Question 1011766: 1.The boys mixed 10 gallons of the 20% pure lemon juice mix and 10 gallons of the 70% pure lemon juice mix, because they wanted 20 gallons of lemonade at a 40% lemon juice mixture. They thought that because 40% was almost halfway between 20% and 70%, they should just mix equal parts of both, but the lemonade turned out too tart. How much of each should they have used to get a final mixture of 20 gallons at 40% lemon juice? Write your answer in complete sentences. Show all work.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The boys mixed 10 gallons of the 20% pure lemon juice mix and 10 gallons of the 70% pure lemon juice mix, because they wanted 20 gallons of lemonade at a 40% lemon juice mixture. They thought that because 40% was almost halfway between 20% and 70%, they should just mix equal parts of both, but the lemonade turned out too tart. How much of each should they have used to get a final mixture of 20 gallons at 40% lemon juice? Write your answer in complete sentences. Show all work.
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Quantity Equation: t + s = 20 gallons
Lemon content Eq: 20t+70s = 40*20 gallons
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Modify for elimination:
2t + 2s = 2*20
2t + 7s = 4*20
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Subtract and solve for "s":
5s = 2*20
s = 8 gallons (amt. of 70% mix needed)
t = 20-8 = 12 gallons (amt of 20% mix needed)
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Cheers,
Stan H.
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