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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?

Respuesta :

Hagrid
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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