3-part alligation--how to do?

[deleted205891]

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The most impractical thing ever, but I need to know how to do it.

For example:

3 solutions have to be mixed: 95%, 60% and 5%, to get a 70% solution. How many parts of each solution are there?

Here's what I have set up:

95 (70-60)+(95-70)=30
70
60 95-70=25
50 x

Don't know how to get x. Says in the book that it's 25, but I have no idea how to get it. Does 95-70 just carry over to the bottom? Seems strange Would it be (50-(95-70))? That's the only thing I can think of, but the other alligation problem doesn't support that.

Other problem:

50% 5

20% 5
10%
5% 10+40=50

As you see: my previous logic didn't work: 50-(10-5) does not equal 5, the answer here.

 

[deleted205891]

Alright, I figured it out (though it took me an hour).

You fix the middle part to be the same as one of the other parts. Since there are 3 parts, you can manipulate one any way you please.

Would delete this post, but I don't see a delete option.

 

[JG9176]

I know you already figured out the answer to this but do you know how to do these using the criss-cross method? It makes it so easy. If not, let me know and I'll explain it or send you an example already worked out. Could be helpful for finals. [email protected]

 

[deleted205891]

I know you already figured out the answer to this but do you know how to do these using the criss-cross method? It makes it so easy. If not, let me know and I'll explain it or send you an example already worked out. Could be helpful for finals. [email protected]

Yeah that's the method I figured out.

Besides, the final's gonna be mult. choice, so I can work backwards from the answers.