4.4 Age, Calendar & Clock Problems

Age Problems

Age problems are linear-equation word problems dressed up. Define one person's current age as a variable, express every other age in terms of that variable, then write the equation the question gives you.

Worked example 1. A father is 4 times as old as his son. In 20 years, he will be twice as old as his son. Find the son's present age.

Let the son's age be x; the father is 4x. In 20 years, son = x + 20 and father = 4x + 20. The condition "twice as old" gives 4x + 20 = 2(x + 20). Expand: 4x + 20 = 2x + 40. So 2x = 20 and x = 10. The son is 10, the father 40. In 20 years, son 30, father 60 — twice, confirmed.

Worked example 2. The sum of the ages of a man and his son is 50. Five years ago, the man was 3 times as old as his son. Find their present ages.

Let the son be x, so the man is (50 - x). Five years ago the son was x - 5 and the man was 45 - x. The condition gives 45 - x = 3(x - 5) → 45 - x = 3x - 15 → 60 = 4x → x = 15. The son is 15 and the man is 35.

Check: five years ago they were 10 and 30 — exactly three times. Note the useful safety signal: RRB age problems are built to give whole-number answers, so a fractional result almost always means you mis-set the equation rather than mis-solved it. Re-read which age the multiple applies to before redoing the arithmetic.

Pattern: define today's ages, shift by ±k for past/future, and equate the multiple or sum. Age ratio questions use the same approach: if A:B = 5:3 today, write A = 5k and B = 3k.

Calendar Problems — The Odd Days Method

An odd day is the remainder when a number of days is divided by 7 (since 7 days complete a week). The day of the week depends only on the total odd days.

• Normal year: 365 days = 52 weeks + 1 odd day.
• Leap year: 366 days = 52 weeks + 2 odd days.

A year is a leap year if divisible by 4; century leap years must be divisible by 400 (so 2000 was a leap year, 1900 was not).

Odd Days per Month

Worked example 3. What day of the week was 26 January 2025?

Take a known reference: 1 January 2025 is a Wednesday. (A quicker method: count odd days from a fixed anchor.) Days from 1 Jan to 26 Jan = 25 days. 25 mod 7 = 4. Wednesday + 4 days = Sunday. So 26 January 2025 was a Sunday.

Worked example 4. If 1 January 2024 was a Monday, what day was 1 January 2025?

2024 is a leap year, so it contributes 2 odd days. Monday + 2 = Wednesday. So 1 January 2025 was a Wednesday (matches example 3).

Century Anchors

In an ordinary century — say 1901 to 2000 counted as 100 years ending on a non-leap century — there are 76 normal years and 24 leap years, because the century year itself is not a leap year unless it is divisible by 400. Odd days = 76 × 1 + 24 × 2 = 124, and 124 mod 7 = 5.

Stack the centuries and reduce mod 7 each time:

The 400-year figure is 0, not 6, and the reason is the extra leap day: a 400-year block contains 97 leap years (100 multiples of 4, minus the 3 century years that are not divisible by 400), giving 303 × 1 + 97 × 2 = 497 odd days, and 497 = 71 × 7 exactly. That is why the calendar repeats identically every 400 years — 1 January 2001 fell on the same weekday as 1 January 1601.

Clock Problems

The minute hand moves 360° in 60 minutes = 6°/min. The hour hand moves 360° in 12 hours = 0.5°/min. So the minute hand gains 5.5° per minute on the hour hand.

Angle between hands at H hours and M minutes:

angle = |30H - 5.5M|

Take the smaller of the angle and 360° minus the angle, since the hands form two angles that sum to 360°.

Worked example 5. Find the angle between the hands at 3:20.

H = 3, M = 20. Angle = |30 × 3 - 5.5 × 20| = |90 - 110| = 20°. The hands are 20° apart.

Worked example 6. At what time between 4 and 5 o'clock are the hands of a clock opposite each other (180°)?

At 4:00 the hour hand is at 120° and the minute hand at 0°, so the minute hand starts 120° behind. To end up 180° ahead of the hour hand it must close that 120° gap and then open a fresh 180° on the other side — a total relative gain of 120 + 180 = 300°. At 5.5° gained per minute that takes 300 / 5.5 = 600/11 = 54 6/11 minutes, so the time is 4:54 6/11, about 4:54:33.

Check it with the angle formula: |30 × 4 − 5.5 × 54.545| = |120 − 300| = 180° ✓.

The general rule is 5.5M = 30H + 180 (or 30H − 180, whichever lands in the 0–60 minute range for the hour in question). The common error is to compute only the 60° that separates 120° from 180° — that gives 10 10/11 minutes, which is the moment the hands are 60° apart, not opposite.

Mirror-Image Clock Times

A clock held in front of a mirror shows a reversed time. For a 12-hour clock, the actual time plus the mirror-displayed time = 11:60 (i.e., 11 hours 60 minutes, which is 12:00). So actual = 11:60 - displayed, with wraparound if the result is negative.

Worked example 7. A clock seen in a mirror shows 4:15. What is the actual time?

11:60 - 4:15 = 7:45. Verify: a clock reading 7:45, when reflected, shows 4:15. Correct.

Clocks Gaining or Losing Time

A clock that gains 5 minutes per day will read 12:05 at the true 12:00 a day later. To find the true time when the faulty clock reads T, scale by the ratio (true minutes / faulty minutes) per day. If a watch gains 10 minutes in 24 hours, then for every 24 hours of true time the watch shows 24 hours 10 minutes = 1450 minutes; to find the true time when the watch reads a given value, multiply by 1440/1450.

Exam Traps

• "x years ago" with the wrong sign: subtract k from current ages, not add.
• Forgetting century leap-year rule: 1900 was not a leap year; assuming it is will throw the day-of-week off by one.
• Angle formula using 6M instead of 5.5M: 6M ignores the hour hand's drift during the minutes, giving wrong answers.
• Mirror wraparound: 11:60 - 11:25 should give 0:35, not a negative number — add 12 hours if needed.
• Mixing 12-hour and 24-hour formats when a question mentions "the 24-hour clock" explicitly.

Speed Strategy

Age and clock problems should each take under 45 seconds once the equation is set up. Calendar questions need 60 seconds for the odd-day count. If a clock question demands "how many times do the hands coincide in a day," remember the answer is 22 (not 24, because the 11:00+ coincidence is the same as 12:00). Hands coincide 11 times in 12 hours and are opposite 11 times in 12 hours; they are at right angles 22 times in 12 hours.