SOLUTION: You have one type of nut that sells for $2.20/lb and another type of nut that sells for $9.50/lb. You would to have 21.9 lbs of a nut mixture that sells for $3.60/lb. How much of e

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: You have one type of nut that sells for $2.20/lb and another type of nut that sells for $9.50/lb. You would to have 21.9 lbs of a nut mixture that sells for $3.60/lb. How much of e      Log On

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Question 1201337: You have one type of nut that sells for $2.20/lb and another type of nut that sells for $9.50/lb. You would to have 21.9 lbs of a nut mixture that sells for $3.60/lb. How much of each nut will you need to obtain the desired mixture?
You will need
Ibs of the cheaper nut
and
lbs of the expensive nut.
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Two types nuts: 2.2 $/#, and 9.5 $/#
Mixture wanted: 21.9 # at price 3.6 $/#


v, how much of $9.5/# nuts to use
21.9-v, how much the other nuts to use

9.5v%2B2.2%2821.9-v%29=%283.6%29%2821.9%29
.
.
.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


An informal solution... if a formal algebraic solution like the one shown by the other tutor is not required....

Look at the 3 prices pr pound on a number line and observe/calculate that 3.60 is (1.40)/(7.30) = 14/73 of the way from 2.20 to 9.50.

That means 14/73 of the mixture is the higher-priced nuts.

(14/73)(21.9) = 14*(21.9/73) = 14*0.3 = 4.2

ANSWER: 4.2 pounds of the higher-priced nuts; the other 21.9-4.2 = 17.7 pounds of the lower-priced nuts