← Sensim Lab Sensim Math

AMC 8 · 2003 · #20

Easy mode Grade 5
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Problem

Picture the face of a clock at 4:20 PM.

The minute hand and the hour hand are pointing in slightly different directions. Between them is a small angle and a large angle that add up to a full circle.

What is the smaller of those two angles (in degrees)?

Pick an answer.

(A)
0
(B)
5
(C)
8
(D)
10
(E)
12
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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

#1 Draw a Diagram 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 $12$.
$$\dfrac{360^\circ}{12} = 30^\circ \text{ between consecutive numbers}$$

💡 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.

#9 Solve an Easier Problem 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 $4$ ($4 \times 30^\circ = 120^\circ$).
$$\text{minute hand} = 20 \times 6^\circ = 120^\circ$$

💡 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.

#9 Solve an Easier Problem 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 $4$.
$$\text{hour hand} = 120^\circ + 20 \times 0.5^\circ = 120^\circ + 10^\circ = 130^\circ$$

💡 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.

#1 Draw a Diagram 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.
$$|130^\circ - 120^\circ| = 10^\circ \;\Rightarrow\; \textbf{(D)}$$

💡 Once both hands have angle addresses on the same dial, the angle between them is just one subtraction — the diagram does the rest.

[1] #1 4.MD.C.5 Set the scale on the dial. The clock face is a $360^\circ$ circle divided into $
[2] #9 4.MD.C.7 Locate the minute hand. The minute hand makes a full $360^\circ$ turn in $60$ mi
[3] #9 4.MD.C.7 Locate the hour hand. The hour hand makes a full $360^\circ$ turn in $12$ hours
[4] #1 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.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 each hand has a measurable angle position.)
  • 4.MD.C.7 Recognize 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.6 Solve 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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