Ten years ago, a father was twelve times as old as his son and ten year hence, he will be twice as old as his will be then. Find their present ages.

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Solution

Let the present age of father be $x,$ and son be $y .$
$∴ x - 10 = 12 (y - 10) \Rightarrow x - 10 = 12 y - 120 \Rightarrow x - 12 y = - 110 ⟶ (i)$
And $,$ $x + 10 = 2 (y + 10) \Rightarrow x - 2 y = 10 \Rightarrow x = 10 + 2 y ⟶ (i i)$
Using $(i i)$ in $(i),$
$x - 12 y = - 110 \Rightarrow 10 + 2 y - 12 y = - 110 \Rightarrow y = 12$

And $,$ $x = 10 + 2 y = 10 + 2 \times 12 = 34$
$∴$ Age of son $= 12$ years $.$
Age of father $= 34$ years $.$

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Ten years ago, a father was twelve times as old as his son and ten year hence, he will be twice as old as his will be then. Find their present ages.

Let the present age of father be x, and son be y.∴x−10=12(y−10)⇒x−10=12y−120⇒x−12y=−110⟶(i)And , x+10=2(y+10)⇒x−2y=10⇒x=10+2y⟶(ii)Using (ii) in (i),x−12y=−110⇒10+2y−12y=−110⇒y=12And , x=10+2y=10+2×12=34∴ Age of son =12 years. Age of father =34 years.