A number is increased by R%. To get back to the original number, it is to be reduced to initial value by what %? A number is increased by R%. To get back to the original number, it is to be reduced to initial value by what %?
What I understand:
Increased value = $P+\frac{PR}{100}$
 A: Hint:
Basically, you want $$P(1+\frac{R}{100})(1-\frac{R_1}{100})= P$$ $$\implies 1-\frac{1}{(1+\frac{R}{100})} = \frac{R_1}{100}$$ $$\implies R_1 = \, ?$$
where $R_1$ is the required rate of reduction.
A: Let $P$ be the initial price.
The increased price $Q$ is $(1+\frac{R}{100})P$.
The equation you want to solve is $P = (1-\frac{x}{100})Q$, with unknown $x$.
That is $P = (1-\frac{x}{100})(1+\frac{R}{100})P$.
You can finish from here. 
A: We know that$$Q=P+P\frac{R}{100}=P\frac{100+R}{100}.$$
We want to have
$$P=Q-Q\frac S{100}=Q\frac{100-S}{100}.$$
Then
$$P=Q\frac{100-S}{100}=P\frac{100-S}{100}\frac{100+R}{100}$$ so that
$$(100-S)(100+R)=10000.$$
From this we draw
$$S=100-\frac{10000}{100+R}.$$
For example, if $R=5\%$, $S\approx4.76\%$.
A: 
A number is increased by $R\%$. To get back to the original number, it is to be reduced to the initial value by what $\%$?

Solution:
Let $N$ be the original number. The statement that it is increased by $R$ percent means that you're going to obtain the new number by doing the following: you're going to take the original number $N$, find out what portion of it constitutes $R$ percent and finally add that portion back to $N$. Mathematically, that process can be expressed like this $N+RN$*. So, we now have two numbers—our original number $N$ and the newly-obtained one $N+RN$.
The statement that the new number that we've got is to be reduced by what percent to get back to the original one means the following: take our new number $N+RN$, figure out what portion of it is taken up by $N$ and find the difference between $100\%$ (or $1$ in mathematical terms) and that portion. Mathematically, this is equivalent to saying $1-\frac{N}{N+RN}$. In other words, it's saying $100\%$ minus the percentage of $N$ in $N+RN$. And that's what we're trying to get. It's now all down to simple algebra to find the answer:
$$
1-\frac{N}{N+RN}=1-\frac{N}{N(1+R)}=1-\frac{1}{1+R}=\frac{1+R-1}{1+R}=\frac{R}{1+R}.
$$
Answer: $\frac{R}{1+R}$.

* Notice that it's simply $R$ and not $\frac{R}{100}$ or anything like that because the actual numerical value of $R$ is hiding in the variable. If I say, for example, that $R$ is equal to $10\%$, then $R$ is going to be the number $\frac{10}{100}=0.1$ (one tenth of a hundred or one tenth of a one). And if I say that I'm going to increase the number $12$ by $10\%$, I'm going to do the following: $12 + 0.1 * 12$ which equals $13.2$.
PS: If there's still something you don't understand, feel free to ask.
