The fathers age one year back is 4 times the age of son, after 6 years the father’s age exceeds twice the age of the son by 9. Wha

1 thought on “The fathers age one year back is 4 times the age of son, after 6 years the father’s age exceeds twice the age of the son by 9. Wha”

1. Question:–

   • The fathers age one year back is 4 times the age of son, after 6 years the father’s age exceeds twice the age of the son by 9. What is fathers age?

   To Find:-

   • Find the fathers age.

   Given:–

   • The fathers age one year back is 4 times the age of son, after 6 years the father’s age exceeds twice the age of the son by 9.

   Solution:–

   Let the present of father be x years

   Present age of son be y years

   According to Question:-

   [tex]\tt\implies \: x – 1 = 4( y – 1 ) [/ex]

   [tex]\tt\implies \: x – 4y = – 3 . . . . ( i ) [/tex]

   Now ,

   [tex]\tt\implies \: x + 6 = 2( y + 6 ) + 9 [/tex]

   [tex]\tt\implies \: x – 2y = 15 . . . . ( ii ) [/tex]

   Now ,

   We have to Subtract ( ii ) from ( i )

   [tex]\tt\implies \: x – 4y – ( x – 2y ) = – 3 – 15 [/tex]

   [tex]\tt\implies \: x – 4y – x + 2y = – 18 [/tex]

   [tex]\tt\implies \: – 2y = – 18 [/tex]

   [tex]\tt\implies \: x = \cancel\dfrac { – 18 } { – 2 } [/tex]

   Substitute in equation ( i ) to get x value:-

   [tex]\tt\implies \: x – 4y = – 3 . . . . ( i ) [/tex]

   [tex]\tt\implies \: x = 33 [/tex]

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