Percentage Change Logic Question

[BossChloe]

Question regarding different percentage changes.

If I have a monthly rental fee of \$100 in July with a discount of 60%. I will pay \$40.

In the month of August the monthly rental fee is \$100 with a discount of 45%. I will pay \$55.

My discount has decreased by %15.

However my percentage increase of what I am paying is 27%. (\$55-\$40)/\$55 = 27%.

I don't quite understand how the percentage increase is 27% and not simply 15%. Could someone explain the logic?

 

[MarkFL]

It's because you are comparing 15 to 2 different numbers. 15 is 15% of 100, however 15 is roughly 27% of 55. :)

 

[Evgeny.Makarov]

First, let's recall some definitions. If a value equals $x_1$ at some point and $x_2$ at a later point, the actual change, denoted by $\Delta x$, is $x_2-x_1$. The relative change, which is often measured in percents and which I'll denote by $\delta x$, is $\dfrac{x_2-x_1}{x_1}=\dfrac{x_2}{x_1}-1$.

My discount has decreased by %15.

This phrase is ambiguous due to the fact that discount itself is measured in percents. Let $d_1=0.6$ and $d_2=0.45$; then indeed $\Delta d=-0.15$, but $\delta d=-0.25$.

However my percentage increase of what I am paying is 27%. (\$55-\$40)/\$55 = 27%.

As written above, the relative change is usually measured relative to the earlier value, so if the amounts payed are $p_1=40$ and $p_2=55$, then $\Delta p=15$ and $\delta p=0.375$.

I don't quite understand how the percentage increase is 27% and not simply 15%.

I don't think that $\delta p$ should equal $\delta d$. Let $t=100$ be the total rent. Then $p_i=t(1-d_i)$ for $i=1,2$. Therefore
\[
\Delta p=p_2-p_1=t(d_1-d_2)=t\Delta d
\]
but
\[
\delta p=\frac{p_2}{p_1}-1=\frac{1-d_2}{1-d_1}-1=\frac{d_1-d_2}{1-d_1}=\frac{\Delta d}{d_1-1}=\frac{\frac{d_2}{d_1}-1}{1-\frac{1}{d_1}}=\frac{\delta d}{1-\frac{1}{d_1}}.
\]