Question 701153: my dad was 10 years old twice the age of me 5 years ago. 5 years later he would be twice the age f me. find current ages
Answer by CRandunu(8)   (Show Source):
You can put this solution on YOUR website! let the present ages of father and me be f and m
f-5=2*(m-5)+10
f-5=2m----------(1)
f+5= 2(m+5)
f+5=2m+10
f-5=2m ---------(2)
these two equations are identical :-)
hence we come to a conclusion anyhow my age should be a positive integer which should be greater than 5 ,because given before 5 years ago I exist and not my age is zero!!
it gives m>5 where m=6 or 7 or 8 or....or n and
from (1) m=(f-5)/2 >0 where m=(f-5)/2 = 6 or 7 or 8 or....or n ------(A)
According to this to get a positive m>5 integer for my age "m" , father's present age can be 17 or 19 or 21 or 23 or 23 or 25 or 27 or 29 or 31 or 33 or 35 or....(2n+5)etc (substitute these ages in (A) and see)
hence there's a group of answers
when m=6 ,f= 17
when m=7, f=19
when m=8, f=21
.............
.............
when m=n-1 ,f= 2n+3
when m=n , f=2n+5 where n>5 and a integer
etc
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