# Analog Clock Problem; Finding Present with Different Time

It is now past 3 o'clock in the afternoon. The minute hand is 3 minutes from now will be directly opposite the hour hand 9 minutes ago. What is the time now?

I recall previous formulas on this:

For every x travel of minute hand:
The hour hand travels = x/12 minute spaces


The correct answer is 3:45; Any hint how to solve this? I know how to use the formulas for other application; but this is kinda hard to visualize

Every minute, the minute hand travels $6^{\circ}$. The hour hand travels $\frac1{12}$th as much, or $\frac12^{\circ}$.

If the minute hand is at $m$ and the hour hand is at $h$, we have $m+18 = h - \frac92 + 180$, or $m = h + 157.5^{\circ}$.

Additionally, we know it's past 3pm, so $h\geq 90^{\circ}$, and $h$ and $m$ have to be consistent - which means the minutes elapsed have to line up with where $h$ is. That is, $\frac{m}{6}\cdot\frac12+90=h$.

So we have two equations:

$$\begin{split} m &= h + 157.5 \\ \frac{m}{12} + 90 &= h \end{split}$$

Solving those gives us $m = 270^{\circ}, h = 112.5^{\circ}$. Which is 3:45pm.

• i came back to this problem; i forgot how you got 18? -9/2 and 180.. – james Jun 8 '15 at 1:04