# Two years ago ,father was three times as old as his son and two years hence,twice his age will be equal to five times that of his son .find their present ages?

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by Shw4mj1nssonas

2016-03-25T18:45:23+05:30

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Let
Present age of Father = f
Present age of Son = s

Two years ago
f-2=s-2
Father was 3 times as old as his son
f-2=3(s-2)
f-2=3s-6
f=3s-6+2
f=3s-4............(1)
and 2 year hence,
f+2=s+2
then twice his age will be equal to 5 times that of his son
2(f+2)=5(s+2)

2f+4=5s+10
2f=5s+10-4
2f=5s+6.............(2)
Put the value of f in (2) from (1)
2(3s-4)=5s+6
6s-8=5s+6
6s-5s=6+8
s=14
Put the value of s in (1)
f=3s-4............(1)
f=3(14)-4
f=42-4
f=38
Present age of Father = f = 38 years old
Present age of Son = s = 14 years old
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2016-03-25T18:45:35+05:30

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Two years ago, Let the age of father be f years
and the age of son be s years
A/q
f= 3s
and , 2(f+4) = 5(s+4)
putting f= 3s
⇒2(3s+4) = 5(s+4)
⇒6s +8 = 5s +20
⇒s = 12
so present age of son = 14 years
and present age of father = 12×3 +2 = 38 years