A coffee mixture has beans that sell for $1.35 per pound and beans that sell for $1.05 per pound. If 120 pounds of beans create a mixture worth $1.15 per pound, how many pounds of each bean is used?

a. 50 lb of $1.35/lb; 70 lb of $1.05/lb
b. 40 lb of $1.35/lb; 80 lb of $1.05/lb
c. 45 lb of $1.35/lb; 75 lb of $1.05/lb

Question

A coffee mixture has beans that sell for $1.35 per pound and beans that sell for $1.05 per pound. If 120 pounds of beans create a mixture worth $1.15 per pound, how many pounds of each bean is used?

a. 50 lb of $1.35/lb; 70 lb of $1.05/lb
b. 40 lb of $1.35/lb; 80 lb of $1.05/lb
c. 45 lb of $1.35/lb; 75 lb of $1.05/lb

ganeshie8 · Answer

let
x = pounds of beans required of $1.35 type
y = pounds of beans required of $1.05 type

anonymous · Answer

The fastest way to do this would be by trying all the multiple choice answers (this is an averages problem):
A - (50*1.35 + 70*1.05)/120 = 1.203
B - (40*1.35 + 80*1.05)/120 = 1.172
C - (30*1.35 + 90*1.05)/120 = 1.142
The closest answer is C (which I assume is correct)
If not, you just solve the formula I used in each answer, except instead of let's say 30 and 90, you write X and 120-X respectively. Sorry I did not do the math here.

ganeshie8 · Answer

since we defined 2 variables, we need to form 2 equations for solving

ganeshie8 · Answer

@T-ara93  can you try putting two equations in x and y