Percentage problems are generally asked in all major exams like SSC, IBPS,CAT, GATE etc.These problems become complex if we do not have basic concept of problems.We will deal with concept of percentage problems and formulas are given at the end of this chapter to help students revise important formulas and concepts.
Concept Of Percentage
By certain percentage we mean many hundredths.It means X % means X hundredths.
There are two things which should be clear to us.

1.Conversion into percentage
 In order to convert some quantity(say X) into percentage.We simply multiply that quantity with 100.
For example:
Convert 5/2 into percentage.
We simply multiply it with 100=(5/20)*100=250%
 2.Conversion into Fraction from given percentage
 If we have given X %.In order to change it into fraction we simply divide it with hundred i.e X/100.
For example:
Convert 20% into fraction
20/100 = 1/5
In percentage, a term ‘OF’ is used.Like, 20% “OF” all students play cricket and rest play football. If there are 50 students in the class, 20% “OF” 50 students play cricket.
It means (20/100) * 50 =10 students play cricket.Replace “of” with a multiplication sign.
Intext Question: 
 Convert 0.6% into fraction
Solution Divide 0.6 with hundred i.e 0.6/100 = 6/1000
 Convert 30/20 into rate percent
SolutionMultiply fraction with 100 = (30/20) * 100 = 150%
 Evaluate 30% of 200 + 50% of 500
Solution (30/100) * 200 + (50/100) * 500 = 60 + 250 = 310
 What percent of 60 is 15
Solution Let that percent be X.
X percent of 60 = 15
X * (60/100) = 15
or, X = (15/60) * 100 =25%

Percentage Important Formulas 

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: %
If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:%

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 =
2. Population n years ago =

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 =
2. Value of the machine n years ago =
3. If A is R% more than B, then B is less than A %
4. If A is R% less than B, then B is more than A by %

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