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Father is aged three times more than his son Ronit. After 8
years, he would be two and a half times of Ronit's age.
After further 8 years, how many times would he be of Ronit's
age?
A.2 times
B.2 1/2times
C.2 3/4times
D.3 times
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Answer Posted / vignesh
You can also solve this problem by taking son's age as x and father's age as y. First, y = 4x as it is 3 times 'more' than that of his son. Then after 8 years son's age will be x+8 and father's age will be y+8=5/2(x+8) after solving you will get x = 8 and y = 32. Then after 16 years let father be n times of son's age i.e.
y + 16 = n(x + 16).
or 32 + 16 = n(8 + 16)
[putting x = 8 and y = 32]
or 48 = 24n,
or n = 2.
Therefore father is 2 times his son's age after 16 years.
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