Question 515292: 1 year ago,a man was 8 times as old as his son.now his age is equal to the square of his son's age.find their ages
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! Let
Present age of man = m
Present age of son = s
now his age is equal to the square of his son's age
m=s^2...........(1)
1 year ago
m-1=s-1
Then
man was 8 times as old as his son
m-1=8(s-1)
m-1=8s-8
m=8s-8+1
m=8s-7
Put the value of m from (1) to above equation
s^2=8s-7
s^2-8s+7=0
s^2-7s-s+7=0
s(s-7)-1(s-7)=0
(s-7)(s-1)=0
s-7=0 or s-1=o
s=7 or s=1
Present age of son is 7 years old or 1 year old, but we have been told that the man age one years ago was 8 times the son's age. If we assume that son is 1 year old it means one years back he is age was 0 and the square of the 0 will remain 0, so present age of son cannot be 1 year, Son is 7 years old.
S=7
Put the value of s in equaton (1)
m=s^2
m=7^2
m=49
Present age of man = m = 49 years old
Present age of son = s = 7 years old
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