# Another one of those time problems

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.

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 $$2x - \frac{x}{6}= 10 + \frac{10}{12} \implies x = \frac{65}{11}$$

Let minute hand be at $y$ degrees from $12$-hour mark initially. Hour hand is at $(60 + \frac{y}{12})^\circ$ initially. After $10$ minutes, it's at $$60 + y$$, hour hand is at $$60 + \frac{y}{12} + \frac{60}{12}$$ So, $60+ \frac{y}{12}-y = 60 + y - (60+ \frac{y}{12}+ \frac{60}{12})$ $\implies 60 + \frac{60}{12} = 2y - \frac{y}{6} = \frac{11y}{6}$ $y = \frac{6}{11} \times 65 ^\circ = \big(35\frac{5}{11}\big)^\circ = \big(30 + \frac{60}{11}\big)^\circ = 5 \frac{10}{11}\text{min}$

• This is a comment, not an answer. – amWhy Jul 4 '18 at 15:41
• Thanks for the answer. However, I got this problem from a textbook I was looking at. I looked at the answer and it said the answer was $5\frac{10}{11}$ minutes past two. Your answer seems to be close, but still off by a small amount. – S. Sharma Jul 4 '18 at 16:22
• @ASilentCat I have edited the answer now, have a look. – ab123 Jul 4 '18 at 18:32
• That makes more sense. Using degrees was a smarter approach than what I was thinking. Thank you. – S. Sharma Jul 4 '18 at 19:39

Let $t$ be the time in minutes after 12:00 and let $\theta_m$ be the angle in radians that the minute hand makes clockwise with respect the vertical, similarly for the hour hand $\theta_h$

$$0 \leq t \leq 720$$

$$\theta_h(t) = 2\pi \frac{t}{720}$$

$$\theta_m(t) = 2\pi \frac{t \mod 60}{60}$$

$$120<t<180$$

$$\theta_h(t+10) - \theta_m(t+10) = \theta_m(t) - \theta_h(t)$$

$$2\pi \frac{t+10}{720} - 2\pi \frac{(t+10) \mod 60}{60} = 2\pi \frac{t \mod 60}{60} - 2\pi \frac{t}{720}$$

$$\frac{t+10}{720} - \frac{(t+10) \mod 60}{60} = \frac{t \mod 60}{60} -\frac{t}{720}$$

Solving for $t$ in the given range:

$$t = \frac{1385}{11} \approx 02:05$$