# Interesting percentage calculation shortcut formulae

I came across the following shortcut formulae while going through a text book:

For questions of the type:
If the price of a commodity increases by R%,
find the % decrease in the consumption given that
the expenditure remains same


One can directly find the value by applying: Decrease = 100R/(100 + R)

e.g. If the price of potato is increased by 20%,
by how much should the consumption be decreased
so as to have no change in the expenditure.
Decrease = 100*20/(100 + 20) = 50/3 = 16.67%


Can someone please help me know why the above shortcut formulae works for such type of problems?

-
There's nothing related to [logic] in the question. I'm not sure about the current tag, however it seems better fitting. I also edited the quotes to eliminate the scrollbars. – Asaf Karagila Jun 12 '11 at 15:15
There is even more shortcut approximate formula that tells that the consumption should be decreased by 20%. :) See my answer math.stackexchange.com/questions/27202/what-does-x-faster-mean/… . – beroal Jun 12 '11 at 16:48

First of all, percentages are just another way of talking about hundredths. So whenever you see a percent sign, you might as well imagine it says "/100" instead.

In the given type of problem, the expenditure $x$ is given by the consumption $c$ times the price $p$ (per unit of consumption), i.e. $x=cp$. Now if $p$ increases by $R\%$ of $p$, that is, by $(R/100)p$, then $c$ needs to change to $c'$ such that $x$ remains the same, i.e. $c'(p+(R/100)p)=cp$. Solving for $c'$ yields

$$c'=\frac{p}{p+(R/100)p}c=\frac{1}{1+R/100}c=\frac{100}{100+R}c\;.$$

Then the relative change in consumption is

$$\frac{c'-c}{c}=\frac{\frac{100}{100+R}c-c}{c}=\frac{100}{100+R}-1=-\frac{R}{100+R}\;.$$

Since the relative change is the percentage change over $100$, the percentage change is $100$ times this, i.e. $-100R/(100+R)$.

-
Thanks a lot @joriki for explaining this so neatly! – peakit Jun 12 '11 at 18:11