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This is an Algebra II question I need help on.?
Question:
A coffee blend contains Sumatra beans, which cost $5/pound, and Kona beans, which cost $13/pound. If the blend costs $10/pound, how much of each type of coffee is in 50 pounds of the blend?
Want to know how to solve and get the answer please!
You need 75/4 pounds of Sumatra beans and 125/4 pounds of Kona beans.
You know two basic things here, the prices of the individual coffees by pound and the blend's price by pound, and the total weight of the resulting blend. This lets you set up two separate equations for the two variables. Then you just solve the pair of simultaneous equations for the answer.
Use:
S = pounds of Sumatra beans
K = pounds of Kona beans
So $5 times S plus $13 times K should equal the $10 blend price times the total number of pounds in the blend so far. That's S plus K. So:
5S + 13K = 10 * (S + K)....multiply the right side out
5S + 13K = 10S + 10K....subtract 10S and 10K from each side
3K - 5S = 0
Now the easy one. You know the total amount of Sumatra and Kona beans is 50 pounds, so:
K + S = 50
Now to put them together, let's multiply the last equation by 3:
3K + 3S = 150
Subtracting the other result (3K - 5S = 0) from this gives you:
8S = 150....divide each side by 8
S = 150/8 = 75/4
and substituting back into the easy equation:
150/8 + K = 50....subtract 150/8 from each side
K = 400/8 - 150/8 = 250/8 = 125/4
So you need 75/4 pounds of Sumatra beans and 125/4 pounds of Kona beans.
Check that:
$5 * (75/4) + $13 * (125/4)
$375/4 + $1625/4
$2000/4
$500
$500 divided by $10/pound is 50 pounds.
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October 19th, 2006
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