Question 244989: One antifreeze solution is 10% alcohol. Another antifreeze solution is 18% alcohol. How many liters of each antifreeze solution should be combined to create 20 liters of antifreeze solution that is 15% alcohol?
Found 2 solutions by Theo, richwmiller:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = liters of first solution.
y = liters of second solution.
formula for amount of alcohol required in the new solution is:
.10 * x + .18 *y = .15 * 20
formula for total amount of solution required is:
x + y = 20
you have 2 equations that have to be solved simultaneously.
since x + y = 20, this means that y = 20-x
substitute in the other equation and solve for x.
.10 * x + .18 *y = .15 * 20 becomes:
.10 * x + .18 * (20-x) = .15 * 20
simplify to get:
.10 * x + .18 * 20 - .18 * x = .15 * 20
simplify further to get:
.10 * x + 3.6 - .18 * x = 3
combine like terms to get:
-.08 * x + 3.6 = 3
subtract 3.6 from both sides of the equation to get:
-.08 * x = -.6
divide both sides of this equation by -.08 to get
x = -.6 / -.08 = 7.5
if x = 7.5, then y = 20-7.5 = 12.5
we have x = 7.5 litres and y = 12.5 liters
substitute in original solution equation to get:
.10 * x + .18 *y = .15 * 20 becomes:
.10 * 7.5 + .18 * 12.5 = .15 * 20
simplify to get:
.75 + 2.25 = 3 which is true so the answer is confirmed to be good.
.1 * 7.5 liters of solution 1 + .18 * 12.5 liters of solution 2 = .15 * 20 liters.
you need 7.5 liters of solution 1 and 12.5 liters of solution 2 to make a total of 20 liters.
The percent alcohol will be .1 * 7.5 + .18 * 12.5 = 3 liters of alcohol.
3 liters / 20 liters = 15% of the new solution.
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! here is another problem all worked out for you.
For your problem just change the numbers
This time, suppose you work in a lab. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?
Let x stand for the number of liters of 10% solution, and let y stand for the number of liters of 30% solution. (The labeling of variables is, in this case, very important, because "x" and "y" are not at all suggestive of what they stand for. If we don't label, we won't be able to interpret our answer in the end.) For mixture problems, it is often very helpful to do a grid:
liters sol'n percent acid total liters acid
10% sol'n x 0.10 0.10x
30% sol'n y 0.30 0.30y
mixture x + y = 10 0.15 (0.15)(10) = 1.5
Since x + y = 10, then x = 10 – y. Using this, we can substitute for x in our grid, and eliminate one of the variables: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
liters sol'n percent acid liters acid
10% sol'n 10 – y 0.10 0.10(10 – y)
30% sol'n y 0.30 0.30y
mixture x + y = 10 0.15 (0.15)(10) = 1.5
When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution. Then:
0.10(10 – y) + 0.30y = 1.5
1 – 0.10y + 0.30y = 1.5
1 + 0.20y = 1.5
0.20y = 0.5
y = 0.5/0.20 = 2.5
Then we need 2.5 liters of the 30% solution, and x = 10 – y = 10 – 2.5 = 7.5 liters of the 10% solution. (If you think about it, this makes sense. Fifteen percent is closer to 10% than to 30%, so we ought to need more 10% solution in our mix.)
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