SOLUTION: Two years ago, a man was eight times as old as his son. In three years time, he will be four and half times his son's age then. Find their present ages.
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Question 498809: Two years ago, a man was eight times as old as his son. In three years time, he will be four and half times his son's age then. Find their present ages. Answer by Maths68(1474) (Show Source):
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Present age of father = f
Present age of son = s
Two years ago means we have to subtract 2 from their present ages
f-2=s-2
then
father was eight times as old as his son
f-2=8(s-2)
f-2=8s-16
f=8s-16+2
f=8s-14..................(1)
In three years time means we have add 3 in their present ages
f+3=s+3
then he will be four and half times his son's age
f+3=(4.5)(s+3)
f+3=4.5s+13.5
f=4.5s+13.5-3
f=4.5s+10.5 ........(2)
Put the value of f from (1) to (2)
f=4.5s+10.5 ........(2)
8s-14=4.5s+10.5
8s-4.5s=10.5+14
3.5s=24.5
Divide both sides of above equation by 3.5
3.5s/3.5 = 24.5/3.5
s=7
plug in the value of s in equation (1)
f=8s-14..................(1)
f=8(7)-14
f=56-14
f=42
Present age of father = 42
Present age of son = 7
