# Formulas Of Percentages Questions

## Percentages Formulas and Application

Go through the entire page to know important Formulas Of Percentages Questions and their applications.

In Aptitude , Percentage is the  number expressed as a fraction of 100. It is calculated with the help of ratio. The symbol for the Percentage is  “% ” . ### Formulas for Percentages & Definitions

• In mathematics, a percentage is a number or ratio expressed as a fraction whose denominator (bottom) is 100.
Thus, x percent means x hundredths, written as x%.
We express x% as a fraction as $\frac{x}{100}$
• For example 10% = $\frac{10}{100}$ = $\frac{1}{10}$

### Formulas & Basic Concept of Percentages

• To calculate a % of b = $\mathbf{\frac{a}{100} × b }$
• To find what percentage of a is b =  $\mathbf{\frac{b}{a} × 100}$
• To calculate percentage change in value
Percentage change = $\mathbf{\frac{Change}{Initial Value }× 100 }$
• Percentage Increase or Decrease
• Percentage increase =  $\mathbf{\frac{R}{100 + R } × 100 }$ %
• Percentage decrease = $\mathbf{\frac{R}{100 – R } × 100 }$ %
• Successive Percentage Change
If there are successive percentage increases of a % and b%, the effective percentage increase is: $\mathbf{ a + b + \frac{ab}{100} }$ %
• Succesive Discount :

If there are successive discount of a % and b%, the effective discount is: $\mathbf{ a + b – \frac{ab}{100} }$ %

• Results on population
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
• Population after n years = $\mathbf{P (1 +\frac{R}{100})^n}$
• Population n years ago = $\mathbf{\frac{P}{(1 +\frac{R}{100})^n}}$
• Results on Depreciation:
Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
• Value of the machine after n years = $\mathbf{P (1 – \frac{R}{100})^n}$
• Value of the machine n years ago = $\mathbf{\frac{P}{(1 -\frac{R}{100})^n}}$
• If A is R% more than B, then B is less than A by $\mathbf{\frac{R}{100 + R } × 100 }$ %
• If A is R% less than B, then B is more than A by$\mathbf{\frac{R}{100 – R } × 100 }$ %
• Increase N by S% : $N ( 1 + \frac{S}{100})$.
• Decrease N by S% : $N ( 1 – \frac{S}{100})$

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