Minute Hand will be as much ahead of the hour Hand as it is Behind it

The time is past 2 o'clock in 10 minutes. The minute hand will be as much as ahead of the hour hand as it is behind it. What time is it?

I am having trouble interpreting " Minute Hand will be as much ahead of the hour Hand as it is Behind it ". Does it mean the hour hand somehow bisects the two times of the minute hand? I am more of a visual learner... but i dont know why clock problems for me are so hard to visualize. maybe there is an easier way?

• Is this the exact wording of the problem? – OnceUponACrinoid Jun 8 '15 at 2:42
• @OnceUponACrinoid yes – james Jun 8 '15 at 4:32

I would suggest that the interpretation given by Annie Carter is correct, and would solve the problem as follows. Suppose that the time now is $m$ minutes past $2$. For convenience we measure angles in units of full circles. Then the angles between "12" on the clock, the centre of the clock, and the minute and hour hands are $$\angle TOM=\frac{m}{60}\ ,\quad \angle TOH=\frac{2+\frac{m}{60}}{12}\ .$$ In ten minutes time it will be $$\angle TOM=\frac{m+10}{60}\ ,\quad \angle TOH=\frac{2+\frac{m+10}{60}}{12} \ .$$ So we have $$\frac{m+10}{60}-\frac{2+\frac{m+10}{60}}{12} =\frac{2+\frac{m}{60}}{12}-\frac{m}{60}\ .$$ Probably the best way to solve this is to write it as $$\frac{2m+10}{60}=\frac{4+\frac{2m+10}{60}}{12}\ ;$$ you then easily get $2m+10=\frac{240}{11}$ and so $m=\frac{65}{11}=5.91$.