Delaney would like to make a 5 lb nut 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) Which of the following is a solution to the system: 2 lb peanuts and 3 lb mixture; 2.5 lb peanuts and 2.5 lb mixture; 4 lb peanuts and 1 lb mixture? Show your work.

Question

Francis

Mathematics

8 July, 02:10

Delaney would like to make a 5 lb nut 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) Which of the following is a solution to the system: 2 lb peanuts and 3 lb mixture; 2.5 lb peanuts and 2.5 lb mixture; 4 lb peanuts and 1 lb mixture? Show your work.

+1

Answers (1)

Know the Answer?

Donavan Middleton · Answer 1

First equation is just the sum of the amounts of peanuts (p) and almonds (m). We know the mixture is 5lb so p+m=5.

The total number of peanuts we will get in the mixture will be p+0.2m and the number of almonds we will get is 0.8m.

We aim to have 0.6*5=3lbs of peanuts and 0.4*5=2lbs of almonds.

So p+0.2m=3 and 0.8m=2. From these we get m=2/0.8=2.5lb and p+0.2*2.5=3, p=3-0.5=2.5.

Note that m=2.5 and p=2.5 satisfies p+m=5.

So we need 2.5 of pure peanuts and 2.5 of mixture.