SOLUTION: I need help setting up this problem. Thanks
Red cards are worth 3 credits, while blue cards are worth 5 credits. You need any combination of 20 cards to play a game.
With 84
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Question 243093: I need help setting up this problem. Thanks
Red cards are worth 3 credits, while blue cards are worth 5 credits. You need any combination of 20 cards to play a game.
With 84 credits to buy cards , how many of each type of card will you have when you play?
Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Red cards are worth 3 credits, while blue cards are worth 5 credits. You need any combination of 20 cards to play a game.
With 84 credits to buy cards , how many of each type of card will you have when you play?
----------------
Equations:
Quantity Equation:: r + b = 20 cards
Credit Equation::: 3r + 5b = 84
----------------------------------------
Multiply thru the Quantity Equation by 3 to get:
3r + 3b = 60
Subtract that from the Credit Equation to get:
2b = 24
b = 12 (# of blue card needed)
---
Since r + b = 20, r = 8 (# of red card needed)
============================================
Cheers,
Stan H.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Let:
r = number of red cards
b = number of blue cards
We are told there are 20 cards, so:
r + b = 20
which can be rearranged as:
b=20-r
...
Since each r has a value of 3 and each b has a value of 5, we can compute the total.
3r + 5b = 84
...
Substituting,
3(20-b) + 5b = 84
60 -3b +5b = 84
2b=24
b=12
Since the total number of cards is 20, then
r=8
Checking the answer...
3r = 24
5b = 60
24+60=84
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