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Aptitude Day 28

Problem on Ages

 1. The sum of the 5 children born at the intervals of 3 years each is 50 years. What's the age of the youngest child?

Solution:
                Let the ages of the 5 children be x,(x+3),(x+6),(x+9) and (x+12)
                x+(x+3)+(x+6)+(x+9)+(x+12)= 50  [Given]
                5x+30=50
                5x=20
                 x=4
                 The age of the youngest child is x = 4 years.


 2. A is aged 3 times more than B. After 8 years, A would be 2 and half times of B's age. After further 8 years, how many times he would be of B's age?

Solution:
               Let the age of B is x and A's age will becomex+3x = 4x.         
               After 8 years, (4x+8) = 5/2(x+8)           [Note: 2 and half becomes 5/2]
                                       2(4x+8) = 5(x+8)
                                         8x+16 = 5x+40
                                                3x = 24
                                                  x =8.
              (4x+8+8) / (x+8+8) = (4x+16) /(x+16) = 48/24 = 2 times.
3. The difference between the ages of 2 persons is 10 years. 15 years ago, the elder one was twice as old as the young one. The present age of the elder person is:

 Solution:
               Let the ages of 2 persons be x and (x+10) respectively.
               15 years ago,
                                 (x+10)-15 = 2(x-15)
                                        x-5 = 2x-30
                                            x = 25
               The present age of the elder one is x+10 = 25+10 = 35 years.

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