Since $x^{2}<x$ and $x^{2} \geq 0$, then $x>0$ and so it cannot be the case that $x$ is negative. Thus, neither (D) nor (E) is the answer. Since $x^{2}<x$, then we cannot have $x>1$. This is because when $x>1$, we have $x^{2}>x$. Thus, (A) is not the answer and so the answer is (B) or (C). If $x=\frac{1}{3}$, then $x^{2}=\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}$ and $\frac{x}{2}=\frac{1 / 3}{2}=\frac{1}{6}$. Since $\frac{1}{6}>\frac{1}{9}$, then $(B)$ cannot be the answer. Therefore, the answer must be (C). Checking, when $x=\frac{3}{4}$, we have $x^{2}=\frac{9}{16}$ and $\frac{x}{2}=\frac{3}{8}$. Since $\frac{x}{2}=\frac{3}{8}=\frac{6}{16}<\frac{9}{16}=x^{2}$, then $\frac{x}{2}<x^{2}$. Also, $x^{2}=\frac{9}{16}<\frac{12}{16}=\frac{3}{4}=x$. This confirms that $x=\frac{3}{4}$ does satisfy the required conditions.
The value of $\frac{x}{2}$ is less than the value of $x^{2}$. The value of $x^{2}$ is less than the value of $x$. Which of the following could be a value of $x$?