SOLUTION:
1.Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 725 mL of a solution that is 60% alcohol. How much (in mL) of each solution should you use?
30
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1.Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 725 mL of a solution that is 60% alcohol. How much (in mL) of each solution should you use?
30
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Question 1145668:
1.Mix a solution that is 30% alcohol with a solution that is 80% alcohol to make 725 mL of a solution that is 60% alcohol. How much (in mL) of each solution should you use?
30% alcohol solution__ml
80% alcohol solution__ml.
2.A welder has 15 pieces of 4 ft steel angle and 18 pieces of 2 ft steel angle. What is the ratio of pieces of 4 ft steel angle to 2 ft steel angle? (Enter your answer in fraction form.) Found 3 solutions by solver91311, josgarithmetic, greenestamps:Answer by solver91311(24713) (Show Source):
First a standard algebraic solution to the mixture problem....
x = mL of 30% alcohol solution
725-x = mL of 80% alcohol solution
.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
ANSWER: 290 mL of the 30% solution and 725-290 = 435 mL of the 80% solution.
That's a perfectly good method for solving mixture problems like this.
But there is a much easier way....
Now a much easier solution, based on where the 60% lies between the 30% and 80%....
(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.
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