1. Use the mean condition

There are $8$ numbers: $3,7,9,12,13,20,x,y$

Given mean $=10$, so total sum is $8\times 10=80$

Now sum of the known numbers: $3+7+9+12+13+20=64$

Hence, $x+y=80-64=16$

2. Use the variance condition

Given variance $=25$.

For a set of $n$ observations, variance is $\sigma^2=\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2$

Here, $25=\frac{\sum x_i^2}{8}-10^2$

So, $25=\frac{\sum x_i^2}{8}-100$ $\frac{\sum x_i^2}{8}=125$ $\sum x_i^2=1000$

Now compute squares of known numbers: $3^2+7^2+9^2+12^2+13^2+20^2$ $=9+49+81+144+169+400=852$

Thus, $x^2+y^2=1000-852=148$

3. Find $xy$ using identity

We know $(x+y)^2=x^2+y^2+2xy$

Substitute values: $16^2=148+2xy$ $256=148+2xy$ $2xy=108$ $xy=54$

4. Final answer

$\boxed{54}$

5. Comparison with stored answer

Stored correct answer = $54$.

This matches the derived answer.