AMC 8 · 1999 · #2

Easy mode Grade 4
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Problem

Picture a clock at 10 o'clock.

The minute hand points straight up at the 12. The hour hand points at the 10.

These two hands make two angles around the center. One is smaller, one is bigger.

What is the smaller angle, in degrees?

Pick an answer.

(A)
30
(B)
45
(C)
60
(D)
75
(E)
90
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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

#7 Identify Subproblems 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.
$$\dfrac{360^\circ}{12} = 30^\circ \text{ per sector}$$

💡 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.

#1 Draw a Diagram 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 $11\to 12$).
$$\text{sectors between } 10 \text{ and } 12 = 2$$

💡 Walking the gap on the dial is just adding sector-sized pieces — a Grade 4 "angle measure is additive" move done by simple counting.

#1 Draw a Diagram 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.
$$2 \times 30^\circ = 60^\circ \;\;(<\; 300^\circ) \;\Rightarrow\; \textbf{(C)}$$

💡 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.

[1] #7 4.MD.C.5 Find the size of one sector. The clock face is a $360^\circ$ circle divided by t
[2] #1 4.MD.C.7 Count the sectors between the two hands. At $10{:}00$ the minute hand sits on th
[3] #1 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.5 Recognize 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.7 Recognize 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.5 Multiply 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 $2 \times 30^\circ = 60^\circ$.