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The Age of a Father is Twice the Square of the Age of His Son. Eight Years Hence, The Age of the Father Will Be 4 Years More than Three Times the Age of the Son. Find Their Present Ages. - ICSE Class 10 - Mathematics

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Question

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

Solution

Let the present age of the son be x years.
∴ Present age of father =` 2x^2 `years
Eight years hence,
Son's age = (x + 8) years
Father's age =` (2x^2 + 8)` years
It is given that eight years hence, the age of the father will be 4 years more than three  times the age of the son. 

`∴ 2x^2 + 8 = 3(x + 8) + 4`
`2x^2 + 8 = 3x + 24 + 4`
`2x^2 − 3x − 20 = 0`
`2x^2 − 8x + 5x − 20 = 0`
`2x(x − 4) + 5(x − 4) = 0`
`(x - 4) (2x + 5) = 0` 

`x = 4, (-5)/2` 

But, the age cannot be negative, so, x = 4.
Present age of son = 4 years
Present age of father = 2(4)2 years = 32 years.

  Is there an error in this question or solution?

APPEARS IN

 Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Ex.6D | Q: 6
Solution The Age of a Father is Twice the Square of the Age of His Son. Eight Years Hence, The Age of the Father Will Be 4 Years More than Three Times the Age of the Son. Find Their Present Ages. Concept: Quadratic Equations.
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