Scott wants to make 6 gal of a 50% sugar solution by mixing together a 60% sugar solution and pure water. How much of each solution must he use?

mixture 1 @ 60% = x gallons
pure water = 100% - x (??? that's my question...is that correct? or is it 0?)
mixture @ 50% = 6 gallons

60x + 100(6-x)= 50 * 6
60x + 600-100x = 300
600 - 40x = 300
-40x = -300
x = 7.5

I would do it this way:

Amount of pure water needed --- x gal
Amount of the 60% stuff ------ 6-x

.6(6-x) + 0x= .5(6)
3.6 - .6x = 3
-.6x = -.6
x = 1

so use 5 gal of the 60% plus 1 gal of pure water

your answer of 7.5 gal should have made no sense to you. You want to end up with only 6 gal, so how could you add 7.5 ????

Thank you. My answer of 7.5 didn't make sense. i was obviously overthinking...thinking pure would = 100%.

Again, thanks for your help.

In order to solve this problem, we need to set up a system of equations based on the given information.

Let's assume that Scott uses x gallons of the 60% sugar solution. In that case, he would use (6 - x) gallons of pure water because the total mixture is 6 gallons.

Now, we can determine the amount of sugar in each solution.

The 60% sugar solution contains 60% of x gallons of sugar, which is equal to: 60/100 * x = 0.6x gallons of sugar.

The pure water contains 0% sugar since it is just water.

The total amount of sugar in the final mixture is 50% of 6 gallons, which is equal to: 50/100 * 6 = 3 gallons.

Now we have our equation: 0.6x + 0 = 3.

Simplifying it, we get 0.6x = 3.

Dividing both sides of the equation by 0.6, we find x = 3/0.6, which is equal to 5.

So Scott needs to use 5 gallons of the 60% sugar solution and (6 - 5) = 1 gallon of pure water to make 6 gallons of a 50% sugar solution.