# Quant Handy Concepts and tricks: "Percentage"

**Percentage**

*The term per cent mean 'for every hundred'.*

*"A per cent is a fraction whose denominator is 100 and the numerator of the fraction is called the rate per cent." Per cent is denoted by the sign '%'*.**Concept to Caluculate Per cent**

*If we have to find y% of x, then*

*y% of x=(x*y)/100***Conversion of Per Cent into Fraction**

**Expression per cent (x%) into fraction.**

**Required fraction=x/100**

**Conversion of fraction into Percentage**

**Expressing a fraction (x/y) in per cent.**

**Required percentage=(x/y)*100)%**

**Expressing One Quanity as a Per Cent with Respect to Other**

**To express a quantity as a per cent with respect to other quantity following formula is used.**

**(The quantity to be expressed in per cent)/ (2nd quantity (in respect of which the per cent has to be obtained))X100%**

**Important Concept and Tricks**

**1. If x% of A is equal to y% of B, then**

**z% of A=(yz/x)% of B**

**2. When a number x is increased or decreased by y%, then the new number will be**

**(100+y)*x/100**

**3. When the value of an object is first changed (increased ) by a% and then changed (increased ) by b%,then**

**Net effect=[a+b +ab/100]%**

**4. Suppose in an examination, x% of total number of students failed in subject A and y% of total number of students failed in subject B and z% failed**

**in both the suject. Then,**

**(i) Percentage of students who passed in both the subjects=[100-(x+y-z)]%**

**(ii) Percentage of students who failed in either subject=(x+y-z)%**

**5. If due to r% decrease in the price of an item, a person can buy A kg more in Rs.x, then**

**Actual price of that item= Rs (rx)/((100-r)A) Per kg**

**Example :If due to 10% decrease in the price of sugar ,Ram can buy 5 kg more sugar in Rs 100 , then find the actual Price of sugar ?**

**solution : Here r = 10 % ,x = 100 and A = 5 kg**

**Actual price of sugar = 10*100/((100-10 )*5) = Rs. 2(2/9)**

**6.If the population of a town is P and it increases at the rate of R% per annum, then**

**(i) Population after n yr= P(1+r/100)^n**

**(ii) Population, n yr ago=P/(1+r/100)^n**

**7. If the present population of a city is P and there is a increment of R1%, R2% ,R3% in first, second and third year respectively, then**

**Population of city after 3 yr=P(1+R1/100)(1+R2/100)(1+R3/100)**

**Example : Population of a city in 20004 was 1000000. If in 2005 there is an increment of 15 % , in 2006 there is a decrement of 35 % and in 2007 there is an increment of 45 %, then find the population of city at the end of the year 2007.**

**solution : Required population = P (1 + R1/100)**

**(1 - R2/100)**

**(1 + R3/100)**

=

**P (1 + 15/100)****(1 - 35/100)****(1 + 45/100)****= 1083875**

*To do the Quizzes based on "Percentages", click on the below links -***1. Quiz 1**

**2. Quiz 2**

**3. Quiz 3**

**4. Quiz 4**

**To do the various Mix Quant Quizzes,**

**Click here**

**To do the Quizzes on D.I.,****Click here**

**To view the above post in Hindi: Click Here**

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