A father is aged three times more than his so | vedclass

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(A) Let the present age of Ronit be $x$ years.Since the father is three times older than his son,the father's age is $x + 3x = 4x$ years.After $8 \, \

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$2$

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(A) Let the present age of Ronit be $x$ years.Since the father is three times older than his son,the father's age is $x + 3x = 4x$ years.After $8 \, 	ext{years}$,Ronit's age will be $(x + 8)$ and the father's age will be $(4x + 8)$.According to the problem,$(4x + 8) = 2.5(x + 8)$.$4x + 8 = 2.5x + 20$.$4x - 2.5x = 20 - 8$.$1.5x = 12$.$x = 12 / 1.5 = 8$.So,Ronit's present age is $8$ years and the father's present age is $4 	imes 8 = 32$ years.After further $8 \, 	ext{years}$ (total $16 \, 	ext{years}$ from now),Ronit's age will be $8 + 16 = 24$ years.The father's age will be $32 + 16 = 48$ years.Ratio of father's age to Ronit's age $= 48 / 24 = 2$.Thus,the father will be $2$ times the age of Ronit.

$A$ father is aged three times more than his son Ronit. After $8 \, \text{years}$,he would be two and a half times of Ronit's age. After further $8 \, \text{years}$,how many times would he be of Ronit's age?

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