Competition · AMC preparation · step 4 of 4

AMC 8 · 2003 · #20

Grade 5 geometry-2d

Archetype Angle Chase in a Synthetic Figure └ Sub-archetype Fixed-Instrument Angle Composition

rateangle-sum-trianglefraction-arithmetic identify-subproblemsdimensional-analysis ↑ Prerequisites: multi-digit-arithmeticfraction-arithmetic

📏 Medium solution 💡 3 insights

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Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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° per minute for the minute hand, 0.5° per minute for the hour hand) finish each sub-problem in one multiplication.

1STEP 1

Find the degrees per number

Set the dial's scale: the 360° face splits into 12 equal sectors, so consecutive numbers sit 30° apart.

360°/12 = 30° between consecutive numbers

2STEP 2

Place the minute hand

The minute hand sweeps 6° per minute, so at 20 minutes it reaches 120° — exactly on the 4.

minute hand = 20 × 6° = 120°

3STEP 3

Place the hour hand

The hour hand creeps 0.5° per minute, so from 4{:}00 it drifts 10° past the 4 to 130°.

hour hand = 120° + 20 × 0.5° = 120° + 10° = 130°

Sub-problem two: the hour hand keeps moving between the hour marks. The drift is small (10°) but it is the whole point of the problem.

At 4:20 the hour hand no longer sits on the 4; it has drifted forward toward the 5 by the same share of the number-gap as the share of the hour that has already passed.

▸ Why?

The hour hand crosses exactly one number-gap each hour, so between 4:00 and 5:00 it travels the whole gap from the 4 to the 5 at a steady pace.

▸ Why?

The twelve numbers split the full turn into twelve equal gaps, and the hand makes one full turn in twelve hours, so each single hour is spent crossing exactly one of those equal gaps.

▸ Why?

Twelve equal gaps that together fill the whole 360° turn must each be one-twelfth of it, a 30° step.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

There are twelve gaps and twelve hours in one full turn, so pairing one gap to each hour uses them up evenly — one gap per hour.

One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.

▸ Why?

Because the pace is steady, the part of that one-hour gap the hand covers equals the part of the hour that has elapsed — 20 of the 60 minutes covers 20 out of every 60 of the gap.

▸ Why?

The hour is just 60 equal one-minute steps, so after 20 of them the hand has taken 20 equal steps into the gap.

▸ Why?

One hour is the same as 60 minutes, a fixed exchange, so the hour breaks cleanly into 60 equal minute-steps.

A bigger unit is a fixed count of smaller unitsA bigger unit always equals the same fixed number of smaller units (1 hour = 60 minutes, 1 km = 1000 m), so a measure can be renamed between them.

▸ Why?

A steady pace moves the hand the same small amount every minute, so 20 minutes is just 20 copies of that one equal amount.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

4.MD.C.7Solve An Easier Related Problem

4STEP 4

Subtract the two positions

Both hands now have dial addresses, so the gap is one subtraction: 130° - 120° = 10°.

|130° - 120°| = 10° → (D)

Answer

The 10° 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°, and 20 minutes is one-third of an hour, so the hour hand has covered one-third of that gap:

\frac{1}{3}

× 30° = 10°. 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).

💡Key takeaway

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° gap between consecutive numbers is the 10° answer.

Find the degrees per number
Place the minute hand
Place the hour hand
Subtract the two positions

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Find the degrees per number

Place the minute hand

Place the hour hand

Subtract the two positions

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