Question

Joanie runs the concession stand for the school's baseball games. At each game, the best selling items are pretzels and hot dogs. Pretzels are $3 each, and hot dogs are $2 each. Today she sold 15 more hot dogs than pretzels and made $195 in total sales.

Use a system of equations to model the situation above, and determine which of the following are possible amounts of pretzels and hot dogs that Joanie sold today.
A.
33 pretzels, 48 hot dogs
B.
15 pretzels, 30 hot dogs
C.
35 pretzels, 45 hot dogs
D.
31 pretzels, 51 hot dogs

Answer 1

Answer: Let's assume the number of pretzels sold is represented by the variable "P," and the number of hot dogs sold is represented by the variable "H."

Based on the information given in the problem, we can create a system of equations:

The total sales equation: $3P + $2H = $195

The relationship between the number of pretzels and hot dogs: H = P + 15

Now, let's evaluate each answer choice by substituting the values into the equations and checking if they satisfy the system:

A. 33 pretzels, 48 hot dogs:

$3(33) + $2(48) = $99 + $96 = $195 (satisfies the total sales equation)

48 = 33 + 15 (satisfies the relationship equation)

So, option A is a possible combination.

B. 15 pretzels, 30 hot dogs:

$3(15) + $2(30) = $45 + $60 = $105 (does not satisfy the total sales equation)

30 ≠ 15 + 15 (does not satisfy the relationship equation)

Therefore, option B is not a possible combination.

C. 35 pretzels, 45 hot dogs:

$3(35) + $2(45) = $105 + $90 = $195 (satisfies the total sales equation)

45 = 35 + 15 (does not satisfy the relationship equation)

Thus, option C is not a possible combination.

D. 31 pretzels, 51 hot dogs:

$3(31) + $2(51) = $93 + $102 = $195 (satisfies the total sales equation)

51 = 31 + 15 (satisfies the relationship equation)

Therefore, option D is a possible combination.

Based on the evaluation, the possible amounts of pretzels and hot dogs that Joanie sold today are:

A. 33 pretzels, 48 hot dogs

D. 31 pretzels, 51 hot dogs

Explanation: