Question 347087: A dietitian arranges a special diet composed of two types of food, A and B. Each ounce of food A contains 10 units of calcium and 4 units of iron; each ounce of food B contains 6 units of calcium and 4 units of iron. How many ounces of foods A and B should be used to obtain a food mix that contains 92 units of calcium and 44 units of iron? Thank You
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A dietitian arranges a special diet composed of two types of food, A and B. Each ounce of food A contains 10 units of calcium and 4 units of iron; each ounce of food B contains 6 units of calcium and 4 units of iron. How many ounces of foods A and B should be used to obtain a food mix that contains 92 units of calcium and 44 units of iron?
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Calcium Equation: 10A + 6B = 92 units
Iron Equation:::: 4A + 4B = 44 units
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Simplify:
5A + 3B = 46
A + B = 11
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Multiply thru 2nd equation by 5:
5A + 3B = 46
5A + 5B = 55
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Subtract 1st from 2nd and solve for "B":
2B = 9
B = 9/2 = 4.5 units (amt. of food B needed)
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Since A + B = 11, A = 11-4.5 = 6.5 units (amt. of food A needed)
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Cheers,
Stan H.
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