2015 AMC 10A Problems/Problem 14

Revision as of 20:24, 4 February 2015 by Suli (talk | contribs) (Solution)


Clock - radius 20 cm Disk - radius 10 cm, externally tangent at 12 o'clock, arrow pointing up

If the disk rolls clockwise, where on the clock will it be when the arrow points up?


The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is $\boxed{\textbf{(C) }4 \ \mathrm{o'clock}}$ .

Solution 2

The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. $\boxed{\textbf{(C)}}$

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