9514 1404 393
Answer:
20 pounds
Step-by-step explanation:
Often, mixture problems are worked by writing two equations: one for the total quantity of constituents in the mix, and one for the total price of the mix. When these equations are solved by substitution, the first step is to write one quantity in terms of the other.
You can save an equation and a step by using one variable for quantity. It usually works best to let the variable represent the highest-cost (or most concentrated) contributor. Here, that is the quantity the problem asks for, so ...
Let c represent the number of pounds of cashews needed to be mixed with the peanuts. Then the revenue from selling the mix is ...
80($1.50) +c($4.00) = (80+c)($2.00)
120 +4c = 160 +2c . . . . . divide by $, simplify
2c = 40 . . . . . . . . . . . . . subtract 120+2c from both sides
c = 20 . . . . . . . . . . . . . divide by 2
The manager should mix 20 pounds of cashews with the 80 pounds of peanuts.
