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PERCENTAGES CONCEPTS,FORMULAS,DEFINITIONS,EXAMPLES,TRICKS

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PERCENTAGES is very important chapter for all competitive exam and everything is included in it like all Problems on percentages, fractions and percentages, price change,result and population,results on depreciation,comparison of percentages,successive increase and decrease etc.






Percentages



The Percentage is a fraction whose denominator is always 100. The percentage sign is %.


Example - 


 7% can be converted to a fraction as 7/100 = 0.07


  If, we want to calculate P% of Q, then



   Percentage Formula:-  


   P% of Q =  Q× P/100



Qes.  If 30% of x = 60, then find the value of P.

Soln. 

    x × 30/100 = 60

⇒ x = 60 × 100/30

⇒ x = 200.


Fractions and Percentages


To express Y % as a fraction,  

 Y% = Y/100 



As -  30% = 30/100 = 3/10

         98% = 98/100 = 49/50



To Express X/Y as a percentage


we can say, 

X/Y = (X/Y × 100) %


As -   1/5 = (1/5 × 100) % = 20 %

         0.7 = 7/10 = (7/10 × 100)% = 70%


          20 % = 20/100 = 1/5 


      Where, 1 is percentage result and 

         5 is original value or number 





        Fraction          –         Percentage


            1/2                –                 50%


           1/3                 –                 33 (1/3)% 

        

           1/4                 –                 25%


           1/5                 –                 20% 


          1/6                  –                16 (2/3)%


          1/7                 –                14(2/7)%


          1/8                 –               12(1/2)%


          1/9                 –               11(1/9)%


         1/10                –                 10%


          1/11               –                9(1/11)%


          1/12               –                8(1/3)%


          1/20               –                       5%


          1/25                –                       4%


           2/3                 –                   66(2/3)%


            3/4                –                    75%


            3/7                –                 37(1/2)%


             1                  –                   100%

      



TRICK - 


31/6     =        5    +    1 /6  =   516(2/3)%

                      |              |

                    500%       16(2/3)%



                    


CONCEPT :  If the price of a object is increases by P%, then reduction in consumption so that the expenditure remains constant can be find by formula.


 =     [(P / 100 ± P) × 100] %


Note: +ve sign means increase and 

           -ve sign means decrease



Qes. If the price of wheat is increased by 10%, then by how much percent consumption should be reduced so that the expenditure remains the same?


Soln, 


Let the price be Rs. a /kg


Consumption be b kg


Therefore,


expenditure = price × consumption

⇒ Expenditure = ab


Price of wheat is increased by 10%


So, new price of sugar = 1.1a /kg


Let new consumption be c kg


So, new expenditure = (1.1a) × c


Now new expenditure = old expenditure


⇒ (1.1a) × c = a × b

⇒ c = b/1.1


  Reduction in consumption = 

   (b - c) = b – (b/1.1) = b/11


Hence, Percentage reduction in consumption 

 =  [(b/11)/b] × 100 = 100/11 = 9.09%




 Alternate method,


 By using formula,


   [(P/100+P)×100]%

  Given P = 10%

   [(10 / 100-10)×100]% = 9.09%



Qes. The price of rice has decreased by 25%. By what percentage can a person   increase the consumption so the there is no change in expenditure?

Soln,

  [(P/100-P)×100]%

 Given ,  P = 25%

  Put the value ,

  [(25/100-25)×100]% = 33.33%




  Population Concept 


 Let the population of a town be x 


 Now Population after t years  


 Population after t years if rate of increment r is same =  X( 1 + r/100)^2


Population after t years if rate of increment r is different =

 X( 1 + r1/100) + ( 1 + r2/100) ............




Qes.The population of a village 2 years ago was 255000. It increased by 20% in the first year and then increased by 30% in the second year. What is the current population of the village?



Soln. The population of a village 2 years ago was 255000


It increased by 20% in the first year


∴ The population after first year will be 

 120/100 × 255000 = 306000


The population then increased by 30% in the second year.


∴ The population after second year will be

  130/100 × 255000 = 331500



Population after t years 

Population t years ago, 

   P / (1+R/100)^t



Qes. The current population of a village is 25000. If the current population of the town  increases by 10% every year, what is the approximate population of the village 4 years ago? 

Soln. 

Current population = 25000


∴ Population of village 7 years ago = 


 25000 / (1+10/100)^4 = 17,123.28



In Depreciation


Let the present value of a object be X. Suppose it depreciates at the rate of r% per annum.


Value of the object after t years =

X(1-r/100)^t



Qes. A electric wire was bought at Rs. 5000. Its value depreciates at the rate of 7% per annum. Its value after one year will be.

Soln.

Actual price of the electric wire = Rs. 5000


⇒ Depreciation rate = 7% 


So, Value after 1 year = 5000 – 7% of 4100 = 5000 – 5000 × (7/100) =  4650 RS



Alternate ,


5000(1 - 7/100) = 4650 rs



Value of the object after t years ago,


      P / (1 - R/100)^t



 Percentages Comparison


X is R% more than Y


If  X is r% more than Y , then Y is less than X by , 


[ (r / 100+r)×100 ] %



 

Qes. If income of x is 60% more than the income of y, then what percentage of income of y is less then income of x?

Soln. 

Let the income of y be 100


∴ Income of x = 160


Y’s income is less than income of x by (160 - 100) = 60


Required percentage = (60/160)×100 

                                       = 37.5%




X is r% less then y ,


If x is r% less than y, then y is more than x by,


 [r / (100 - r)] %


Qes. If x is 60% less than y, then y is how much percentage more than x?

Soln.

X is 60% less then y 


Let y be 100


X = y - 60% of y = 100 - 60% of 100

                             =  40


Required percentage,


 [(100 - 40)/40]×100 =( X×Y)/100



Successive Increase and Decrease


Successive Increase 


If some value is increasing at A% and then B%, then combined be ,


 [A + B + (A×B)/100]%



Qes. The population of a city is 20000. It increases by 20% during first year and by 40% during the second year. The population after 2 years will be.

Soln.

A = +20%

B = +40%


Net increase = 20 + 40 + ( 20×40/100)

                        = 68%


Population after 2 years,

20000 + 20000×68/100 = 20000 + 13600

                                         = 33600



Successive Decrease


[A + B – A×B/100] %


Qes. A shop gives 2 successive discounts of 20% and 40%. What is the overall discount given?

Soln. 


Net discount = 20 + 40 – 20×40/100 = 52%






 



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