ADF – Interview Aptitude Questions – Payilagam – Part – V

Here in this blog we are discussing about ADF – Interview Aptitude Questions – Payilagam – Part – V . Some of them are :

1. The profit earned by selling an article for Rs. 832 equal to the loss incurred when            the same article is sold for Rs. 448 .  What should be the sale price for making a         profit of 50%  ?

  a) 920      b) 960          c) 1060      d) 1200

        SOLUTION :

       Let the cost price of that article be Rs. X then the

          Selling price – Cost price =   Cost price – Loss

  •                            832   –   x       =        x   –   448   
  •                                      =>   2x  =        832 + 448            =  Rs.1280
  •                        X = cost price = Rs.640
  •                      Now the profit = 50% 0f Rs. 640 = 320 
  •   Thus the new cost price = Rs. 640 + 320 = Rs.960

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  2.A train of 108 m long moving at the speed of 50 kmph crosses a train 112 m long  coming from the opposite direction in 6 seconds. The speed of the second train is ?

  a)48 kmph                 b)54 kmph                       c)66 kmph                  d)82 kmph  

      SOLUTION:

      Total distance = 108 + 112 m =  220 m

      Speed of the first train = 50 kmph and the second train = x kmph

           Relative speed = x + 50 kmph = (x + 50) * 5 / 18   m/sec

       Total time taken = 6 seconds

          (X + 50 ) 5 /18   =   220 / 6

                    (X + 50 ) 5 = 3 * 220     

                            X +  50 =      132  =>    x = 82 kmph

          Speed of the second train = 82 kmph

     

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3 . In an examination it is required to get 36 % of maximum marks to pass. A student got 113 marks and failed by 85 marks. The maximum marks are

        a) 500 b)        550                 c)       640                d) 1008

    SOLUTION :

      From the given data we get the maximum marks to pass =  113 + 85 = 198 marks

          36 % of X marks = 198 marks

        100 % of X marks = (198 / 36 )   * 100 

                                            =  11 *50 =  550 marks

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4. If the diameter of the cylindrical jar is increased by 25 % by what percent must the height be decreased so that the volume remains unaltered ?

   a) 10                   b) 20                 c) 25               d) 36

    SOLUTION :

            Let the diameter of a cylinder =  8 cm    => radius = r1 = 4 cm

              Increase in 25 % of diameter = 8cm + 2 cm = 10 cm

                                                          => New radius =   r2 = 5 cm

                      If the height of the cylinder = h cm

                                          Volume = π r12h1 = π r22 h2

                                                                   16 h1 = 25 h2

                                         Change in height =  h –  (16 / 25)  h   = 9 /25 h

               Percentage in change in height =( 9 /25 )  * 100    = 9 * 4

                                                                             =   36 %

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 5.The number of ordered pairs of intergers ( m,n) such that the product ‘mn’ is non negative and m3 + n3 + 99 mn = 333

  a) 30 b)    32                      c)   33                         d) 34

ANSWER :  33 because the product ( 99 )is a multiple of 33

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6. A father is 30 years older than his son however he will be only thrice as old as the son after 5 years, what is the father’s present age ?

    a) 40 years b) 30 years                       c) 50 years           d) None of these

      SOLUTION :

             Let the son’s present age =   x years  

   Then the father’s present age =   ( x + 30) years

             After 5 years ,  Son’s age =    x  + 5    and father’s age =  x + 35

                       Given ,     ( x  +  35 ) =  3 ( x  + 5 ) 

  •                                       X + 35 = 3 X + 15
  •                                              2X =   20 years      =>     x = 10 years.
  •            
  •       Thus the father’s present age = 40 years

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