SOLUTION: how much of an 80% orange juice drink must be mized with 20 gallons of a 20% orange juice drink to obtain a mixture that is 50% orange juice

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: how much of an 80% orange juice drink must be mized with 20 gallons of a 20% orange juice drink to obtain a mixture that is 50% orange juice      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 503047: how much of an 80% orange juice drink must be mized with 20 gallons of a 20% orange juice drink to obtain a mixture that is 50% orange juice
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
--------percent ---------------- quantity
Orange juice 20 ---------------- 20 gal
Orange juice 80 ---------------- x gal
Mixture 50 ---------------- 20 + x gal

20 * 20 + 80 x = 50 ( 20 + x )
400 + 80 x = 1000 + 50 x
80 x + -50 x = 1000 - -400
30 x = 600
/ 30
x = 20
20 gal of 80 % Orange juice