Question 1197844
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Here are two variations of a method that can be used to solve any 2-part "mixture" problem like this without formal algebra.<br>
(1) The 40% of the mixture is 5% away from the 35% of one ingredient and 10% away from the 50% of the other ingredient.  So 40% is "twice as close" to 35% as it is to 50%.  That means the amount of the 35% solution must be twice the amount of the 50% solution.  Since there are 35 gallons of the 35% solution, there must be 35/2 = 17.5 gallons of the 50% solution.<br>
ANSWER: 17.5 gallons of the 50% alcohol<br>
(2) Look at the three percentages on a number line -- 35, 40, and 50 -- and observe/calculate that 40 is one-third of the way from 35 to 50.  That means 1/3 of the mixture is the ingredient with the higher percentage.  So the 35 gallons of 35% alcohol is 2/3 of the total; that means the 1/3 of the total that is the 50% alcohol is half of 35 gallons, which is 17.5 gallons.<br>
ANSWER: 17.5 gallons of the 50% alcohol<br>
CHECK:
.35(35)+.50(17.5) = 12.25+8.75 = 21
.40(35+17.5) = .40(52.5) = 21<br>