If it’s 4 o clock now, how many minutes will pass before the minute and hour hand are in the exact same place?
In the solution, they come up with the formula:
x=20+5(x/60), where x is the number of minutes that have passed. However, I am unsure how they derived this formula. Can anyone help?
The position of the two hands has to be equal. The relative position of the hour hand can be described as $$\frac{4+x/60}{12},$$ where $x$ is the number of minutes that have passed. This is a relative number (which is between $0$ and $1$ and represents the position; think about why this is the case by plugging in some numbers). Now, this has to be equal to the relative position of the minute hand, which is simply $x/60$.
Now, these two expressions have to be equal, which results in
$$4+\frac{x}{60}=\frac{12}{60}x\implies x=240/11.$$
