A man has a daughter and son. The son is three times older than the daughter. In one year the man will be six times as old as the daughter is now. In ten years the man will be fourteen year older than the combined ages of his children at that time. What is the mans present age?

Answer 1

s = 3d
m+1 = 6d
m+10 = 14+(s+10)+(d+10)
solve to find m.

I get non-integer values for s and d.

If "three times older" means d+3d=4d, then I get integer values, but the man is 149. Not likely.