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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}