AMC 8 · 2003 · #20
Grade 5 geometry-2dProblem
What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: At 4:20 PM, what is the measure of the acute angle between the hour hand and the minute hand of a standard 12-hour clock?
Givens: Standard clock with $12$ equally-spaced numbers around a full $360^\circ$ circle; The angle between consecutive numbers is $360^\circ \div 12 = 30^\circ$; The minute hand makes one full turn ($360^\circ$) in $60$ minutes; The hour hand makes one full turn ($360^\circ$) in $12$ hours; Time shown: $4{:}20$ PM; Answer choices: (A) $0$, (B) $5$, (C) $8$, (D) $10$, (E) $12$
Unknowns: The acute angle (in degrees) between the two hands at $4{:}20$
Understand
Restated: At 4:20 PM, what is the measure of the acute angle between the hour hand and the minute hand of a standard 12-hour clock?
Givens: Standard clock with $12$ equally-spaced numbers around a full $360^\circ$ circle; The angle between consecutive numbers is $360^\circ \div 12 = 30^\circ$; The minute hand makes one full turn ($360^\circ$) in $60$ minutes; The hour hand makes one full turn ($360^\circ$) in $12$ hours; Time shown: $4{:}20$ PM; Answer choices: (A) $0$, (B) $5$, (C) $8$, (D) $10$, (E) $12$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Problem
A clock face is already a labeled circular diagram — Tool #1 (Draw a Diagram) lets us mark each hand's position on the dial and read the gap directly instead of juggling formulas. The trap is assuming the hour hand sits exactly on the $4$ at $4{:}20$; the diagram makes the small drift visible. Tool #9 (Solve an Easier Problem) supports it by splitting the question into two simpler sub-problems we already know how to do: first find where the minute hand is at $4{:}20$, then find where the hour hand is, and subtract. Direct rates ($6^\circ$ per minute for the minute hand, $0.5^\circ$ per minute for the hour hand) finish each sub-problem in one multiplication.
Execute — Answer: D
4.MD.C.5 Step 1 - Set the scale on the dial.
- The clock face is a $360^\circ$ circle divided into $12$ equal sectors by the numbers, so the gap between any two consecutive numbers is $360^\circ \div 12 = 30^\circ$.
- Measure both hand positions clockwise from the $12$.
💡 Grade 4 says a full turn is $360^\circ$ and the $12$ equal numbers split that turn into $30^\circ$ pieces — the dial is a built-in protractor.
4.MD.C.7 Step 2 - Locate the minute hand.
- The minute hand makes a full $360^\circ$ turn in $60$ minutes, so it sweeps $360^\circ \div 60 = 6^\circ$ per minute.
- At $20$ minutes past the hour it has moved $20 \times 6^\circ = 120^\circ$ clockwise from the $12$.
- That is exactly on the $4$ ($4 \times 30^\circ = 120^\circ$).
💡 Sub-problem one: a steady angular rate times the number of minutes gives the swept angle — a Grade 4 "add angle pieces" idea, here done as one multiplication.
4.MD.C.7 Step 3 - Locate the hour hand.
- The hour hand makes a full $360^\circ$ turn in $12$ hours $= 720$ minutes, so it sweeps $360^\circ \div 720 = 0.5^\circ$ per minute.
- At $4{:}00$ it sits on the $4$ at $120^\circ$.
- In the $20$ minutes from $4{:}00$ to $4{:}20$ it drifts an extra $20 \times 0.5^\circ = 10^\circ$ past the $4$.
💡 Sub-problem two: the hour hand keeps moving between the hour marks. The drift is small ($10^\circ$) but it is the whole point of the problem.
4.MD.C.7 Step 4 - Subtract the two positions on the diagram.
- The minute hand is at $120^\circ$ and the hour hand is at $130^\circ$, both measured from the $12$, so the gap between them is the difference.
💡 Once both hands have angle addresses on the same dial, the angle between them is just one subtraction — the diagram does the rest.
4.MD.C.5 Set the scale on the dial. The clock face is a $360^\circ$ circle divided into $ 4.MD.C.7 Locate the minute hand. The minute hand makes a full $360^\circ$ turn in $60$ mi 4.MD.C.7 Locate the hour hand. The hour hand makes a full $360^\circ$ turn in $12$ hours 4.MD.C.7 Subtract the two positions on the diagram. The minute hand is at $120^\circ$ and Review
Reasonableness: The $10^\circ$ answer matches the dial picture. At $4{:}20$ the minute hand is exactly on the $4$ and the hour hand sits a short way past the $4$ toward the $5$. The full gap from $4$ to $5$ is $30^\circ$, and $20$ minutes is one-third of an hour, so the hour hand has covered one-third of that gap: $\tfrac{1}{3} \times 30^\circ = 10^\circ$. That matches choice (D) and rules out (A) $0$ (would require both hands on the same spot), (B) $5$ and (C) $8$ (too small for a one-third drift past the $4$), and (E) $12$ (too large).
Alternative: Tool #9 (Solve an Easier Problem) in fraction form: in $20$ minutes the hour hand moves $\tfrac{20}{60} = \tfrac{1}{3}$ of the way from the $4$ to the $5$. Since the gap from one number to the next is $30^\circ$, the hour hand sits $\tfrac{1}{3} \times 30^\circ = 10^\circ$ past the $4$. The minute hand is on the $4$, so the angle between them is exactly that drift: $10^\circ$, choice (D).
CCSS standards used (min grade 5)
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 each hand has a measurable angle position.)4.MD.C.7Recognize angle measure as additive; solve addition and subtraction problems to find unknown angles (Computing the minute-hand and hour-hand positions by adding swept angles, then subtracting to get the angle between the hands.)5.NF.B.6Solve real world problems involving multiplication of fractions and mixed numbers (Multiplying the hour-hand rate $0.5^\circ$ per minute by $20$ minutes (equivalently $\tfrac{1}{3} \times 30^\circ$) to get the $10^\circ$ drift past the $4$.)
⭐ At $4{:}20$ the minute hand is exactly on the $4$, but the hour hand has already drifted one-third of the way toward the $5$ — and one-third of the $30^\circ$ gap between consecutive numbers is the $10^\circ$ answer.
⭐ At $4{:}20$ the minute hand is exactly on the $4$, but the hour hand has already drifted one-third of the way toward the $5$ — and one-third of the $30^\circ$ gap between consecutive numbers is the $10^\circ$ answer.
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