After $t$ hours, the hour hand was where the minute hand had been, and vice versa

On Saturday, Jimmy started painting his toy helicopter between 9:00 a.m. and 10:00 a.m. When he finished between 10:00 a.m. and 11:00 a.m. on the same morning, the hour hand was exactly where the minute hand had been when he started, and the minute hand was exactly where the hour hand had been where the hour hand had been when he started. Jimmy spent $t$ hours painting. Determine the value of $t$.

Let $h$ represent the initial position of the hour hand.

Let $m$ represent the initial position of the minute hand.

Note that in this solution position will be represented in “minutes”. For example, if the hour hand was initially at $9$, its position would be $45$.

$$45\le h \le 50$$
$$50 \le m \le 55$$
$$\implies 0 \le (m-h) \le 10$$

Time can be represented as the number of minutes passed since 12 a.m. (for example 1 am = 60, 2 am = 120 etc.)

Then:

$$60(\frac{h}{5}) + m + t = 60 (\frac{m}{5}) + h$$
$$\implies 12 h + m + t = 12m + h$$
$$\implies t = 11 m – 11 h$$
$$\implies t = 11(m-h)$$

$$0 \le t \le 120$$
$$\implies 0 \le 11(m-h) \le 120$$
$$\implies 0 \le m -h \le \frac{120}{11}$$

That was as far as I got. Could someone point me on the right path to complete the above solution? (if possible). I am not simply looking for a solution but instead a way to complete the above solution. Any help is appreciated, thank you.

Solutions Collecting From Web of "After $t$ hours, the hour hand was where the minute hand had been, and vice versa"

I’m not sure what you mean by “complete the above solution”. The above attempt is actually pretty far from actually finding out what $m$ and $h$ are. However, you do get $t$ as a function of $m$ and $h$, which will be needed to compute the time elapsed.

To actually solve this problem, you have to make use of the fact that $m$ and $h$ encode the same information. If I know the exact position of the hour hand, then I know the exact time because the minute hand’s information is encoded in the position of the hour hand between the two integer hours.

At the beginning, we have
$$ \frac{h-45}{5} = \frac{m}{60} $$
Then, at the end, we have
$$ \frac{m-50}{5} = \frac{h}{60} $$
Now, it’s just a matter of solving a system of two equations/unknowns: http://www.wolframalpha.com/input/?i=h-45+%3D+m%2F12,+m-50+%3D+h%2F12

Plugging in these values for $h$ and $m$ into $t = 11(m-h)$ yields approximately $t=$ 50.8 minutes, or $t = 0.846$ hours.