Check my work for motion problem

Question

This is the problem I am trying to solve:

An employee drove to work on Monday at $45$mi/h and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving $40$m/h and arriving one minute late.

a. How far does the employee live from work?

b. At what speed should the employee drive to arrive five minutes early?

My work: $$45(x-1)=40(x+1)$$ $$45x-45=40x+40$$ $$5x=85$$ $$x=17$$ Now it starts to get a little shady. I think this is the time it will take the employee to arrive on time, so I substituted it into the equation $d=rt$ with $r$ being $45$ and $t$ being $16$. This gave me the result of $720$, which seems unreasonable for the distance.

Am I doing anything wrong? If so, please explain.

Note: Please do not offer the solution; I want to solve this myself.

Answer 1

You're mixing units. The speeds are in miles per hour, but the time difference is in minutes.

As @thunderbolt's comment indicates, yes, you can fix this by dividing the final answer by $60$ to give $12$ miles, which is the correct answer for part a). But I think the working has to be clearer.

(The reason why this works here is the factor of $\frac{1}{60}$ on both sides cancels out, so the equation still holds. However, I would start by writing the equation properly to begin with, otherwise it looks like the student is not aware of the different units of time involved here.)

Answer 2

I would use more meaningful symbols and write down the data first:

${d \over 45} = T-{1 \over 60}$

${d \over 40} = T+{1 \over 60}$.

From this you can compute $d$.

From this you can compute $T$.

Now compute $v$ such that

${d \over v} = T-{5 \over 60}$.

Answers:

$d=12$, $T = {17 \over 60}$, $v=60$.