【Question 7】
Among the 95 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the numbers with an odd digit in the tens place are $\qquad$ in total.
【Analysis and Solution】
Number Theory, Perfect Squares.
$$
\begin{array}{l}
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2} ; \\
0^{2}=0,1^{2}=1,2^{2}=4,3^{2}=9,4^{2}=16,5^{2}=25,6^{2}=36,7^{2}=49,8^{2}=64,9^{2}=81 ;
\end{array}
$$

If the tens digit of a perfect square is odd, then its units digit must be 6;
Conversely, if the units digit of a perfect square is 6, then its tens digit must be odd.
Therefore, if the tens digit of a perfect square \( n^{2} \) is odd, then the units digit of \( n^{2} \) is 6, and the units digit of \( n \) is 4 or 6; there are 19 numbers between 1 and 95 whose units digit is 4 or 6;
Thus, among the 95 numbers \( 1^{2}, 2^{2}, 3^{2}, \ldots, 95^{2} \), there are 19 numbers whose tens digit is odd.