1.

The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son. Find their present ages.

Answer» Let, the present age of the son be x years.
Hence, the age of father `=2x^(2)`
8 years hence, the age of son =(x+8) years
Accoding to given statement
`2x^(2)+8=3(x+8)+4`
`implies2x^(2)+8=3x+24+4`
`implies2x^(2)-3x-20=0`
`implies2x^(2)-8x+5x-20=0`
`implies2x(x-4)+5(x-4)=0`
`implies(2x+5)(x-4)=0`
`impliesx=(-5)/(2)and x=4` `impliesx=4" "(because"age cannot be negative")`
Therefore, present age of son is 4 years and present age of father is `2xx4^(2)=32` years.


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