Two Clock Hands on Same Position

Geometry Level 2

The clock has the hour and minute hands. The hour hand is the shortest, whereas the minute hand is the longest.

Assume that all hands continuously and smoothly rotate throughout the whole time at the constant speed. How many total different ways are there, such that both hands are exactly at the same position?

Note: Both hands do not have to point exactly at the marks or numbers for this to count. Only consider position for this problem.

11 12 13 15 14 10 None of the answers. 1

2 solutions

Chew-Seong Cheong
Oct 10, 2017

Let the angle between the marking for 12 and the position of either hour or minute hand be $\theta$ , and the angular velocities of the hour and minute hands be $\omega_h = 0.5$ °/min and $\omega_m = 6$ °/min respectively. Then the angles made after time $t$ minutes are $\theta_h= \omega_ht = 0.5t$ and $\theta_m =\omega_mt = 6t$ .

Let $t=0$ when the two hands meet at 00:00. They meet again, when $\theta_h = \theta_m - 360n$ , where $n$ is a positive integer. Or $t=\dfrac{720n}{11}$ I. Therefore, within 12 hours, $12 \times 60 = \dfrac{720n}{11}$ , $\implies n = \boxed{11}$ .

Edwin Gray
Oct 11, 2017

The hour hand makes 12revolutions in 12 hours. The minute hand gains 1/11 of an hour on each revolution. So the hands are alignd every 12/11 hours, and they are together 11 times in any 12 hour period. Ed Gray

Two Clock Hands on Same Position

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