SOLUTION: 5 years ago, a man was 7 times as old as his son. 5years hence, father will be 3 times as old as his son. Find their present ages.

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Question 8046: 5 years ago, a man was 7 times as old as his son. 5years hence, father will be 3 times as old as his son. Find their present ages.
Found 2 solutions by gokhalesujata, kritikapd:
Answer by gokhalesujata(4) About Me  (Show Source):
You can put this solution on YOUR website!
Sons age 5 years ago=x father's age = 7x
their present age = x+5 and 7x+5 resp
Sons age 5 years hence = x+10(since x is sons age 5 years ago from present age and we are talking of his age 5 years hence from his present age so 5+5=10)
fathers age 5 years hence =3(x+10)
hence the fathers present age will be 3(x+10)-5
also fathers present age =7x+5
therefore 7x+5=3(x+10)-5
7x+5=3x+30-5
7x-3x=25-5
4x=20
x=5
sons age 5 years ago=5years fathers age =5*7=35years
present age of the son=10years(5+5) fathers age=35+5=40 years
sons age 5 years hence= 5+10=15years fathers age =3*(5+10)=45 years

Answer by kritikapd(4) About Me  (Show Source):
You can put this solution on YOUR website!
Let present age of son = X
Age of son before 5 years = X-5
So, Present age of father = 7(X-5)+5 ……… (i)
Age of son after 5 years = X+5
Age of father after 5 years = 3(X+5)
So, present age of father = 3(X+5)-5 ……….(ii)
Equating age of father i.e. (i) and (ii)
7(X- 5)+5 = 3(X+ 5)-5
7X – 35+5 = 3X + 15-5
7X-3X = 15 - 5 + 35 - 5
4X = 40
X = 40/4 . X (son)= 10 , Father (3(10+5)-5)= 40 yrs