Question 1142452: A chemist has two alcohol solutions of different strengths, 30% alcohol and 45% alcohol solutions respectively. how many cubic cm of each must be used so as to make a mixture of 30cc that will contain 39% alcohol?
Found 3 solutions by Theo, greenestamps, josgarithmetic:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = amount of 30% solution.
y = amount of 45% solution.
x + y = 30
.3 * x + .45 * y = .39 * 30
in the first equation, solve for y to get y = 30 - x.
in the second euation, replace y with 30 - x to get .3 * x + .45 * (30 - x) = .39 * 30.
simplify to get .3 * x + 13.5 - .45 * x = 11.7.
combine like terms and subtract 13.5 from both sides of the equation to get -.15 * x = -1.8
solve for x to get x = -1.8 / -.15 = 12
y = 30 - x = 30 - 12 = 18.
you have x = 12 and y = 18.
x + y = 30
.3 * x + .45 * y = .3 * 12 + .45 * 18 = 3.6 + 8.1 = 11.7.
11.7 / 30 = .39
solution looks good.
solution is 12 cc of 30% solution and 18 cc of 45% solution are combined to make 30 cc of 39% solution.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The standard algebraic approach for mixture problems like this would be something like this:
x cc of 30% alcohol, plus (30-x) cc of 45% alcohol, yields 30 cc of 39% alcohol:
You can solve that using basic algebra.
That method gives good practice with algebra; you should understand why and how it works.
But mixture problems like this can be solved much faster informally, with only logical reasoning and basic arithmetic.
Here is the solution using this alternate method.
(1) 39% is 3/5 of the way from 30% to 45%. (Easy to determine if you are good with mental math; otherwise, picturing the three percentages on a number line might help.)
(2) That means 3/5 of the mixture must be the 45% alcohol.
So...
ANSWER: 3/5 of 30 cc = 18cc of 45% alcohol; the rest, 12cc, of 30% alcohol.
Answer by josgarithmetic(39617)  (Show Source):
|
|
|
