SOLUTION: At a fish fry fundraiser, the Math Club raised $277.50. Child plates were sold for $2.25 and adult plates were sold for $4.75. If there were twice as many child plates sold than

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Question 1152162: At a fish fry fundraiser, the Math Club raised $277.50. Child plates were sold for $2.25 and adult plates were sold for $4.75. If there were twice as many child plates sold than adult plates, how many adult plates were sold?
Found 3 solutions by MathLover1, MathTherapy, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let the number of adult plates be x and the number of child plates y
the Math Club raised $277.50.
child plates were sold for $2.25 : 2.25y
adult plates were sold for $4.75 : 4.75x
=>4.75x%2B2.25y=277.50.........eq.1
If there were twice as many child plates sold than adult plates,we have
y=2x.......eq.2...substitute in eq.1

4.75x%2B2.25%2A2x=277.50.........solve for x
4.75x%2B4.50x=277.50
9.25x=277.50
x=277.50%2F9.25
x=30
go to eq.2
y=2x..........substitute x
y=2%2A30
y=60


how many adult plates were sold?
answer: x=30
how many child plates were sold?
answer: x=60

check:
4.75x%2B2.25y=277.50
4.75%2A30%2B2.25%2A60=277.50
142.5%2B135=277.50
277.50=277.50



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

At a fish fry fundraiser, the Math Club raised $277.50. Child plates were sold for $2.25 and adult plates were sold for $4.75. If there were twice as many child plates sold than adult plates, how many adult plates were sold?
Let number of adults' plates be A

Then number of children's plates = 2A

We then get: 2.25(2A) + 4.75A = 277.5

4.5A + 4.75A = 277.5

9.25A = 277.5

A, or adults' plates sold = highlight_green%28matrix%281%2C3%2C+%22%24277.5%22%2F9.25%2C+%22=%22%2C+30%29%29

That's IT!! Nothing MORE, nothing LESS!

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Take a good look at now much work tutor @MathLover1 did to solve the problem using two variables; then look how little work tutor @MathTherapy did to solve the problem using a single variable.

The lesson there is that it is nearly always easier to solve a problem if you can set it up using one variable instead of two (or more).

In this problem, that is definitely the case, since the information that the number of child plates was twice the number of adult plates makes it easy to use a single variable.

REMEMBER THIS LESSON!!