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Question
A solution containing 12% alcohol is to be mixed with a solution containing 4% alcohol to make 20 gallons of solution containing 9% alcohol. How much of each solution should be used?
Solution
Let x gallons of 12% alcohol and y gallons of 4% alcohol be mixed.
Then, we have
x + y = 20 ....(i)
And, 12% of x + 4% of y = 9% of 20
⇒ `(12)/(100) xx +(4)/(100)"y"`
= `(9)/(100) xx 20`
⇒ 12x + 4y = 180
⇒ 3x+ y = 45 ....(ii)
Subtracting eqn. (i) from eqn. (ii), we get
2x = 25
⇒ x = 12.5
⇒ 12.5 + y = 20
⇒ y = 7.5
Hence, 12.5 gallons of 12% alcohol and 7.5 gallons of 4% alcohol should be used.
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