The percentage concept has a major role in almost all the chapters of arithmetics.
So mastering this concept helps in solving questions of topics from sale, purchase, profit& loss, discount, interest, allegation and many more. Generally, the percentage is considered as per hundred or for every hundred.
The percentage is represented by the symbol %.
Calculate the percentage of any fraction by multiplying it with 100.
Some useful Formula in Percentage Concept
- When A is X% more than B then B is less than A by \begin{align*}
\left( \dfrac {x}{100+x}\times 100\right) \% \end{align*}
if A is X% less than that of B then B is more than A by \begin{align*}
\left( \dfrac {x}{100-x}\times 100\right) \% \end{align*} - When A is considered as X% of C and B is Y% of C then, \begin{align*}
A=\dfrac {x}{y}\times 100\% \ of \ B \end{align*} - When two numbers are X% and Y% respectively more than the third number then,
first number is
\begin{align*}\left( \dfrac {100+x}{100+y}\times 100\right) \% \ of \ second \ number \end{align*} and
second number is
\begin{align*}\left( \dfrac {100+y}{100+x}\times 100\right) \% \ of \ first \ number \end{align*}
similarly when the two numbers are X% and Y% respectively less than third number then,
first number is
\begin{align*}\left( \dfrac {100-x}{100-y}\times 100\right) \% \ of \ second \ number \end{align*} and
second number is
\begin{align*} \left( \dfrac {100-y}{100-x}\times 100\right) \% \ of \ first \ number \end{align*}
- If an amount is sucessively increased fist by X% and then reduced by X% then
the resultant is always reduced by
\begin{align*}\dfrac {x^{2}}{100}\% \end{align*}
Similarly, if a number is successively increased/decreased first by X% and then increased/decreased by Y% then
the net increase/decrease in percentage is
\begin{align*}\left( a+b+\dfrac {ab}{100}\right) \% \end{align*} here consider negative sign before X for decrease otherwise positive sign only
same here to put negative sign before Y for decrease otherwise positive sign - Let present population of a town is P and the population increases every year at a rate of r1, r2,r3,……rn up to n years then
population after n years is
\begin{align*}p\left( 1+\dfrac {r_{1}}{100}\right) \left( 1+\dfrac {r_{2}}{100}\right) \ldots \left( 1+\dfrac {r _{n}}{100}\right) \end{align*} put negative sign for decrease in population
when the population will increases at same rate r% for n years then
populaion after n years is
\begin{align*} p\left( 1+\dfrac {r}{100}\right) ^{n} \end{align*}
populaion n years ago is
\begin{align*} \dfrac {p}{\left( 1+\dfrac {r}{100}\right) ^{n}} \end{align*}
- If a number A is successively increased by X% followed by Y% and then by Z% then,
the final value of A is
\begin{align*} A\left( 1+\dfrac {x}{100}\right) \left( 1+\dfrac {y}{100}\right) \left( 1+\dfrac {z}{100}\right) \end{align*}
- consider a product whose price increases by P%, to maintain the same expenditure (without increase of expenditure) the reduction in consumption of that product is \begin{align*}
\left( \dfrac {p}{100+p}\times 100\right) \%
\end{align*}
if the value of the commodity decreases then, increase in consumption is \begin{align*}
\left( \dfrac {p}{100-p}\times 100\right) \%
\end{align*}