How Do You Calculate Percentage Difference Between Two Numbers?

  • Thread starter Thread starter SimpleSoCal
  • Start date Start date
  • Tags Tags
    Work
AI Thread Summary
To calculate the percentage difference between two numbers, first find the ratio of the two values. For example, dividing 0.4088 by 1.235 yields approximately 0.331, indicating that 0.4088 is about 66.9% smaller than 1.235. Conversely, dividing 1.235 by 0.4088 shows that 1.235 is about 202.1% larger than 0.4088. The percentage difference can be expressed as a percentage of either number, depending on the context. Understanding these calculations helps clarify how much larger or smaller one value is compared to another.
SimpleSoCal
Messages
5
Reaction score
0
So my white collar job demands lots of number crunching as of lately. Now this is most likely simple, and I guess I should know this but... here it goes.

I just need to know how to calculate a percentage difference on the calculator.

For example. I take 1.46 * .28 which equals .4088. I then take 2.47 * .5 which obviously halves it to 1.235.

I know how to find percentages, like a simple X * whatever percent I want. What I'm asking is, what simple thing would I do on the calculator to get the percentage difference between .4088 and 1.235?

Thanks, hope I made sense!
 
Mathematics news on Phys.org
Hey SimpleSoCal.

If you want to find how much larger one thing is in terms of another then you need to look at the ratio of the two objects.

For example with your numbers, the ratio of the two numbers 0.4088/1.235 = 0.3310121 and the difference between the two is found by subtracting 1 from the answer which gives -0.6689879 or in other words 0.4088 is -66.89879% larger than 1.235.

Now if you reverse the order you get 1.235/0.4088 = 3.021037 and minus 1 gives 2.021037 or 1.235 is 202.1037% larger than 0.4088.

You can interpret negative numbers like -66.89879% to be 66.89879% smaller than the larger value (so if you talk about the positive % then you are talking about much smaller something is).

If two things are equal then the ratio of both is equal to 1 and 1-1 = 0 which is a 0% difference which is what we expect.
 
It depends what you want the difference as a percentage of. 1.235 - 0.4088 = 0.8262. If you want the difference as a percentage of 1.235, it is
100*\frac{0.8262}{1.235} = 66.9\%
If you want the difference as a percentage of 0.4088, it is
100*\frac{0.8262}{0.4088} = 202.1\%
And similarly to get the difference as a percentage of any other number.
 
Thanks, guys! That helps tons. I hope I made sense on my end. What I really meant to say was percentage increase I guess. Like how much more percent bigger is 1.235 than .4088
 
SimpleSoCal said:
Thanks, guys! That helps tons. I hope I made sense on my end. What I really meant to say was percentage increase I guess. Like how much more percent bigger is 1.235 than .4088

You'll want to use the formula I posted in that case.

Remember that an increase or decrease means you have to subtract 100% from the answer which just means subtracting 1.
 
Gotcha. Man, this place is great. More than I need but it's all useful. Thank God for the internet.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top