00:01
Problem 72 reads how much water should be added to 20 ounces of a 15 % solution of alcohol to dilute it to a 10 % alcohol solution.
00:08
So to start this problem, i made a chart that has a solution, the water, the mix, and then i'm using the equation amount time strength is equal to the amount of alcohol.
00:18
So i'm going to start by listing out the information for the solution.
00:22
So we know that there's 20 ounces of solution and that it is a 15 % alcohol strength.
00:28
So as a decimal, that's 0 .15.
00:34
We don't know how much water needs to be mixed with the solution, and water is 0 % alcohol.
00:42
We know that the mix is 10 % alcohol, so as a decimal, that's 0 .10.
00:47
And then to figure out what the amount is for the mix, it's going to be 20 plus x.
00:55
To figure out the amounts of alcohol in each liquid, i'm going to multiply the amount times of strength.
01:00
So for the solution, 20 times 0 .15 is 3.
01:04
For the water x times zero is zero and then for the mix it's going to be 0 .10 times 20 plus x so from here i'm going to write an equation that combines the amounts of alcohol for the solution and water and makes it equal to the amount of alcohol in the mix so we have three and i'm not going to write the plus 0 just to simplify the equations.
01:37
We have 3 is equal to 0 .10 times 20 plus x.
01:48
So from here, i want to get rid of this decimal.
01:52
So i'm going to multiply both sides of the equation by 100.
02:04
You could also distribute the 0 .10 to the 20 and the x and then afterwards multiplied by 100, but i'm just choosing to do it before.
02:12
So 100 times 3 is 300.
02:19
And then 0 .10 times 100...