SOLUTION: 3) Jordan has 1 pound of brand 110 nuts that sell for $2.46 a pound. She also has brand 403 nuts that sell for $1.75 a pound. If she wants a mixture of nuts that sell for $1.84 a p

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: 3) Jordan has 1 pound of brand 110 nuts that sell for $2.46 a pound. She also has brand 403 nuts that sell for $1.75 a pound. If she wants a mixture of nuts that sell for $1.84 a p      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 765964: 3) Jordan has 1 pound of brand 110 nuts that sell for $2.46 a pound. She also has brand 403 nuts that sell for $1.75 a pound. If she wants a mixture of nuts that sell for $1.84 a pound, how many pounds of brand 403 should she add?

3) Peanut butter costs sixteen cents an ounce. Jelly costs two hundred twenty four cent per pound. Haliey wants to make 1 1/2 pounds of peanut butter and jelly. If she wants the entire mix to cost $3.52, how many pounds of jelly should she use?

6) Rachel has 6 pounds of brand 188 nuts that sell for $2.50 a pound. SHe also has brand 482 that sell for $2 a pound. If she wants a mixture of nuts that sells for $2.20 a pound, how many pounds of brand 482 should she add?

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
#3(first) and #6 are the same general problem. #3(second) is slightly different but still a related general problem, in that it uses TWO unknown numbers.

Some help for #3(second):

L = 0.16*16 (dollars per ounce)(ounce per pound)
H = 0.24 dollars per pound
T = 3.52 dollars per pound
u = pounds of peanut butter, UNKNOWN
v= pounds of jelly, UNKNOWN
M = 1.5 pounds of peanut butter & jelly mixture or combination

Watch carefully that units for the solution process have been changed to pounds, dollars, but you could convert to whichever units you need or want to work with.

Formulate the equations:
%28Lu%2BHv%29%2FM=T and u%2Bv=M
Remember, u and v are in POUNDS. L and H and T are in dollars per pound.
Solve for u and v.

The process can continue like this,
Lu%2BHv=MT, and maybe use the M equation as v=M-u and substitute.
Lu%2BH%28M-u%29=MT
Lu%2BHM-Hu=MT
Lu-Hu=MT-HM
%28L-H%29u=MT-MH
u=%28MT-MH%29%2F%28L-H%29
.
. and watch the signs carefully as you go.
.
continue..
.