Question

A food store makes a 11-lb mixture of oatmeal, crispies, and chocolate chips. The cost of oatmeal is $1.50 per pound, crispies cost $2.00 per pound, and chocolate chips cost $1.00 per pound. The mixture calls for twice as much oatmeal as crispies. The total cost of the mixture is $17.00. How much of each ingredient did the store use?

Answer 1

let's let o = # of lbs of oatmeal, r = # of lbs of crispies, and c = # of lbs of chocolate chips

(I'm using r instead of c for crispies because we have another c, chocolate chips)

we have three unknowns so we'll need three equations

first equation:

if the mixture is 11 lbs total, then o + r + c = 11

Answer 2

second equation:

"The mixture calls for twice as much oatmeal as crispies."

therefore, o = 2r

Answer 3

third equation:

The cost of oatmeal is $1.50 per pound, crispies cost $2.00 per pound, and chocolate chips cost $1.00 per pound. The total cost of the mixture is $17.00.

if oatmeal costs $1.50 per lb, then for (o) lbs of oatmeal, it costs 1.50*o

same logic for the crispies and chocolate chips: 2r and 1c

so the total cost = 1.5o + 2r + 1c = 17

Answer 4

now, putting the equations together:

1) o + r + c = 11
2) o = 2r
3) 1.5o + 2r + 1c = 17

if you plug in o = 2r into equations 1 and 3 you can reduce both equations to two variables. from there, it's a typical 2-equation system where you can solve via elimination or substitution.