Question 1145668
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First a standard algebraic solution to the mixture problem....<br>
x = mL of 30% alcohol solution
725-x = mL of 80% alcohol solution<br>
.30(x) = amount of alcohol in x mL of 30% solution
.80(725-x) = amount of alcohol in (725-x mL of 80% solution
.60(725 = amount of alcohol in 725 mL of 60% solution<br>
{{{.30(x)+.80(725-x) = .60(725)}}}
{{{30(x) + 80(725-x) = 60(725)}}}
{{{30x + 58000-80x = 43500}}}
{{{14500 = 50x}}}
{{{x = 290}}}<br>
ANSWER: 290 mL of the 30% solution and 725-290 = 435 mL of the 80% solution.<br>
That's a perfectly good method for solving mixture problems like this.<br>
But there is a much easier way....<br>
Now a much easier solution, based on where the 60% lies between the 30% and 80%....<br>
(1) 60 is 3/5 of the way from 30 to 80 (look at the 3 numbers 30, 60, and 80 on a number line; 30 to 60 is 30; 30 to 80 is 50; 30/50 = 3/5)
(2) Therefore, 3/5 of the mixture should be the higher percentage solution.<br>
ANSWER: 3/5 of 725 mL, or 435 mL, of 80% alcohol solution; 2/5 of 725 mL, or 290 mL, of 30% alcohol solution