There are $522$ people at a beach, each of whom owns a cat, a dog, both, or neither. If $20$ percent of cat-owners also own a dog, $70$ percent of dog-owners do not own a cat, and $50$ percent of people who don’t own a cat also don’t own a dog, how many people own neither type of pet?
1. Let \( x \) be the number of people who only own a cat.
2. Let \( y \) be the number of people who own both a cat and a dog.
3. Let \( z \) be the number of people who only own a dog.
4. Let \( w \) be the number of people who own neither a cat nor a dog.

From the given information, we can set up the following equations:

1. \( 20\% \) of cat-owners also own a dog:
   \[
   \frac{y}{x + y} = \frac{1}{5}
   \]
   Solving for \( y \):
   \[
   y = \frac{1}{5}(x + y) \implies 5y = x + y \implies 4y = x \implies x = 4y
   \]

2. \( 70\% \) of dog-owners do not own a cat:
   \[
   \frac{z}{y + z} = \frac{7}{10}
   \]
   Solving for \( z \):
   \[
   z = \frac{7}{10}(y + z) \implies 10z = 7y + 7z \implies 3z = 7y \implies z = \frac{7}{3}y
   \]

3. \( 50\% \) of people who don’t own a cat also don’t own a dog:
   \[
   \frac{w}{w + z} = \frac{1}{2}
   \]
   Solving for \( w \):
   \[
   w = \frac{1}{2}(w + z) \implies 2w = w + z \implies w = z
   \]

4. The total number of people is 522:
   \[
   x + y + z + w = 522
   \]

Substituting \( x = 4y \), \( z = \frac{7}{3}y \), and \( w = z \) into the total number of people equation:
\[
4y + y + \frac{7}{3}y + \frac{7}{3}y = 522
\]
Combining like terms:
\[
4y + y + \frac{14}{3}y = 522 \implies 5y + \frac{14}{3}y = 522 \implies \frac{15y + 14y}{3} = 522 \implies \frac{29y}{3} = 522
\]
Solving for \( y \):
\[
29y = 1566 \implies y = \frac{1566}{29} \implies y = 54
\]

Now, substituting \( y = 54 \) back into the expressions for \( x \), \( z \), and \( w \):
\[
x = 4y = 4 \times 54 = 216
\]
\[
z = \frac{7}{3}y = \frac{7}{3} \times 54 = 126
\]
\[
w = z = 126
\]

Finally, the number of people who own neither type of pet is:
\[
w = 126
\]

The final answer is \(\boxed{126}\).