AMC 8 · 1999 · #2
Grade 4 geometry-2d
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The clock face given in the problem is already a labeled circular diagram — Tool #1 (Draw a Diagram) lets us mark both hand positions on the dial and read the gap straight off, no formula needed. Tool #7 (Identify Subproblems) splits the work into two clean Grade 4 steps: first find the size of one sector between consecutive numbers, then count how many of those sectors sit between the 10 and the 12. Multiplying gives the answer, and the diagram confirms it is the smaller of the two possible angles.
The dial splits a full 360° turn into 12 equal wedges, so one number-to-number wedge is 360° ÷ 12 = 30°.
Grade 4 angle thinking: a full turn is 360°, and splitting that turn into 12 equal pieces makes each piece a 30° wedge — the dial itself is a built-in protractor.
4.MD.C.5Identify SubproblemsAt 10/:00 the minute hand is on the 12 and the hour hand on the 10; sweeping 10 → 11 → 12 clockwise crosses exactly 2 wedges.
Walking the gap on the dial is just adding sector-sized pieces — a Grade 4 "angle measure is additive" move done by simple counting.
4.MD.C.7Draw A DiagramTwo 30° wedges give 2 × 30° = 60°; the other angle is 360° - 60° = 300°, so 60° is the smaller one — choice (C).
Repeated addition of equal angles becomes one multiplication. The diagram confirms which of the two angles is the smaller, so no second check is needed.
4.MD.C.7Draw A DiagramEach number-to-number gap on a clock is 30° (360° split into 12 equal pieces), and the 10 sits two gaps away from the 12 — so the smaller angle at 10 o'clock is 2 × 30° = 60°.