5. Tickets to a school dance cost $\$ 4$ and the projected attendance is 300 persons. For every $\$ 0.10$ increase in ticket price, the dance committee projects that attendance will decrease by 5 people. What ticket price will generate $\$ 1237.50$ in revenue?

Step 1 ：Let x be the number of $0.10$ increases in ticket price, and let y be the number of people attending the dance.

Step 2 ：The ticket price is $4 + 0.10x$, and the attendance is $300 - 5x$.

Step 3 ：The revenue is the product of the ticket price and the attendance: $(4 + 0.10x)(300 - 5x)$.

Step 4 ：We want to find the value of x when the revenue is $1237.50$: $(4 + 0.10x)(300 - 5x) = 1237.50$.

Step 5 ：Expanding the equation: $1200 - 20x + 0.10x^2 - 0.50x^2 = 1237.50$.

Step 6 ：Simplifying the equation: $-0.40x^2 - 20x + 1200 = 1237.50$.

Step 7 ：Subtract 1237.50 from both sides: $-0.40x^2 - 20x - 37.50 = 0$.

Step 8 ：Divide both sides by -0.40: $x^2 + 50x + 93.75 = 0$.

Step 9 ：Using the quadratic formula, we find that $x \approx -1.87$ or $x \approx -50.13$. Since x represents the number of $0.10$ increases, it must be a positive integer.

Step 10 ：Thus, $x = 1$ (rounded to the nearest integer).

Step 11 ：Substitute x back into the ticket price equation: $4 + 0.10(1) = 4.10$.

Step 12 ：\(\boxed{\$4.10}\)