MATHS PERCENT

                                                       PERCENTAGE

Percent means some part of its hundred parts.It is always represented as %.
 ie. 

   If we say 35% 
   meaning that 35/100
   means 35 part out of 100.
   
                                SOME STRATEGY FOR PERCENTAGE PROBLEMS

1.To Express x% in form of numerator and denominator 
            x%=x/100

   e.x   20%=20/100=1/5

2.    To express number X as percentage= In This case simply multiply given no to 100.

                    i.e.   x%=x*100

3. These are two formulas which are important enough

 a). [ (S/100+S)*100 ]  %

  b).[ (S/100-S)*100 ]

                these two formula are useful in following case.

         ** If Price of things are increase by S% , so that reduce in consumption  as not       increase  expenditure . In this case formula a) will be applied.[TRCICK TO REMEMBER    FORMULAS IS: increase S% then denominator must be (100+S).

         ** If Price of things are decrease by S% , so that increase in consumption  as not                   decrease  expenditure . In this case formula b) will be applied.[TRCICK TO  REMEMBER             FORMULAS IS: decrease S% then denominator must be (100-S).

          

         ** If A is S% more than B then B is less than A by is represented by formula a).

         ** If A is S% less then B then B is more than A by is represented by formula b).

4.POPULATION:  Any  country /state/village/town  have some population .so examiner want to check your calculation and thinking power.

               In this type of question Population (P) and Rate of increase or decrease (R) is given

a)       Population after/before n year=[ P(1 + R/100)ⁿ ]                

                    Here +R for Rate for increase in population.

                             -R for Rate for decrease in population.

                                         OR

      

                Population after/before n year=[ P(1+R/100)ⁿ ]                

  Put + for Rate if increase . Put - for Rate if Decrease

              

         b)    Population  n year ago=  [P/(1 +  R/100)ⁿ ]

                                                    Here +R for Rate for increase in population

                                                   -R for Rate for decrease in population.

                                    

                                   OR

      

                Population  n year ago=  [P/(1 +  R/100)ⁿ ]

  Put + for Rate if increase . Put - for Rate if Decrease.

5.DEPRECIATION: Everything(Machine/person) will be depreciate(decrease) time to time.

                               In this type of question Machine present value (P) and Rate of depreciation (R) is  given

 a)          Machine value after n year=[ P*(1 - R/100)ⁿ ]

         b)          Machine value before n year=[P/(1 - R/100)ⁿ ]

      

6. IF SOME TWO THING INCREASE/DECREASE  BY  A% AND B% respectively then net % change =A+ B + (XY/100)

                                                Put + for Rate if increase . Put - for Rate if Decrease.

                                                    END OF THEORY

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                                                                                                QUESTION

QUESTION 1.  IF THE INCOME OF MOHAN IS 20% MORE THAN OF THE RADHEMOHAN INCOME.HOW MUCH %  RADHEMOHAN INCOME IS LESS THAN THAT OF MAOHAN INCOME.

Solution . by using direct formula…

       [ (S/100+S)*100 ]  %

   Let  S=20

Less %= 20/120*100  %

             =50/3 %

QUESTION 2. IF MOHAN SPENDS  10%  FOR PETROL, 20%  FOR FOOD , 15 % FOR ROOM RENT, 20 % FOR  LIC DEPOSIT , 5% EDUCATION  OF HIS IMCOME. AFTER ALL THESE EXPENDITURE MOHAN SAVE RS. 9000 .THEN FIND THEIR  AMOUNT SPEND ON PETROL.

Solution .  Total expense =(10+20+15+20+5)%=70%

                   Since we know that

                             Saving =income – expense.

     Letus suppose if income 100  the expense 70. Therefore total saving 30. 

Given that he saves 9000 . Now

 30 is equal to == 9000

10 will equal to ==9000*10/30

                             ==3000 Rs.