Question

How many gallons of 40% antifreeze should be mixed with 10 gallons of 80% antifreeze to obtain a 65% antifreeze mixture?

Answer 1

Equation:

alcohol + alcohol = alcohol

0.40x + 0.85*10 = 0.65(x+10)

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40x + 85*10 = 65x + 65*10

20*10 = 15x

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x = (2/3)20

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x = 13 1/3 gallons (amt. of 40% solution needed)