Mathematics
bagga
22

Delaney would like to make a 5lb mixture that is 60% peanuts and 40% almonds. She has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds. Let p represent the number of pounds of peanuts needed to make the new mixture,and let m represent the number of pounds of the 80% almond-20% peanut mixture. (a) What is the system that models this situation? (b) How many pounds of peanuts and how many pounds of the 80% almond-20% peanut mixture will she need?

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(1) Answers
Czarny333

This is a mixture problem. Let's call the mixture of 20%peanuts - 80% almonds as mixture A Let's call the 100% peanuts as mixture B and the 60% peanuts - 40% almonds as mixture C Since we get mixture C by adding mixture A and B together, we know that the amount in pounds of mixture A and mixture B is 5 pounds. Hence we let m be the amount in pounds of 20%peanuts-80% almonds mixture, or mixture A and 50-m be the amount in pounds of mixture B So, the system of equations that models this situation is: 0.20m + 1(50-m) = 0.60(5) 0.80m + 0(50-m) = 0.40(5) Solving for m using equation 2 gives us m = 2.5 lb. That means it will take 2.5 lb of mixture A (20%peanuts-80%almonds) and 2.5 lb of mixture B (100%peanuts) to form 5 lb of mixture C (60%peanuts-40%almonds).

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