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Arithmetic Aptitude :: Percentage

  1. Concept of Percentage:

    By a certain percent, we mean that many hundredths.

    Thus, x percent means x hundredths, written as x%.

     

    To express x% as a fraction: We have, x% = \( \frac { X } { 100} \)   .
     

     

        Thus, 20% =    \( \frac { 20 } { 100} \)\( \frac { 1 } { 5 } \)   .
       

     

    To express   as a percent: We have, \( \frac { a } { b } \)   = \( \frac { a } { b } \) x 100 %.
    \( \frac { a } { b } \)    

     

        Thus,   =   \( \frac { 1 } { 4 } \) 100 % = 25%.
     \( \frac { 1 } { 4 } \)  
  2. 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:

     

       \( \frac { R } { (R*100) } \)x 100 %
     

     

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

     

      x 100 %
    \( \frac { R } { (R-100) } \)

     

  3. 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 + \( \frac { R } { 100 } \) n
     

     

    2. Population n years ago =  
    1 + \( \frac { R } { 100 } \)   n
     
  4. 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- \( \frac { R } { 100 } \)   n
     

     

    2. Value of the machine n years ago =  

     

     1- \( \frac { R } { 100 } \)   n
     

     

    3. If A is R% more than B, then B is less than A by    \( \frac { R } { 100+R} \)x 100 %.
     

     

    4. If A is R% less than B, then B is more than A by    \( \frac { R } { 100-R} \)x 100 %.