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One years ago , a man …

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One years ago , a man was 8 times as old as his son . At present , his age is equal to the square of his son's age in years . Find their present age .

Posted by Uzzwal Upadhyay 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Saurabh Rajput 6 years, 10 months ago

Let the present age of man be x Let " " " " " of son be y Case 1 Present age =x=y2........1 Case 2 One year ago = x-1=8(y-1). ......2 When, we put the value of x in eq' no 2 Y2-1=8y-8 Y2-8y+7=0 Y2-7y-y+7=0 Y(y-7)-1(y-7)=0 (Y-1) (y-7)=0 The age of be 7 Then ,In eq no ..1 X=(7)2 X=49 Here the age of man is 49 Y=7,x=49

Gaurav Seth 6 years, 10 months ago

Let the age of the son one year ago be x years.
Age of the father one year ago = 8x years.

Present age of the father = 8x+1
Present age of the son = x+1
Given that (8x+1) = (x+1)²
(8x+1) = (x² + 2x + 1)
x² - 6x = 0
x(x - 6) = 0
x = 0 or x = 6

Son's age cannot be zero.

So, the present age of son is (x+1) i.e. 7 years.
Present age of the father = (8x+1) i.e. 49 years.

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