Why does flipping the order of numbers in a basic percent problem work?

[starrynight108]

If I am looking for 88% 50, I can put the percent sign on the other number and flip the order of the numbers:

88% of 50 = 50% of 88 = 44.

I can solve this numerically but I don't understand conceptually why this works. Why does this work?

 

[rootone]

Conceptually it's the same thing as (2 x 3) = (3 x 2)
Percentages are ratios, but are equivalent to a real number and can be treated in exactly the same way.

 

[billy_joule]

This may help:

0.88*50 = (10/10) * 0.88*50 = (10*0.88) * (50/10) = 88 * 0.50

 

[Mark44]

This may help:

0.88*50 = (10/10) * 0.88*50 = (10*0.88) * (50/10) = 88 * 0.50

Let me fix that for you ...
0.88*50 = (100/100) * 0.88*50 = (100*0.88) * (50/100) = 88 * 0.50

 

[starrynight108]

Thank you everyone. Mark44, thanks for clearing that up! This does make sense. It's basically the Commutative Property of Multiplication.

 

[Mark44]

Thank you everyone. Mark44, thanks for clearing that up! This does make sense. It's basically the Commutative Property of Multiplication.

It's a bit more than that. Your question was why 88% of 50 is the same as 50% of 88. Writing the percent figures as decimal fractions, we have
$.88 * 50 = \frac{88}{100} \cdot 50 = 88 \cdot \frac{50}{100} = \frac{50}{100} \cdot 88$
The latter expression is the same as 50% of 88.

In the two middle expressions in my equation, I am using the idea that $\frac a b \cdot c$ is equal to $a \cdot \frac c b$. IOW, it doesn't matter which fraction has the denominator. $\frac a b \cdot c = a \cdot \frac 1 b \cdot c$ and I can group the 1/b factor with either the first number or the last.

 

[billy_joule]

Let me fix that for you ...

Oops, thanks!

 

[starrynight108]

So, they are equivalent expressions like was mentioned before. The form a/b * c = a * c/b clears it up. Thank you again.