: 100 hot dogs and 200 pretzels.