Question

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.

What is the system that models this situation?

Which of the following is a solution to the situation?
a) 2lb peanut and 3lb mixture
b) 2.5lb peanuts and 2.5lb mixture
c) 4lb peanuts and 1lb mixture

Answer 1

add 60 + 20

Answer 2

Let p equal number of pounds of peanuts needed to add to the mixture.
Let m be the number of the 80% almond, 20% peanut mixture.

The objective is to come up with a new mixture, that is 60% peanuts and 40% almonds. We accomplish this by adding peanuts to the 80% almond-20%peanut mixture.
.2m = the number lbs of peanuts old mixture
p + .2m = the number of pounds in the new mixture.
p + .2m is also .6 of the new mixture.
The new mixture will contain .8m of almonds
.8m + p + .2m=new mixture
.8m = .4 new mixture
p + .2m = .6 new mixture
Somehow, one should be able to come up with a solution. Myself I need to ponder about this.

Answer 3

Since the new mixture will be 5 pounds we now can do the following:
p + .2m + .8m= 5 or m + p = 5
p + .2m = .6(5)= 3
m = 5 - p (from first) now substitute for m in 2nd getting
p + .2(5 - p) = 3
p + 1 - .2p = 3
.8p = 3 - 1 = 2
p=2/.8= 2.5 lbs of peanute added 2.5 lbs of the orig mix. Looks like b is good.