Let the number of gold coins in the original collection be $g$ and the number of non-gold coins be $n$. According to the problem, the ratio of gold coins to non-gold coins is 1:3, so:

$g = \frac{1}{3} \cdot n$ or $n = 3 \cdot g$.

Next, 10 more gold coins are added to the collection, so the number of gold coins becomes $g + 10$, and the number of non-gold coins remains $n$. The new ratio of gold coins to non-gold coins is given as 1:2, so:

$\frac{(g + 10)}{n} = \frac{1}{2}$.

Now, substitute $n = 3 \cdot g$ into the equation:

$\frac{(g + 10)}{(3 \cdot g)} = \frac{1}{2}$.

To solve for $g$, cross-multiply:

$2 \cdot (g + 10) = 3 \cdot g$.

Expand the equation:

$2 \cdot g + 20 = 3 \cdot g$.

Subtract $2 \cdot g$ from both sides:

$20 = g$.

So, there are initially 20 gold coins. Using $n = 3 \cdot g$, the number of non-gold coins is:

$n = 3 \cdot 20 = 60$.

Thus, the total number of coins before adding the 10 gold coins is:

$g + n = 20 + 60 = 80$.

After adding the 10 gold coins, the total number of coins is:

$g + 10 + n = 20 + 10 + 60 = 90$.

Therefore, the total number of coins in the collection now is:

$90$.