What is the formula to calculate Profit Percentage? Let cost price of an item be $C$, selling price be $S$. Assume the seller makes a profit. 
Then profit would be: $P = S - C$.
Now, what is the formula for calculating Profit Percentage?


*

*$P \% = \dfrac{P}{C} \times 100$

*$P\% = \dfrac{P}{S} \times 100$


Which one is right and why?
 A: They are both right.
The first expresses the profit as a percentage of the cost price. This is the profit mark-up.
The second expresses the profit as a percentage of the sales price. This is the profit margin.
Percentage profit on its own doesn't mean anything unless you are talking mark-up or margin. By convention, in a module I teach, percentage profit means profit mark-up.
A: The percentage profit $X$ is defined by
$$X = \left(\frac{\textrm{Amount of money you have at the end}}{\textrm{Amount of money you had at the start}} - 1\right) \times 100$$
Since you have $S$ at the end and $C$ at the start (because that's the money you needed to buy the item) then
$$X = \left( \frac{S}{C} - 1\right)\times 100 = \left(\frac{S-C}{C}\right)\times 100 = \frac{P}{C} \times 100$$

To see why your second decision has to be wrong, consider the case where you buy something for \$1 and sell it for \$1,001, so that $P$=1000. With your first definition,
$$X = 100\times \frac{1000}{1} = 100,000\%$$
which makes sense - you clearly made a huge profit, so you expect your percentage profit to be huge. With your second definition,
$$X = 100\times \frac{1000}{1001} = 99.9\%$$
which is nowhere near big enough.
A: $C : S = 100 : (100+x) \Rightarrow 100\cdot S=100 \cdot C +C \cdot x \Rightarrow x=\frac {100(S-C)}{C}\Rightarrow x= \frac{100\cdot P}{C}$
A: You have stumbled on one of the tools that journalists and politicians use to manipulate statistical evidence: selection of a divisor when calculating percentage change.  Using the starting value or the ending value as a divisor results in a difference between an increase or a decrease.  For example, if you buy something for 100 and sell it for 200, you have a profit of 100.  But if you buy for 200 and sell for 100, you have a loss of only 50%.   That is total nonsense that should be purged from public education.  A much better divisor would be the average of the starting and ending values; that would make the percent change the same in both directions.  
But what if something is bought by a distributor for 100 and sold to a retailer for $200.  Then the profit % is 100/150 = 67%.   Next, the retailer sells the product for 400.  His profit % is also 67%.   The sum of these two profits is 133%.   But the total profit is the increase from 100 to 400.  And the percent profit, using the average as a divisor is 300/250 = 120%.  Note the discrepancy!
The discrepancy can be reduced by using the geometric mean as a divisor, and can be totally eliminated by using the modified AGM from Borchardt's algorithm.  That leads to the fair and objective caluculation of %P as 100*Ln(S/C).  Try this with the extreme example given by Chris Taylor above, and you will get reasonable results!
I am under no illusion that such a practice could ever find its way into the educational establishment.  Use this for your own purposes, but when you do school work, always give the teacher the incorrect answer that he expects.
A: answer two is right because :
Suppose
Cost of your product is 5400
Selling price is 8000
then profit is 2600
According to 2nd formula P%=P/S×100
profit % = 2600/8000 X 100 = 32.5%
Let's check 
profit according to 32.5%
profit = Selling price X profit % /100
         8000 X 32.5% / 100 = 2600
hence formula 2nd is right
