AMC 8 · 1999 · #2
Grade 4 geometry-2dProblem
What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: On a standard 12-hour clock, the hour hand points at the $10$ and the minute hand points at the $12$ when the time is $10{:}00$. What is the measure (in degrees) of the smaller of the two angles between the hands?
Givens: Standard clock face: $12$ equally-spaced numbers around a full $360^\circ$ circle; Time shown: exactly $10$ o'clock; At $10{:}00$ the minute hand is on the $12$ and the hour hand is on the $10$; Answer choices: (A) $30$, (B) $45$, (C) $60$, (D) $75$, (E) $90$
Unknowns: The smaller of the two angles (in degrees) formed by the two hands
Understand
Restated: On a standard 12-hour clock, the hour hand points at the $10$ and the minute hand points at the $12$ when the time is $10{:}00$. What is the measure (in degrees) of the smaller of the two angles between the hands?
Givens: Standard clock face: $12$ equally-spaced numbers around a full $360^\circ$ circle; Time shown: exactly $10$ o'clock; At $10{:}00$ the minute hand is on the $12$ and the hour hand is on the $10$; Answer choices: (A) $30$, (B) $45$, (C) $60$, (D) $75$, (E) $90$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
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.
Execute — Answer: C
4.MD.C.5 Step 1 - Find the size of one sector.
- The clock face is a $360^\circ$ circle divided by the $12$ numbers into $12$ equal sectors, so each sector spans $360^\circ \div 12 = 30^\circ$.
- That is the angle between any two consecutive numbers on the dial.
💡 Grade 4 angle thinking: a full turn is $360^\circ$, and splitting that turn into $12$ equal pieces makes each piece a $30^\circ$ wedge — the dial itself is a built-in protractor.
4.MD.C.7 Step 2 - Count the sectors between the two hands.
- At $10{:}00$ the minute hand sits on the $12$ and the hour hand sits on the $10$.
- Sweeping clockwise from the $10$ to the $12$ passes through the $11$ on the way, crossing exactly $2$ sectors ($10\to 11$ and $11\to 12$).
💡 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.7 Step 3 - Multiply sectors by sector size to get the angle between the hands.
- Two sectors of $30^\circ$ each add to $60^\circ$.
- The other angle the hands make is $360^\circ - 60^\circ = 300^\circ$, so $60^\circ$ is clearly the smaller one.
💡 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.5 Find the size of one sector. The clock face is a $360^\circ$ circle divided by t 4.MD.C.7 Count the sectors between the two hands. At $10{:}00$ the minute hand sits on th 4.MD.C.7 Multiply sectors by sector size to get the angle between the hands. Two sectors Review
Reasonableness: Sanity check the answer against the diagram. The hands point at the $10$ and the $12$, only two numbers apart on the clock face — a small slice of the circle, nothing close to a half turn. A $90^\circ$ angle would mean three sectors apart (like $12$ and $3$), and a $30^\circ$ angle would mean just one sector apart (like $12$ and $1$). Two sectors must sit between those, giving $60^\circ$, which matches choice (C). The reflex angle on the other side is $360^\circ - 60^\circ = 300^\circ$, far larger, confirming $60^\circ$ is the smaller angle the problem asks for.
Alternative: Tool #9 (Solve an Easier Problem): rotate the clock in your head so that the minute hand points at the $12$ as usual and the hour hand points at the $2$ instead of the $10$ — the picture is the same by symmetry because $10$ and $2$ are both two numbers away from $12$. At $2{:}00$ the angle is well-known to be $60^\circ$ (one-sixth of a full turn, since $2$ out of $12$ sectors is $\tfrac{2}{12} = \tfrac{1}{6}$ and $\tfrac{1}{6} \times 360^\circ = 60^\circ$). Same answer, choice (C).
CCSS standards used (min grade 4)
4.MD.C.5Recognize angles as geometric shapes formed by two rays and understand concepts of angle measurement (Reading the clock face as a $360^\circ$ circle divided into $12$ equal $30^\circ$ sectors so the angle between consecutive numbers has a definite size.)4.MD.C.7Recognize angle measure as additive; solve addition and subtraction problems to find unknown angles (Adding two $30^\circ$ sectors (or multiplying $2 \times 30^\circ$) to get the $60^\circ$ angle between the hour hand at the $10$ and the minute hand at the $12$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Doing the arithmetic $360 \div 12 = 30$ and $2 \times 30 = 60$ to turn the sector picture into a degree measure.)
⭐ Each number-to-number gap on a clock is $30^\circ$ ($360^\circ$ split into $12$ equal pieces), and the $10$ sits two gaps away from the $12$ — so the smaller angle at $10$ o'clock is $2 \times 30^\circ = 60^\circ$.
⭐ Each number-to-number gap on a clock is $30^\circ$ ($360^\circ$ split into $12$ equal pieces), and the $10$ sits two gaps away from the $12$ — so the smaller angle at $10$ o'clock is $2 \times 30^\circ = 60^\circ$.
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