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18 votes
18 votes
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

Chamblee High School is selling Valentine's Day gifts as a fundraising event. One-example-1
User Jit B
by
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1 Answer

20 votes
20 votes

Answer: Choice B) 60 roses and 10 carnations

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Step-by-step 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.

User Rafl
by
3.0k points