SOLUTION: a grocer mixes cookies worth 80 cents a kg with cookies worth 95 cents a kg making a mixture selling at 85 cents a kg. If he mixes 60 kilograms , how many kilograms of each kind d

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: a grocer mixes cookies worth 80 cents a kg with cookies worth 95 cents a kg making a mixture selling at 85 cents a kg. If he mixes 60 kilograms , how many kilograms of each kind d      Log On


   

Question 138707: a grocer mixes cookies worth 80 cents a kg with cookies worth 95 cents a kg making a mixture selling at 85 cents a kg. If he mixes 60 kilograms , how many kilograms of each kind does he use?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = the kgs of $.80/kg cookies needed
Let b = the kgs of $.95/kg cookies needed
In words, the problem is:
(kgs $.80/kg cookies) x ($.80/kg) + (kgs $.95/kg cookies) x ($.95/kg) /
(kgs $.80/kg cookies + kgs $.95/kg cookies) = $.85/kg
or in symbols,
%28.8a+%2B+.95b%29+%2F+%28a+%2B+b%29+=+.85 cents/kg
Also given is:
a+%2B+b+=+60 kg
%28.8a+%2B+.95b%29+%2F+60+=+.85
.8a+%2B+.95b+=+60%2A.85
.8a+%2B+.95%2860+-+a%29+=+51
.8a+%2B+57+-+.95a+=+51
-.15a+=+-6
a+=+40
b+=+60+-+a
b+=+60+-+40
b+=+20
The grocer needs to mix 40 kg of $.80/kg cookies and 20 kg of $.95/kg
cookies
check answer:
%28.8a+%2B+.95b%29+%2F+%28a+%2B+b%29+=+.85
%28.8%2A40+%2B+.95%2A20%29+%2F+60+=+.85
%2832+%2B+19%29%2F+60+=+.85
51%2F60+=+.85
.85+=+.85
OK