A father's present age is three times the sum of the present ages of his two children. Five years from now, his age will be twice the sum of their ages at that time. Find the father's present age.

Let the father be $x$ years and the two children's ages total $y$ years. $x=3y$. In five years each child is 5 years older, so their total is $y+10:x+5=2(y+10)$, that is $x-2y=15$. So $3y-2y=15,y=15$ and $x=45$. Check: $45=3\times 15$, and $50=2\times 25$. 15 years is the children's combined age, and it is also what you get from the common slip $x+5=2(y+5)$, which adds only one lot of 5 years for two children; 50 years is the father's age five years from now; 60 years is not reached by any correct step.