Monday, 27 February 2017

Important Formulas and Shortcuts to solve Percentage

Important Formulas and Shortcuts to solve Percentage

Today we will be covering a very important topic from the quantitative aptitude section that is - Percentage.These formulas and shortcuts  will be helpful for your upcoming Exams like Railway and SSC 2016. If you like it let us know. 
Percentage
  • Percentage is per‑cent which means parts per hundred.
 Percent sign
  • The percent sign is the symbol: %
  • It is written to the right side of the number: 50%
Percentage Definition
  • Percentage is a value that represents the proportion of one number to another number.
  • 1 percent represents 1/100 fraction.
 If we have to convert percentage into fraction than it is divide by 100.
Example 1:‑ if we write 45% then its equal to 45/100 or in fraction 9/20 or in decimal 0.45
If we have to convert fraction into percentage we have to multiple with 100.
Example 2:‑ if we write 3/5 in fraction it is equal to 60% =3/5x100=60.
Convert Percentage into Decimal:
  • 20% = 20/100 = 0.5
Convert Decimal Into Percentage:
  • 0.25 = (0.25 × 100) % = 25%
  • 1.50 = (1.50 × 100) % = 150%

Here is a table of commonly used values shown in Percent, Decimal and Fraction

picture 1234

Types of Formulas and Short Tricks 

Type 1: Percentage Increase/Decrease:
  • If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: [R/ (100 + R)] x 100%
  • If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: [R/ (100 - R)] x 100%
Type 2: Results on Population:
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
  1. Population after n years = P(1 + R/100)n
  2. Population n years ago =P/(1 + R/100)n
 Type 3: Results on Depreciation:
Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
  1. Value of the machine after n years = P(1 - R/100)n
  2. Value of the machine n years ago = P/[(1 - R/100)]n
  3. If A is R% more than B, then B is less than A by= [R/ (100 + R)] x 100%
  4. If A is R% less than B, then B is more than A by= [R/ (100 - R)] x 100%
Note: For two successive changes of x% and y%, net change = {x + y +xy/100}%

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