Question

Chamblee High School is selling Valentine's Day gifts as a fundraising event. One long stemmed rose costs $3.00 while one long stemmed carnation costs $1.50. If 50 orders
were placed and they totaled $195, how many roses and carnations were ordered?
100 roses and 100 carnations
60 roses and 10 carnations
125 roses and 75 carnations
50 roses and 145 carnations

Answer 1

Answer:  Choice B) 60 roses and 10 carnations

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Explanation:

• r = number of roses
• c = number of carnations

r and c are positive whole numbers.

r+c = total number of flowers = 50, since 50 orders are made.

The first equation to set up is r+c = 50.

This equation can be solved to get r = 50-c.

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3r = cost of all the roses only, in dollars

1.5c = cost of all the carnations only, in dollars

3r+1.5c = total cost of all the flowers = 195 dollars

3r+1.5c = 195

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Let's apply substitution to solve

3r+1.5c = 195

3(50-c)+1.5c = 195

150-3c+1.5c = 195

-1.5c+150 = 195

-1.5c = 195-150

-1.5c = 45

c = 45/(-1.5)

c = -30

That's not good. We shouldn't get a negative value.

It turns out that the condition r+c = 50 should be ignored. Notice how none of the answer choices listed have r+c leading to 50.

So we'll only focus on the equation 3r+1.5c = 195

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If we plugged in r = 100 and c = 100, then we get

3r+1.5c = 195

3(100)+1.5(100) = 195

300+150 = 195

450 = 195

Which is false. So we can rule out choice A

Let's repeat those steps for choice B

3r+1.5c = 195

3(60)+1.5(10) = 195

180 + 15 = 195

195 = 195

So that works out. I have a feeling your teacher meant to say "70 orders" instead of "50 orders". If so, then the equation r+c = 50 would be r+c = 70 and everything would lead to choice B as the final answer.

Choices C and D are similar to that of choice A, so they can be ruled out.