If $x$ is a real number and $\lceil x \rceil = 11,$ how many possible values are there for $\lceil x^2 \rceil$?
Let's think about the possible values of $x$ that would make the ceiling of $x$ be 11. Since $\lceil x \rceil = 11$, we know that $x$ can be any number between $10$ and $11$. Let's write down the possible values of $x^2$. $x^2$ has to be between $100$ and $121$. Now let's write down the possible values of $\lceil x^2 \rceil$. $\lceil x^2 \rceil$ has to be between $101$ and $121$. How many integers are there between $101$ and $121$? There are $121 - 101 + 1 = 21$ integers between $101$ and $121$. That means there are $21$ possible values of $\lceil x^2 \rceil$.

