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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

Persons Present age  8 years hence  Future age 
Father  2x2  + 8 years 2x2+8
Son  x  +8 years    x+8

Now follow below

Let the present age of the son be x years.
Present age of father = 2xblank squared years [according to the given condition ] 
Eight years hence,
Son’s age = (x + 8) years
Father’s age = (2xblank squared+ 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.
2 x squared space plus space 8 space equals space 3 left parenthesis x space plus space 8 right parenthesis space plus 4 space 2 x squared space plus space 8 space equals space 3 x space plus space 24 space plus 4 space 2 x squared space � space 3 x space � space 20 space equals space 0 space 2 x squared space � space 8 x space plus space 5 x space � space 20 space equals space 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)blank squared years = 32 years


Mathematics
Concise Mathematics
Standard X

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