Introduction
Swap problems ask for a time when the hour hand and minute hand have exchanged positions compared to another time. These require writing exact angular positions for both hands at two times and solving the two swap equations simultaneously. This pattern is important for advanced clock reasoning because it combines angle formulas, linear equations and normalization.
Pattern: Minute and Hour Hand Swap
Pattern
Key concept: Represent hour and minute hand positions in degrees (or minute-scale out of 60), write swap conditions, then solve the two linear equations.
Useful formulas (angles in degrees):
Hour hand at H hours and m minutes: Anglehour = 30H + 0.5m.
Minute hand at m minutes: Anglemin = 6m.
Swap condition between times T₁ and T₂ (hour/min positions interchanged):
Angle_hour(T₂) = Angle_min(T₁) and Angle_min(T₂) = Angle_hour(T₁).
Step-by-Step Example
Question
At what time between 3 and 4 o’clock will the hour and minute hands interchange positions with those they would have between 4 and 5 o’clock? (i.e., the swap between the 3-4 and 4-5 hour slots)
Solution
-
Step 1: Define unknowns and write angles
Let T₁ = 3 + x minutes (so x minutes past 3). Let T₂ = 4 + y minutes (y minutes past 4).
Hour angle at T₁ = 30×3 + 0.5x = 90 + 0.5x.
Minute angle at T₁ = 6x.
Hour angle at T₂ = 30×4 + 0.5y = 120 + 0.5y.
Minute angle at T₂ = 6y. -
Step 2: Write swap equations
Swap means hour angle at T₂ equals minute angle at T₁, and minute angle at T₂ equals hour angle at T₁. So:
(1) 120 + 0.5y = 6x
(2) 6y = 90 + 0.5x -
Step 3: Solve the linear system
From (2): y = 15 + x/12. Substitute in (1):
120 + 0.5(15 + x/12) = 6x
120 + 7.5 + x/24 = 6x
127.5 = 6x - x/24 = x(144/24 - 1/24) = x(143/24)
x = 127.5 × 24 / 143 = 3060 / 143 = 21 57/143 minutes (≈ 21.3993 min). -
Step 4: Compute the swapped time T₁ and T₂
T₁ = 3 : x = 3:21 57/143 (exact).
For completeness, y = 15 + x/12 = 15 + (3060/143)/12 = 15 + 255/143 = (2145 + 255)/143 = 2400/143 = 16 112/143 minutes (≈ 16.7832 min), so T₂ = 4:16 112/143. -
Final Answer:
3:21 57/143 -
Quick Check:
Compute angles numerically (approx): At 3:21.399 → hour ≈ 90 + 0.5×21.399 = 100.6995°; minute ≈ 6×21.399 = 128.394°. At 4:16.783 → hour ≈ 120 + 0.5×16.783 = 128.3915°; minute ≈ 6×16.783 = 100.698°. The hour angle at T₂ ≈ minute angle at T₁ and minute angle at T₂ ≈ hour angle at T₁ (matching to within rounding) ✅
Quick Variations
1. Swaps can be set up between any adjacent hours H and H+1 - follow the same two-equation approach.
2. Some problems ask for the swapped time in the same hour (interpret carefully - that usually reduces to coincidence).
3. For quick estimates, swaps often occur near the 21-22 minute mark past the hour (varies with H).
Trick to Always Use
- Step 1 → Let T₁ = H + x and T₂ = (H+1) + y (adjacent hours), write angles for both times.
- Step 2 → Use swap equations: Angle_hour(T₂)=Angle_min(T₁) and Angle_min(T₂)=Angle_hour(T₁).
- Step 3 → Solve the 2×2 linear system; normalize fractional minutes and verify numerically.
Summary
Summary
- Key takeaway 1: Model hour and minute angles precisely: Hour = 30H + 0.5m, Minute = 6m.
- Key takeaway 2: Swap conditions give two linear equations - solve simultaneously for minutes past each hour.
- Key takeaway 3: For a swap between H and H+1 the solution usually gives T₁ near 21-22 minutes past H (exact fractional form given above).
- Key takeaway 4: Always perform a numerical quick check of angles to confirm the swap.
Example to remember:
Between 3 and 4 the swap occurs at 3:21 57/143 (≈ 3:21.399).
Hint: Set hour(T₂)=minute(T₁) and minute(T₂)=hour(T₁), then use y = 5 + x/12.
Common Mistakes: Using right-angle formulas instead of swap equations.