Mathematics10 years ago father was 12 times as old as his son and 10 years hence he will be twice as old as his son. Find their present ages.

10 years ago father was 12 times as old as his son and 10 years hence he will be twice as old as his son. Find their present ages.


  1. A
    15 years and 37 years
  2. B
    12 years and 34 years
  3. C
    10 years and 30 years
  4. D
    13 years and 40 years 

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    Solution:

    Given that 10 years ago father was 12 times as old as his son and after 10 years he will be two times as old as his son. We need to find the present age of father and son.
    Let x be the present age of the son and y be the present age of the father.
    10 years ago father was 12 times old as his son, therefore, we have,
    y10=12(x10) y10=12x120 12xy=12010 12xy=110.(1)  
    After 10 years father will be twice as old as his son, thus, we have,
    y+10=2(x+10) y+10=2x+20 2xy=20+10 2xy=10.(2)  
    Subtract equations (2) from (1), we have,
    12xy(2xy) 12xy2x+y 110(10) =110+10 10x =120 x =12  
    Substitute the value of x in equation (2), we have,
    2(12)y =10 24y =10 y =34  
    Thus, the present age of son is 12 years and the present age of father is 34 years.
    Therefore, option 2 is correct.
     
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