1. Clock Angles
A clock's hour and minute hands move at different, constant speeds — the entire chapter is Week 9's relative-speed idea, applied to angular motion instead of distance.
Minute hand: 360 degrees / 60 min = 6 degrees per minute
Hour hand: 360 degrees / 720 min (12hrs x 60) = 0.5 degrees
per minute
Relative speed of minute hand vs. hour hand:
6 - 0.5 = 5.5 degrees per minute
(Week 9's "same direction" relative speed -- both hands
move the same way around the clock face)Method: compute each hand's position from 12 o'clock
independently, then find the difference.
Minute hand at 40 min: 40 x 6 = 240 degrees from 12
Hour hand at 3:40: at 3:00 exactly, hour hand is at
3 x 30 = 90 degrees. It's now 40 minutes further, and
moves 0.5 deg/min: 90 + 40x0.5 = 90+20 = 110 degrees
Angle between hands = |240 - 110| = 130 degrees
(if this exceeds 180, subtract from 360 -- the smaller angle
is the conventional answer)2. Odd Days & the Calendar
A week has 7 days, so any span of days, reduced modulo 7 (Week 1's remainder idea, reused), tells you how many weekdays the calendar has "shifted" — the odd days.
Normal year (365 days): 365 mod 7 = 1 odd day
Leap year (366 days): 366 mod 7 = 2 odd days
A leap year: divisible by 4, EXCEPT century years (ending in
00) must be divisible by 400 to still count as leap.
2000: leap (divisible by 400)
1900: NOT leap (divisible by 100 but not 400)
2024: leap (divisible by 4, not a century year)
Odd days for a 100-year block: 76 normal years (1 odd day
each) + 24 leap years (2 odd days each)
= 76 + 48 = 124 -> 124 mod 7 = 5 odd days per century
(with an adjustment for century years' leap-year exception
at 200, 300, 400-year marks)The core mechanism: each non-leap year shifts what weekday a given calendar date falls on by exactly 1 (since 365 = 52 weeks + 1 day), and each leap year shifts it by 2 — tracking this shift, accumulated across however many years a problem asks about, is the entire technique.
3. Finding the Day of the Week
Combining Section 2's odd-days logic with a reference day lets you find the weekday for any date, given one known anchor point.
0 odd days -> Sunday 4 odd days -> Thursday
1 odd day -> Monday 5 odd days -> Friday
2 odd days -> Tuesday 6 odd days -> Saturday
3 odd days -> Wednesday
Worked: January 1, 2001 was a Monday. What day was
January 1, 2005?
Years between: 2001, 2002, 2003, 2004 (4 years elapsed)
2001: normal (1 odd day)
2002: normal (1 odd day)
2003: normal (1 odd day)
2004: LEAP (2 odd days)
Total odd days = 1+1+1+2 = 5
Starting day: Monday (1 odd day reference)
Add 5 odd days: Monday + 5 = Saturday
January 1, 2005 was a Saturday.This worked example only needed to track full elapsed years because both dates are January 1st — a question asking about a date mid-year would additionally need the odd-days contribution of the partial year up to that specific date, using the same day-counting principle at a finer grain.
4. Age Problems
Age problems are Week 10's linear-equation translation exercise, specifically about ages — the "new" content here is really just applying that same translation process to a familiar-sounding but occasionally tricky phrasing.
The ratio of a father's age to his son's age is 5:2. After
8 years, the ratio will be 2:1. Find their current ages.
Step 1 (Week 6's single-multiplier ratio trick):
Father = 5x, Son = 2x
Step 2: translate "after 8 years" into an equation:
(5x+8) / (2x+8) = 2/1
Step 3: solve
5x+8 = 2(2x+8)
5x+8 = 4x+16
x = 8
Father = 5x = 40, Son = 2x = 16
Check: after 8 years, father=48, son=24 -- ratio 48:24 = 2:1
correctThis combines Week 6's ratio setup with Week 10's translation process directly — nothing in this chapter requires a genuinely new technique, which is exactly why age problems appear late in the syllabus, after both prerequisite skills are already solid.
5. Practice Problems
30 problems, no calculator — Easy, Medium & Tough
Set up every age problem with Week 6's ratio technique before translating to an equation. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.
Easy (1–10)
- Find the angle between the hour and minute hands at 3:00.
- Find the angle between the hour and minute hands at 6:00.
- How many times do the hands of a clock coincide in 12 hours?
- If today is Monday, what day will it be after 7 days?
- If today is Wednesday, what day was it 3 days ago?
- Is the year 2024 a leap year?
- Is the year 1900 a leap year?
- A is 5 years older than B. If B is 20, find A's age.
- Find the number of odd days in 1 (non-leap) year.
- Find the number of odd days in a leap year.
1) 90°. 2) 180°. 3) 11 times. 4) Monday (7 days = 1 full week). 5) Sunday. 6) Yes. 7) No (divisible by 100, not 400). 8) 25. 9) 1. 10) 2.
Medium (11–20)
- Find the angle between the hour and minute hands at 4:20.
- Find the angle between the hour and minute hands at 7:45.
- If January 1 is a Monday, what day is January 15 of the same year?
- Find the number of odd days in 3 normal (non-leap) years.
- A father is 4 times as old as his son. If the father is 40, find the son's age.
- The sum of the ages of a father and son is 60. If the father is 4 times as old as the son, find their ages.
- A is twice as old as B. Five years ago, A was 3 times as old as B. Find their current ages.
- If March 15 is a Tuesday, what day is April 15 of the same year (March has 31 days)?
- Find the time between 3 and 4 o'clock when the hands are exactly opposite each other.
- A is 6 years older than B, and their ages are in ratio 4:3. Find their ages.
11) Minute=120°, hour=4×30+20×0.5=130° → angle=10°. 12) Minute=270°, hour=7×30+45×0.5=232.5° → angle=37.5°. 13) 14 days elapsed, mod 7=0 → Monday. 14) 3×1=3 odd days. 15) 40/4=10. 16) F+S=60, F=4S → S=12, F=48. 17) A=2B; A-5=3(B-5) → B=10, A=20. 18) 16 days later, mod 7=3 → Tuesday+3=Friday. 19) |30h-5.5m|=180 at h=3: 5.5m=270 → m≈49.1 → about 3:49. 20) A-B=6, A=4B/3 → B=18, A=24.
Tough (21–30)
- Find the angle between the hour and minute hands at 5:40.
- At what time between 2 and 3 o'clock will the hands of a clock first be at right angles (90°)?
- If January 1, 2023 was a Sunday, what day was January 1, 2024? (2023 was not a leap year.)
- Find the number of odd days in 400 years.
- A father's age is currently 5 times his son's age. In 6 years, the father will be 3 times as old as the son. Find their current ages.
- The ages of A, B and C are in the ratio 3:5:7. If C is 20 years older than A, find B's age.
- Find the angle between the hour and minute hands at 10:10.
- In a leap year, if January 1 falls on a Thursday, find the day of the week for December 31 of the same year.
- A is currently three times as old as B. In 10 years, A will be twice as old as B. Find their current ages.
- A clock loses 15 minutes every 24 hours. If it's set correctly at noon, find the time it shows after exactly 24 real hours have passed.
21) Minute=240°, hour=5×30+40×0.5=170° → angle=70°. 22) |30(2)-5.5m|=90 → 5.5m=150 → m≈27.3 → about 2:27. 23) 2023 not leap → 1 odd day → Sunday+1=Monday. 24) 0 (the calendar repeats exactly every 400 years). 25) F=5S; F+6=3(S+6) → S=6, F=30. 26) 3:5:7 in x-multiples; C-A=4x=20 → x=5; B=5×5=25. 27) Minute=60°, hour=10×30+10×0.5=305° → angle=|305-60|=245°, take smaller=115°. 28) Leap year=366 days=2 odd days; Thursday+2=Saturday. 29) A=3B; A+10=2(B+10) → B=10, A=30. 30) Shows 23h45m elapsed → 11:45am the next day (15 min slow).
6. Knowledge Check
Four quick questions. Expand each to check your answer.
Q1
Why is the relative speed between a clock's minute and hour hands 5.5 degrees per minute, not 6?
Why is the relative speed between a clock's minute and hour hands 5.5 degrees per minute, not 6?
Both hands move in the same direction around the clock face, so their relative speed — how fast the minute hand gains on the hour hand — is the DIFFERENCE of their individual speeds (Week 9's same-direction relative speed rule), not the minute hand's speed alone. 6 degrees/min minus the hour hand's 0.5 degrees/min gives 5.5 degrees/min.
Q2
Why does a normal (365-day) year shift the weekday of a fixed date by exactly 1 day, year over year?
Why does a normal (365-day) year shift the weekday of a fixed date by exactly 1 day, year over year?
365 days is exactly 52 full weeks (364 days) plus 1 extra day — the 52 full weeks bring the calendar back to the same weekday, and that leftover single day is what pushes a fixed date's weekday forward by one position compared to the previous year.
Q3
Why is 1900 not a leap year despite being divisible by 4, while 2000 is?
Why is 1900 not a leap year despite being divisible by 4, while 2000 is?
The leap year rule has a special exception for century years (ending in 00): they must be divisible by 400, not just 4, to count as leap. 1900 is divisible by 100 but not 400, so the century exception disqualifies it; 2000 is divisible by 400, so it remains a leap year despite being a century year.
Q4
Why does setting up an age problem as "5x and 2x" (using Week 6's ratio technique) work better than using two separate unrelated unknowns?
Why does setting up an age problem as "5x and 2x" (using Week 6's ratio technique) work better than using two separate unrelated unknowns?
A stated current ratio (like 5:2) already fixes the relationship between the two ages up to one shared unknown scaling factor — using two independent unknowns would need an extra equation just to reconstruct that known ratio, which is redundant. Starting from 5x and 2x directly encodes the given ratio, leaving only the single new piece of information (the future ratio) to translate into one equation.