It is between 2 and 3 o'clock, and in 10 minutes the minute-hand will be as much before the hour-hand as it is now behind it. What is the time?
Ok, so here is my approach to solving this problem:
When the clock first strikes 2 o'clock, the hour hand is 10 minutes further than the minute hand. Therefore, when the minute hand moves $x$ minutes, the hour hand will move $\frac{x}{12}$ minutes. Because the hour hand is already 10 minutes ahead of the minute hand to start with, the hour hand will be a total of $10 + \frac{x}{12}$ minutes ahead. So, the distance between the 2 hands is $(10 + \frac{x}{12}) – x$ minutes. Now, 10 minutes later, the minute hand has moved 10 minutes, so its total distance moved is $x + 10$ minutes. Meanwhile, the hour hand has moved a twelfth of those minutes. So, it has moved $\frac{10}{12}$ minutes. Therefore, the total distance the hour hand has covered is $10 + \frac{x}{12} + \frac{10}{12}$ or $10 + \frac{x + 10}{12}$ minutes. So, the distance between these 2 times should be the same as the distance between the original times. Therefore,
$$10 + \frac{x + 10}{12} – (x + 10) = (10 + \frac{x}{12}) – x$$
However, as you may have realized, the quantities $x$ and $\frac{x}{12}$ cancel, leaving a wrong equation.
I'm not sure whether I'm understanding something wrong, or I have constructed the equation wrong. If anyone can provide insight as to how to solve the problem, I would appreciate it.
Thanks.
Best Answer
You are assuming that the hour hand will still be ahead of the minute hand after $10$ minutes from the initial position, which will not be the case. This is because of the faster relative speed of minute hand w.r.t. the hour hand. The minute hand cannot be behind the hour hand by the same angle after another $10$ minutes. But it can go ahead with the amount of angle by which it was behind
In that case, you will get $$(x + 10) - (10 + \frac{x + 10}{12}) = (10 + \frac{x}{12}) - x$$ giving $$2x - \frac{x}{6}= 10 + \frac{10}{12} \implies x = \frac{65}{11}$$