one year ago a man was 8 times as old as his son. now his age to the square of his sons age. find their present age.
Answers
Answer:
Let x be the age of the son one year ago, then the age of the man was 8x.
The present age of the son is (x+1) and that of the man is (8x+1). Then,
8x+1=(x+1)
2
8x+1=x
2
+1+2x
x
2
=6x
x=0,6
As the age cannot be 0, so the value of x is 6.
So, present age of son =(x+1)=7 years and present age of man =(8x+1)=49 years
Answer:
Present age is 49
Step-by-step explanation:
LET THE PRESENT AGE OF SON BE = X
Therefore, present age of man= x^2 years one year ago
son's age = (x^2 - 1) years
man's age = (x^2 - 1) years
It is given that one year ago; a man was 8 times older than his son
therefore, (x^2 - 1) = 8 (x - 1)
= x^2 - 8x - 1 + 8=0
= (x - 7) (x - 1) = 0
= x = 7,1
If x =1, then x^2 = 1 which is not possible as father's age cannot be equal to son's age.
so x= 7
present age of son = x years = 7
present age of man = x^2 years = 49 years