A Birthday:
A young boy, his father and his mother were celebrating a birthday.
This got the young boy to thinking about their ages. Upon asking
his father about this, the father told him: "Well, right now I am
six times your age, and all together the sum of our ages (yours,
mine, and your mother's) is 70. Later on, when I am only twice
your age, that sum will be twice what it is now."
Who's birthday is it?
SPOILER:
Let s (Son), f (Father), m (Mother) there ages in years today.
We know: s + f + m = 70. Later on (after t year) they will be
(s+t) + (f+t) + (m+t) = 2*70. That means 3t = 70. Therefore we
will expect non integral solutions for the ages.
s : f = 1 : 6
s+t : f+t = 1 : 2
This gives the linear equations 6s = f and 2s+2t = f+t and
subtraction gives 4s = t. Backsubstitution gives the solution
(t, s, f, m) = (70/3, 35/6, 35, 175/6) = (23 1/3, 5 5/6, 35, 29 1/6).
They are celebrating father's birthday. The mother had two months
ago and the son will have in two months.