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 March 1st, 2016, 09:06 PM #1 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Percentage Transformation I discovered ages ago that any percentage value has a counterpart. Let us say we have a number family, X, Y, and Z. Z is what percentage Y is of X. The thing with this number family, is that while knowing any two, the third is easy discern, X and Y can not be swapped. Well, it is quite often that I might know Z and one of the others, I might also want to know Z if Y and X were reversed. Basically, if Y is 125% of X, then X is what percentage of Y? If you know X or Y, you can find the other, flip them, then find Z again. But I have found the relationship between the two percentages. So now, there is a number family of four, X, Y, Z, and A. Z is the percentage Y is of X, and A is the percentage X is of Y. Z = 1- (A-1/A) No need for a reversal as you can swap A and Z and it still works (which makes sense of course). So what do you all think of this? Neat right? March 3rd, 2016, 12:06 AM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Your formula doesn't make a lot of sense. Say $A = 80%$. Then $Z = 1 - 80 + \dfrac{1}{80} < 0$. How does that work? March 4th, 2016, 06:20 AM #3 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 How exactly did you derive that from my formula? What it should be, Z=1-(A-1/A) Z=1-(80-1/80) z=1-(79/80) z=1-(.9875) z=.0125 If you get something else, then a mistake was made somewhere. March 4th, 2016, 06:27 AM #4 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 Hmm, I see two mistakes, one mine one yours. I normally do programming math, so I forgot that division would be done before subtraction outside of programming. Your mistake was adding rather than subtracting, but could have been a typo. March 4th, 2016, 06:00 PM   #5
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Quote:
 Originally Posted by MystMage I discovered ages ago that any percentage value has a counterpart. Let us say we have a number family, X, Y, and Z. Z is what percentage Y is of X. The thing with this number family, is that while knowing any two, the third is easy discern, X and Y can not be swapped.
That is true of any "non-commutative" binary operation.

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 Well, it is quite often that I might know Z and one of the others, I might also want to know Z if Y and X were reversed. Basically, if Y is 125% of X, then X is what percentage of Y?
You have swapped Y and Z here! If Y is 125% of X Then Y= 1.25X. To find solve for X, divide by 1.25= 5/4. X= Y/1.25= (4/5)Y= 0.80Y. X is 80% of X.

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 If you know X or Y, you can find the other, flip them, then find Z again. But I have found the relationship between the two percentages.
Yes, if Y is a% of X then then X is 1/a % of Y.

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 So now, there is a number family of four, X, Y, Z, and A. Z is the percentage Y is of X, and A is the percentage X is of Y. Z = 1- (A-1/A)
I'm not quite sure what you mean here. Are you saying then that Y is 50% of X then X is 1- (.5- 1/.5)= 1- .5+ 2= 3- .5= 2.5 or 250% of X? That is clearly not true. If Y is 50% of X then Y= 0.50X so X= (1/.50)Y= 2.00Y or X is 200% of Y.

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 No need for a reversal as you can swap A and Z and it still works (which makes sense of course). So what do you all think of this? Neat right?
If you mean what I suggested above, it is not correct. If you do not mean that what do you mean by "Z= 1- (A-1/A)"? If Y is 30% of X, what is your "A"? What is "Z"?

Last edited by Country Boy; March 4th, 2016 at 06:02 PM. March 4th, 2016, 07:01 PM #6 Senior Member   Joined: Nov 2015 From: USA Posts: 107 Thanks: 6 The error I refered to I guess wasn't clear, I didn't include a pair parenthesis, so try this, Z = 1- ((A-1)/A) Of course, you doing Y=ZX works, if you know two of those values. The equation I came up with is good if you don't know X or Y, just Z but want to see the percentage in the reverse direction. Tags percentage, transformation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jmbyrd Elementary Math 22 May 5th, 2015 05:09 PM BahamutZaero Algebra 7 March 19th, 2014 04:09 PM dawgphysics Elementary Math 0 March 6th, 2014 01:28 PM Rfeynman16 Algebra 5 January 27th, 2014 10:31 PM maths_illiterate Elementary Math 1 March 28th, 2011 10:21 PM

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