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 Of Percentages Questions

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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