This is a question on an archived UIL test.
Suppose on Friday a man drove to work at $x_1$ mph and arrived $y_1$ minutes late, and on Saturday he drove $x_2$ mph and arrived $y_2$ minutes early. What speed must the man drive to arrive to work on time? For example, what if the man drove at an average speed of 45mph on the first day and arrived 16 minutes late, and drove at 90mph on the second day and arrived 16 minutes early. What speed must he drive to arrive on time?
I have tried using the formula for average speed which is essentially (total distance/total time), however I am not sure that is the proper route for the solution, since the only time information is relative to being "on time" and because of that, I do not know the distance traveled either. I have also tried using a system of equations with $$45x = -16$$ and $$90x= 16,$$ but that essentially takes $(45+90)/(2)$ which doesn't provide a valid answer. The first sentence of the question states traffic can "vary" and I do not know if that holds any relevance. The answer choices are either $57.5$mph, $60$mph, $62.5$mph, or $65$mph.
Let $T_a$ be the time that the men should arrive at the workplace, and $T_s$ be the time where the men starts driving. (We will have to assume that $T_s$ is the same everyday; if not you do not have sufficient information to solve the problem).
Let $T = T_a- T_s$.
On the first day he arrives $y_1$ minutes late. So the time spend on driving is $T+y_1/60$ (measured in hours) and thus the distance traveled is
$$\tag{1} D = x_1(T+y_1/60).$$
Similarly, since he arrives $y_2$ minutes early the second day, we have
$$\tag{2} D = x_2 (T-y_2/60).$$
Solving the above gives $$ T = \frac{x_1y_1+x_2y_2}{60(x_2-x_1)}$$
and one can use this and either (1), (2) to find $D$. Hence the speed the men should drive is $D/T$.