qbguy said:
i have 8 problems like this and i cant get started
i am not sure how to translate this problem:
part 1
10 gallons of 20% alcohol solution must be achieved by mixing some 18% solution
with 30% solution. How many gallons of each must be mixed.
part 2
in part 1, the other ingredient in each solution is water. do part 1 again, this time
centering the model around the amount of water in each part.
(hint, 20% alcohol means 80 % water)
thanks for any light you can provide for me
Ok...let's start by defining variable expressions to use in the problem
Part 1
Let x = number of gallons of 18% solution
Then, since there must be a total of 10 gallons,
let 10 - x = number of gallons of 30% solution
If we have x gallons of a solution that is 18% alcohol, then the AMOUNT of alcohol is 18% of x, or 0.18x.
If we have (10 - x) gallons of a solution that is 30% alcohol, then the AMOUNT of alcohol is 30% of (10 - x), or 0.30(10 - x)
At the end, we have 10 gallons of a 20% solution. The amount of alcohol in 10 gallons of 20% solution is 20% of 10 gallons, or 0.20(10)
alcohol in 18% solution + alcohol in 30% solution = alcohol in final mixture
0.18x + .30(10 - x) = .20(10)
Solve that for x, and you'll have the number of gallons of 18% solution...and (10 - x) will be the number of gallons of 30% solution.
Now...suppose you want to deal with the amount of WATER rather than the amount of alcohol. Will it make a difference in your answer?
Part 2
Let x = number of gallons of 18% alcohol solution. If 18% of the solution is alcohol, then 100% - 18%, or 82% of the solution is water. In x gallons of solution which is 82% water, the AMOUNT of water is 82% of x, or 0.82x
As above, if we want to end up with 10 gallons, and we use x gallons of that 18% alcohol solution, then we need (10 - x) gallons of 30% solution. If 30% is alcohol, then 100% - 30%, or 70% of the solution is water. (10 - x) gallons of a solution that is 70% water contains 0.70(10 - x) gallons of water.
And at the end, we're to have 10 gallons of a 20% alcohol solution. If 20% of the solution is alcohol, then 100% - 20%, or 80% of that 10 gallons is water.
water in 18% solution + water in 30% solution = water in final mixture
.82x + .70(10 - x) = .80(10)
Solve THAT equation...you should get the same answer for x that you got in the first instance.
