6.3 Clock Hand Angles (Saat Problemleri), Modular Calendar Arithmetic & Compass Bearings

Clock Hand Angles (Saat Problemleri), Modular Calendar Arithmetic & Compass Bearings

Quick Method: For any analog clock time $H\text{:}M$, compute $\theta = \lvert 30H - 5.5M \rvert$ (and take $360^\circ - \theta$ if $\theta > 180^\circ$). For calendar and recurring shift problems, find total elapsed days $\Delta D$ using $\text{LCM}$ and evaluate $\Delta D \pmod{7}$. For compass walks (Yön Problemleri), sum net $\text{East} - \text{West}$ and $\text{North} - \text{South}$ coordinates and apply a standard Pythagorean triple.

The final cluster of classic quantitative reasoning problems on the TR-YÖS Sayısal Yetenek Testi covers circular and modular geometry: analog clock hands (akrep ve yelkovan), modular calendar cycles (gün, ay, yıl ve nöbet problemleri), and orthogonal grid navigation (yön ve konum problemleri). Every problem in this group reduces to a deterministic formula.

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1. Clock Hand Angle Mathematics (Saat ve Açı Problemleri)

An analog clock face is a $360^\circ$ circle divided into $12$ hour sectors (saat aralığı) of $360^\circ / 12 = 30^\circ$ each, and $60$ minute ticks (dakika aralığı) of $360^\circ / 60 = 6^\circ$ each.

Angular Speeds of the Two Hands

• Minute Hand (Yelkovan): Sweeps $360^\circ$ in $60$ minutes, so its angular speed is:

$$v_{\text{min}} = \frac{360^\circ}{60\text{ min}} = 6^\circ\text{ per minute}$$

• Hour Hand (Akrep): Sweeps $1$ hour sector ($30^\circ$) in $60$ minutes, so its angular speed is:

$$v_{\text{hour}} = \frac{30^\circ}{60\text{ min}} = 0.5^\circ\text{ per minute} = \frac{1}{2}^\circ\text{ per minute}$$

• Relative Angular Speed (Bağıl Açısal Hız): Every minute, the minute hand gains:

$$v_{\text{rel}} = 6^\circ - 0.5^\circ = 5.5^\circ = \frac{11}{2}^\circ\text{ per minute}$$

The Universal Clock-Angle Formula

At $H$ hours and $M$ minutes (where $H \in \{0, 1, \dots, 11\}$, replacing $12$ with $0$, and converting any 24-hour time like 15:40 to $H = 3, M = 40$), the angle between the hour hand and minute hand is:

$$\theta = \left\lvert 30H - \frac{11}{2}M \right\rvert = \lvert 30H - 5.5M \rvert$$

• If $\theta \le 180^\circ$, then $\theta$ is the smaller interior angle (küçük açı).
• If $\theta > 180^\circ$, then $\theta$ is the reflex outer angle (büyük açı), and the smaller interior angle is:

$$\theta_{\text{interior}} = 360^\circ - \theta$$

Coincidence ($0^\circ$), Opposition ($180^\circ$), and Broken Clocks

• Hands Overlapping (Üst Üste Gelme, $\theta = 0^\circ$): Set $30H = \frac{11}{2}M \implies M = \frac{60H}{11}$ minutes past hour $H$. (Occurs $11$ times in $12$ hours, every $\frac{720}{11} = 65\frac{5}{11}$ minutes.)
• Hands Opposite (Doğrusal / Zıt Yönlü, $\theta = 180^\circ$): Set $\lvert 30H - \frac{11}{2}M \rvert = 180^\circ$.
• Fast / Slow Broken Clocks (İleri / Geri Kalan Saatler): Because an analog clock repeats its display every $12\text{ hours} = 720\text{ minutes}$, two clocks whose net drift apart is $\Delta v$ minutes per hour (or per day) will show the exact same time again when their accumulated time gap reaches $720\text{ minutes}$:

$$\text{Synchronization Time} = \frac{720\text{ minutes}}{\Delta v}$$

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2. Modular Calendar & Time-Cycle Arithmetic (Takvim ve Periyodik Tekrar Problemleri)

Day-of-the-Week Modulo 7 (Gün Hesaplama)

Since the seven days of the week repeat strictly with period $7$, adding or subtracting any multiple of $7$ days returns you to the same weekday:

$$\text{Target Day} = (\text{Current Day} + \Delta\text{Days}) \pmod{7}$$

• Moving into the future (sonra): Add the remainder $R = \Delta\text{Days} \bmod 7$ forward through the week.
• Moving into the past (önce): Subtract the remainder $R = \Delta\text{Days} \bmod 7$ backward through the week.

Ordinary vs. Leap Years (Normal Yıl ve Artık Yıl)

• Ordinary Year (Normal Yıl, $365$ days): Since $365 = 52 \times 7 + 1 \equiv 1 \pmod{7}$, any fixed calendar date advances by $+1$ weekday per ordinary year.
• Leap Year (Artık Yıl, $366$ days, February has $29$ days): Occurs when the year is divisible by $4$ (and for century years like $2000$ or $2100$, divisible by $400$). Since $366 = 52 \times 7 + 2 \equiv 2 \pmod{7}$, any interval that crosses February 29 shifts the weekday forward by $+2$ weekdays.

Periodic Duty Shifts (Nöbet Problemleri)

If Worker $A$ is on duty every $p_1$ days and Worker $B$ is on duty every $p_2$ days:

1. Their joint duty period is $P = \text{LCM}(p_1, p_2)$ days.
2. Indexing Fencepost Rule: If their 1st joint shift is already given on Day $0$, then their $k$-th joint shift occurs after $(k - 1) \times P$ elapsed days (not $k \times P$!).

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3. Compass Bearings & Grid Displacement (Yön ve Konum Problemleri)

When a person walks along a sequence of cardinal directions (Kuzey = North $+y$, Güney = South $-y$, Doğu = East $+x$, Batı = West $-x$) or makes $90^\circ$ left/right turns:

1. Track Net Coordinates $(\Delta x, \Delta y)$:

$$\Delta x = \sum \text{East} - \sum \text{West}, \qquad \Delta y = \sum \text{North} - \sum \text{South}$$

1. Shortest Straight-Line Distance (En Kısa / Kuş Uçuşu Uzaklık):

$$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

Useful practice values for $(\lvert \Delta x \rvert, \lvert \Delta y \rvert)$ include classic Pythagorean triples: $(3, 4, 5)$, $(6, 8, 10)$, $(5, 12, 13)$, $(8, 15, 17)$, or $(7, 24, 25)$.

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Summary Reference Table: Clocks, Calendars & Bearings

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Worked Examples

Worked Example 6.5: Clock Hand Angle at a 24-Hour Digital Time

Problem: What is the smaller interior angle (küçük açı) between the hour hand (akrep) and minute hand (yelkovan) of an analog clock at 16:48?

Solution:

1. Convert 24-hour time to 12-hour analog coordinates: $16\text{:}48 \implies H = 16 - 12 = 4$ and $M = 48$.
2. Substitute into the universal clock-angle formula:

$$\theta = \lvert 30H - 5.5M \rvert = \lvert 30(4) - 5.5(48) \rvert$$

Compute $5.5 \times 48 = \frac{11}{2} \times 48 = 11 \times 24 = 264^\circ$:

$$\theta = \lvert 120^\circ - 264^\circ \rvert = 144^\circ$$

1. Verify interior range: Since $144^\circ \le 180^\circ$, the smaller interior angle between the hands at 16:48 is $144^\circ$.

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Worked Example 6.6: Recurring Hospital Duty Schedule (Nöbet Problemi)

Problem: Two doctors, Dr. Arda and Dr. Burcu, work recurring night shifts every $6$ days and every $8$ days, respectively. They hold their 1st joint night shift on a Tuesday. On what day of the week will they hold their 10th joint night shift?

Solution:

1. Find the joint shift cycle via LCM:

$$P = \text{LCM}(6, 8) = 24\text{ days}$$

1. Count elapsed days from the 1st shift to the 10th shift: Because the 1st joint shift is already completed on Tuesday, there are $10 - 1 = 9$ intervals between the 1st and 10th shifts:

$$\Delta\text{Days} = 9 \times 24 = 216\text{ days}$$

1. Reduce modulo 7:

$$216 \div 7 = 30\text{ weeks with remainder } 6 \quad (216 = 7 \times 30 + 6 \equiv 6 \pmod{7})$$

(Shortcut: reduce before multiplying: $9 \equiv 2 \pmod{7}$ and $24 \equiv 3 \pmod{7}$, so $9 \times 24 \equiv 2 \times 3 = 6 \pmod{7}$!) 4. Advance from Tuesday: Adding $+6$ days (which is equivalent to subtracting $-1$ day) from Tuesday lands on Monday.

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Worked Example 6.6B: Multi-Year Leap Calendar Shift & Orthogonal Compass Walk

Problem (Part A — Leap Calendar): If 10 February 2024 falls on a Saturday, on what day of the week does 10 February 2028 fall?

Solution (Part A):

1. Check each 1-year transition for February 29 (Artık Yıl):
   • 10 Feb 2024 $\to$ 10 Feb 2025: crosses 29 February 2024 (since $2024$ is divisible by $4$), so it shifts by $+2$ weekdays.
   • 10 Feb 2025 $\to$ 10 Feb 2026: ordinary year ($365$ days), shifts by $+1$ weekday.
   • 10 Feb 2026 $\to$ 10 Feb 2027: ordinary year ($365$ days), shifts by $+1$ weekday.
   • 10 Feb 2027 $\to$ 10 Feb 2028: stops on 10 Feb 2028 before 29 February 2028, so it contains only $365$ days and shifts by $+1$ weekday!
2. Sum total weekday shifts modulo 7:

$$\Delta\text{Weekdays} = 2 + 1 + 1 + 1 = +5\text{ days}$$

Advancing $+5$ days from Saturday (or stepping $-2$ days backward) lands on Thursday.

Problem (Part B — Compass Bearing Walk): A courier starts at origin $O(0,0)$ facing North (Kuzey), walks $9\text{ km}$ forward, turns $90^\circ$ left (sola) and walks $5\text{ km}$, turns $90^\circ$ left again and walks $21\text{ km}$, and finally turns $90^\circ$ right (sağa) and walks $4\text{ km}$ to point $P$. Find the straight-line distance $\lvert OP \rvert$.

Solution (Part B):

1. Convert each turn into a cardinal direction:
   • Step 1: $9\text{ km}$ North ($+9$ on $y$-axis).
   • Step 2: $90^\circ$ left from North is West ($-5$ on $x$-axis).
   • Step 3: $90^\circ$ left from West is South ($-21$ on $y$-axis).
   • Step 4: $90^\circ$ right from South is West ($-4$ on $x$-axis).
2. Compute net displacements $(\Delta x, \Delta y)$:

$$\Delta x = -5 - 4 = -9\text{ km (9 km West)}, \qquad \Delta y = +9 - 21 = -12\text{ km (12 km South)}$$

1. Apply the $(9, 12, 15)$ Pythagorean triple ($3 \times (3, 4, 5)$):

$$\lvert OP \rvert = \sqrt{(-9)^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ km}$$

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Common TR-YÖS Clock, Calendar & Bearing Traps to Avoid

• Trap 1 — Forgetting the Reflex Angle Check ($\theta > 180^\circ$): At 11:10, the formula gives $\lvert 30(11) - 5.5(10) \rvert = \lvert 330^\circ - 55^\circ \rvert = 275^\circ$. Never pick $275^\circ$ when asked for the smaller angle (küçük açı); subtract from $360^\circ$ to get $360^\circ - 275^\circ = 85^\circ$.
• Trap 2 — Multiplying by $k$ Instead of $(k - 1)$ on Duty Problems: When the 1st shared shift is already given, the $k$-th shared shift is $(k - 1)$ cycles later.
• Trap 3 — Counting a Leap Year Without Crossing Feb 29: A leap year only adds $+2$ weekdays if the date interval actually passes through February 29.

Review the Method

Compass bearing and grid walk puzzles collapse into net horizontal displacement Delta_x (East minus West) and net vertical displacement Delta_y (North minus South), with straight-line distance d = sqrt(Delta_x^2 + Delta_y^2).