A solution of 54% alcohol is to be mixed with a solution of 22% alcohol to form 64 liters of a 42% solution. How many liters of the 54% solution must be used?

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Answer 1 · 2018-03-06T11:48:59

$40$ $40$ liters of $54 %$ $54%$ alcohal must be used.

Let $x$ $x$ liters of $54 %$ $54%$ alcohal is mixed with #(64-x) liters of

$22 %$ $22%$ alcohal is mixed to get $64$ $64$ liters of $42 %$ $42%$ alcohal.

$:. x*0.54+(64-x)*0.22=64*0.42$

$:. 0.54x-0.22x=64*0.42-64*0.22$ or

$0.32x=64*(0.42-0.22)$ or

$0.32x=64*0.2 :. x =(64*0.2)/0.32$ or

$x=40 :. 64-x= 64-40=24$ liters. Hence $40$ liters of $54%$

alcohal is mixed with $24$ liters of $22%$ alcohal is mixed to

get $64$ liters of $42%$ alcohal. [Ans]