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### Formulas for calculating Percentage - Aptitude Questions and Answers.

TIPS FOR SOLVING QUESTIONS RELATED TO PERCENTAGE:

Percentage: A fraction whose denominator is 100 is called percentage. The numerator of the fraction is called the rate percent.

1. To express x% as a fraction:
We have, x% = \begin{aligned} \frac{x}{100} \\
Thus, 30\% = \frac{30}{100} = \frac{3}{10} \\
\end{aligned}

2. To express fraction as percentage, we have
\begin{aligned}
\frac{a}{b} = \left(\frac{a}{b}\times100\right)\%
\end{aligned}

3. If A is R% more than B, then B is less than A by
\begin{aligned}
\left[ \frac{R}{(100+R)}\times 100 \right]\%
\end{aligned}

4. If A is R% less than B, then B is more than A by
\begin{aligned}
\left[ \frac{R}{(100-R)}\times 100 \right]\%
\end{aligned}

5. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:
\begin{aligned}
\left[ \frac{R}{(100+R)}\times 100 \right]\%
\end{aligned}

6. If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:
\begin{aligned}
\left[ \frac{R}{(100-R)}\times 100 \right]\%
\end{aligned}

7. Let the population of a town be P now and suppose it increases at the rate of R% per annum, then
\begin{aligned}
1. & \text{Population after n years = }P\left(1+\frac{R}{100}\right)^n \\

2.& \text{Population before n years =} \frac{P}{\left(1+\frac{R}{100}\right)^n} \\
\end{aligned}

8. Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum.

1. Value of the machine after n years =
\begin{aligned} P\left(1-\frac{R}{100}\right)^n \end{aligned}
2. Value of the machine n years ago =
\begin{aligned} \frac{P}{\left(1-\frac{R}{100}\right)^n} \\
\end{aligned}

9. For two successive changes of x% and y%, net change =
\begin{aligned}
\left(x +y + \frac{xy}{100}\right)\%\\
\end{aligned}