SOLUTION: how much of Brand A fruit punch (24% fruit punch) must be mixed with 6 gal. of Brand B fruit punch (19% fruit juice) to create a mixture containing 22% fruit juice?

Question 776036: how much of Brand A fruit punch (24% fruit punch) must be mixed with 6 gal. of Brand B fruit punch (19% fruit juice) to create a mixture containing 22% fruit juice?
Found 2 solutions by stanbon, amalm06:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how much of Brand A fruit punch (24% fruit punch) must be mixed with 6 gal. of Brand B fruit punch (19% fruit juice) to create a mixture containing 22% fruit juice?
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Equation:
fruit + fruit = fruit
0.24a + 0.19*6 = 0.22(a+6)
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24a + 19*6 = 22a + 22*6
2a = 3*6
a = 9 gallons (amt. of Brand A to add)
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Cheers,
Stan H.
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Answer by amalm06(224) (Show Source): You can put this solution on YOUR website!
Use the method of alligation to solve the problem.

Let B=19 and A=24. Then 22-19=3 and 24-22=2

A/B=3/2

So that (3/2)(6)=9 gallons (Answer)
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