Finding the correct time At a certain time between 3pm and 4pm, the hour and the minute hands are at equal angles from the 6 mark, what time will it be exactly?
My approach is at the time t(minutes) the following should hold
 180 - 0.5t = 6t -180
Thus we get the time in minutes. I don't have the answer to this problem. It would be great if someone could verify this
 A: At $3:x$ pm, the minute hand is at $6x$ degrees from the $12$ hour mark.
Note that for each hour, the hour hand moves $30$ degrees. Hence, for each minute, the hour hand moves $0.5$ degrees. Therefore, at the end of $x$ minutes, the angle with the $12$ hour mark is $90 + 0.5x$.
Since we are measuring from the $6$ hour mark, the equation would be $180 - 6x = 180 - (90 + 0.5x) = 90-0.5x$. Solving, we get $5.5x = 90 \implies x = 16.\overline{36}$ minutes.
Now, it is also possible that the minute hand will cross the $30$ minute mark, and on the other side of $30$, it will create the same angle. That is to say, we have to solve the equation $(90 + 0.5x) + 6x = 360$. Solving this, we get $x = 41.538$ minutes, hence even at $3:41.538$, the hour and minute hand will have the same angle with the 6 hour mark.
A: After the minute hand has moved $x$ of the circle around, the hour hand is $\frac14+\frac1{12}x$ of the circle around (the first term corresponding to its initial position, and the second to its movement since 3pm). Then we have two cases:


*

*The two hands are on the same side of 6 o'clock, hence overlapping. This entails $x=\frac14+\frac1{12}x$, hence $\frac{11}{12}x=\frac14$ or $x=\frac{3}{11}$ and the time will be $3:16\frac4{11}$.

*The two hands are on opposite sides. This entails $x+\frac14+\frac1{12}x=1$, which simplifies to $\frac{13}{12}x=\frac34$ or $x=\frac9{13}$. The time will be $3:41\frac7{13}$.

A: There are two different positions:

A: The hour and minute hands are at the same angle at $12\!:\!00\!:\!00=0\!:\!00\!:\!00$. The hour hand moves at $1$ hour mark per hour and the minute hand moves $12$ hour marks per hour. Their distance from the $6$ mark is $6-t\pmod{12}$ and $6-12t\pmod{12}$. Thus, we want
$$
6-t\equiv6-12t\pmod{12}\tag{1}
$$
or
$$
t-6\equiv6-12t\pmod{12}\tag{2}
$$
Equation $(1)$ becomes
$$
11t\equiv0\pmod{12}\tag{3}
$$
while equation $(2)$ becomes
$$
13t\equiv12\pmod{12}\tag{4}
$$
The solutions between $3$ and $4$ are
$$
t=\frac{36}{11}=3\!:\!16\!:\!21\tfrac9{11}\tag{5}
$$
and
$$
t=\frac{48}{13}=3\!:\!41\!:\!32\tfrac4{13}\tag{6}
$$
