← General Intelligence & Reasoning · IOCL Graduate Engineer Civil

Chapter 12 of 12

Clocks, Calendars, Cubes & Dice

In the IOCL Graduate Engineer Civil syllabus under General Intelligence & Reasoning · 3 parts

Clocks · Calendars · Cubes & Dice

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📑 Contents (20 sections)

Part 1 of 3

Clocks

Last reviewed 22 Sept 2026 · 5 min read

How fast the hands move

Hand	In 60 minutes	Per minute
Minute hand	$360°$	$6°$
Hour hand	$30°$	$0.5°$
Relative (minute gains on hour)	$330°$	$5.5°$

FormulaThe angle between the hands

At $h$ hours and $m$ minutes:

θ = ∣ 30 h - 5.5 m ∣

If the result exceeds $180°$ , the smaller angle is $360° - θ$ .

The $30 h$ is the hour hand's position from 12, and $5.5 m$ is how much the minute hand has gained on it.

The standard positions

FormulaCoincidence, opposition and right angles

The hands coincide when $θ = 0$ , i.e. $30 h = 5.5 m$ , so $m = \frac{60 h}{11}$ .
They are opposite ( $180°$ apart) when $∣ 30 h - 5.5 m ∣ = 180$ .
They are at right angles when the expression equals $90$ — which happens twice in most hours.

Counts in 12 hours: coincide 11 times, opposite 11 times, at right angles 22 times. In 24 hours: 22, 22 and 44.

Because the hands coincide 11 times in 12 hours, successive coincidences are $\frac{12 \times 60}{11} = 65 \frac{5}{11}$ minutes apart — not 65 minutes, which is what a "too fast" clock question usually hinges on.

Clocks that gain or lose

A clock that shows the hands coinciding every 65 minutes of true time is gaining, because a correct clock takes $65 \frac{5}{11}$ minutes. The gain in a day is

(\frac{5}{11} \div 65) \times 24 \times 60 minutes

More generally: if a clock gains $g$ minutes in 24 hours, then in $t$ hours of true time it shows $t (1 + \frac{g}{1440})$ hours.

Mirror images

The mirror image of a clock time is found by subtracting it from 11:60 (that is, from 12:00) when the time is between 1 and 11 o'clock; for times between 12:00 and 1:00, subtract from 23:60.

Example: the mirror of 4:20 is $11 : 60 - 4 : 20 = 7 : 40$ .

Worked examples

Worked ExampleExample 1 — angle at a given time

Find the angle between the hands at 3:40.

Solution. $θ = ∣30 \times 3 - 5.5 \times 40∣ = ∣90 - 220∣ = 130°$ .

Worked ExampleExample 2 — an angle over 180

Find the angle at 8:10.

Solution. $∣240 - 55∣ = 185°$ → the smaller angle is $360 - 185 =$ 175°.

Worked ExampleExample 3 — coincidence

At what time between 4 and 5 o'clock do the hands coincide?

Solution. $m = \frac{60 \times 4}{11} = \frac{240}{11} = 21 \frac{9}{11}$ → at 4:21 9/11.

Worked ExampleExample 4 — opposite hands

At what time between 5 and 6 are the hands opposite?

Solution. $∣150 - 5.5 m ∣ = 180 \Rightarrow 5.5 m = 330 \Rightarrow m = 60$ — that is 6 o'clock exactly, so between 5 and 6 they are never opposite; the opposition happens at 6:00. (Between 5 and 6 the solution $5.5 m = - 30$ is negative and rejected.)

Worked ExampleExample 5 — right angles

At what times between 2 and 3 are the hands at right angles?

Solution. $∣60 - 5.5 m ∣ = 90$ gives $5.5 m = 150$ or $5.5 m = - 30$ . The first gives $m = 27 \frac{3}{11}$ → 2:27 3/11; the second is rejected, so there is one such time in this hour.

Worked ExampleExample 6 — counting

How many times in 24 hours are the hands at right angles?

Solution. 44 times.

Worked ExampleExample 7 — a gaining clock

A clock's hands coincide every 64 minutes of correct time. How much does it gain in a day?

Solution. A correct clock takes $65 \frac{5}{11} = \frac{720}{11}$ minutes. The clock runs $\frac{720}{11} - 64 = \frac{720 - 704}{11} = \frac{16}{11}$ minutes fast in every 64 minutes shown. In 24 hours of true time: $\frac{16/11}{64} \times 1440 = \frac{16 \times 1440}{11 \times 64} = \frac{360}{11} =$ 32 8/11 minutes gained.

Worked ExampleExample 8 — the true time

A clock gains 5 minutes a day. It was set right at 8 a.m. on Monday. What will it show at 8 a.m. on Wednesday?

Solution. Two days → 10 minutes fast → it shows 8:10 a.m.

Worked ExampleExample 9 — a mirror image

What is the mirror image of 7:25?

Solution. $11 : 60 - 7 : 25 =$ 4:35.

Worked ExampleExample 10 — angle at a round hour

What is the angle at 5 o'clock?

Solution. $30 \times 5 =$ 150°.

Part 2 of 3

Calendars

Last reviewed 22 Sept 2026 · 6 min read

Odd days

Every calendar question reduces to counting odd days — the remainder when a number of days is divided by 7.

Odd days = total days mod 7

The remainder is then read as a day of the week, counting from a known reference:

Odd days	0	1	2	3	4	5	6
Day	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday

Leap years

FormulaThe rule

A year is a leap year if it is divisible by 4, except a century year, which must be divisible by 400.

So 1996, 2004, 2024 and 2000 are leap years; 1900, 2100 and 2200 are not.

An ordinary year has 365 days $= 52$ weeks $+ 1$ odd day. A leap year has 366 days $= 52$ weeks $+ 2$ odd days.

Odd days in longer periods

Period	Odd days
Ordinary year	1
Leap year	2
100 years	5
200 years	3
300 years	1
400 years	0

Because 400 years contain no odd days, the calendar repeats every 400 years — 1 January 1601, 2001 and 2401 all fall on the same weekday.

Odd days in each month

Month	Days	Odd days
January	31	3
February	28 / 29	0 / 1
March	31	3
April	30	2
May	31	3
June	30	2
July	31	3
August	31	3
September	30	2
October	31	3
November	30	2
December	31	3

Finding the day of a date

Count the odd days in the complete centuries before the year.
Add the odd days of the complete years since: one for each ordinary year, two for each leap year.
Add the odd days of the complete months of the given year, plus the date itself.
Take the total modulo 7 and read off the day.

Repeating calendars

Add odd days year by year until the running total is a multiple of 7 — and then check one more thing: the repeating year must have the same leap status as the original, or the two calendars agree only until the end of February. An ordinary year's calendar therefore returns after 6 or 11 years, and a leap year's after 28 (away from the century exceptions).

Worked examples

Worked ExampleExample 1 — the day of a date

What day was 15 August 1947?

Solution. 1600 years → 0 odd days. 300 more years (to 1900) → 1 odd day. 46 complete years (1901–1946): leap years are 1904, 1908, … 1944 → 11 leap years, 35 ordinary → $11 \times 2 + 35 = 57$ odd days → $57 mod 7 = 1$ . In 1947 to 15 August: Jan 31 + Feb 28 + Mar 31 + Apr 30 + May 31 + Jun 30 + Jul 31 + 15 = 227 → $227 mod 7 = 3$ . Total $= 1 + 1 + 3 = 5$ → Friday.

Worked ExampleExample 2 — a year's first day

1 January 2020 was a Wednesday. What day was 1 January 2021?

Solution. 2020 was a leap year → 2 odd days → Wednesday + 2 = Friday.

Worked ExampleExample 3 — the last day of a year

If 1 January of a non-leap year is a Monday, what day is 31 December of the same year?

Solution. An ordinary year has 1 odd day, so it ends on the same weekday it began → Monday.

Worked ExampleExample 4 — days between dates

How many days are there from 3 March to 17 May in a non-leap year?

Solution. March: $31 - 3 = 28$ remaining; April 30; May 17 → $28 + 30 + 17 =$ 75 days after 3 March.

Worked ExampleExample 5 — a leap year check

Was 1900 a leap year?

Solution. It is a century year and $1900 \div 400$ is not whole → no.

Worked ExampleExample 6 — same day next month

If 5 March is a Tuesday, what day is 5 April of the same year?

Solution. March has 31 days → 3 odd days → Tuesday $+ 3 =$ Friday.

Worked ExampleExample 7 — a date some days later

If today is Thursday, what day will it be after 100 days?

Solution. $100 mod 7 = 2$ → Thursday $+ 2 =$ Saturday.

Worked ExampleExample 8 — counting back

If 26 January 2024 was a Friday, what day was 26 January 2023?

Solution. From January 2023 to January 2024 spans 365 days (the leap day of 2024 falls later, in February) → 1 odd day → Friday $- 1 =$ Thursday.

Worked ExampleExample 9 — a repeating calendar

In which year will the calendar of 2026 repeat? (2028, 2032 and 2036 are leap years.)

Solution. Add the odd days year by year from 2026: 1, 1, 2 (2028), 1, 1, 1 — a total of 7 after 2031, so 1 January 2032 falls on the same weekday as 1 January 2026.

But 2032 is a leap year and 2026 is not, so the two calendars part company at the end of February. Keep going: 2032 → 2, 2033 → 1, 2034 → 1, 2035 → 1, 2036 → 2, giving 14 in all after 2036. So the whole calendar of 2026 returns in 2037, an ordinary year — the usual 11-year gap.

Part 3 of 3

Cubes & Dice

Last reviewed 22 Sept 2026 · 6 min read

A painted cube, cut up

A cube painted on all six faces and cut into $n^{3}$ smaller cubes of equal size gives:

FormulaThe four counts

3 faces painted (corners) = 8

2 faces painted (edges) = 12 (n - 2)

1 face painted (face centres) = 6 (n - 2)^{2}

0 faces painted (inside) = (n - 2)^{3}

They must total $n^{3}$ , which is the check to run before answering.

For $n = 4$ : $8 + 24 + 24 + 8 = 64$ ✓. For $n = 5$ : $8 + 36 + 54 + 27 = 125$ ✓.

When only some faces are painted

If only the top and bottom are painted, no small cube can have three painted faces; the counts must be rebuilt from the picture. The safe method is to think of the big cube as $n$ layers and count layer by layer, rather than to reuse a formula that assumed all six faces.

Dice

A dice has six faces in three opposite pairs. In a standard dice the numbers on opposite faces add to 7 (1-6, 2-5, 3-4); in an ordinary dice the pairs must be worked out from the views given.

FormulaFinding opposite faces from two views

One face common to both views: the remaining four faces (two from each view) are adjacent to the common face; the face not appearing in either view is opposite to the common one.
Two faces common to both views: the two faces that are not common are opposite to each other.
Any two faces seen together in a single view are adjacent, never opposite — this single fact answers most questions.

A useful corollary: if a number appears in a view together with four different numbers across several views, the fifth unseen number is its opposite.

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