word problem with percentages

[Rutgers24]

Hi, I've been stumped on this question:

A chemist with two mixtures of a chemical and water, of 20% and 40% concentrations, wishes to mix the two so as to obtain 10 ounces of 32% concentrations. How much of each must he use?

I've been using X as amount of mix 1, and so X - 10 as mix 2. And using these variables I've been trying to make an equation, but am having no luck. The last equation I tried was: X/0.2 + X-10/0.4 = 10/.32

Help!?

 

[pka]

A chemist with two mixtures of a chemical and water, of 20% and 40% concentrations, wishes to mix the two so as to obtain 10 ounces of 32% concentrations. How much of each must he use?
I've been using X as amount of mix 1, and so X - 10 as mix 2. And using these variables I've been trying to make an equation, but am having no luck. The last equation I tried was: X/0.2 + X-10/0.4 = 10/.32

Staying with your notation, can you solve \(\displaystyle 0.2X+0.4(10-X)=10(0.32)~?\)

 

[JeffM]

Your book should have told you how percentage concentration is defined, but, if not, you should have looked it up. Concentration is measured as

[MATH]\dfrac{\text {mass of solute}}{\text {mass of solution}}.[/MATH]
So

[MATH]\dfrac{\text {desired mass of chemical in oz.}}{10 \text { oz.}} = 32\% = 0.32.[/MATH]
[MATH]\therefore \text {desired mass of chemical in oz.} = 3.2.[/MATH]
[MATH]\text {required quantity of 20% solution in oz.} = x.[/Math]
[MATH]\text {oz. of chemical supplied by x oz of 20% solution} = 0.2x.[/MATH]
[MATH]\text {required quantity of 40% solution in oz.} = (10 - x).[/MATH]
[MATH]\text {oz. of chemical supplied by (x - 10) oz. of 40% solution} = 0.4(10 - x) = 4 - 0.4 x.[/MATH]
If you name things and remember definitions, much will become clear. So for amount desired to equal amount supplied.

[MATH]3.2 = 0.2x + 0.4 - 0.4x \implies \text {WHAT?}[/MATH]
Definitions are critical in math.

 

[Rutgers24]

Got it thanks for your help.