Veda Skills Logical Reasoning Unit 16: Clock and Calendar

$|30 \times 3 - 5.5 \times 40| = |90 - 220| = 130$°.

$|30 \times 7 - 5.5 \times 20| = |210 - 110| = 100$°.

$|30 \times 4 - 5.5 \times 20| = |120 - 110| = 10$°.

$|30 \times 2 - 5.5 \times 30| = |60 - 165| = 105$°.

$|30 \times 9 - 5.5 \times 15| = |270 - 82.5| = 187.5$°; it is over 180°, so the angle is $360 - 187.5 = 172.5$°.

$|30 \times 5 - 5.5 \times 25| = |150 - 137.5| = 12.5$°.

$|30 \times 11 - 5.5 \times 10| = |330 - 55| = 275$°; it is over 180°, so the angle is $360 - 275 = 85$°.

$|30 \times 8 - 5.5 \times 45| = |240 - 247.5| = 7.5$°.

$|30 \times 6 - 5.5 \times 35| = |180 - 192.5| = 12.5$°.

$|30 \times 10 - 5.5 \times 25| = |300 - 137.5| = 162.5$°.

$|30 \times 1 - 5.5 \times 50| = |30 - 275| = 245$°; it is over 180°, so the angle is $360 - 245 = 115$°.

$|30 \times 12 - 5.5 \times 36| = |360 - 198| = 162$°.

$|30 \times 4 - 5.5 \times 54| = |120 - 297| = 177$°.

$|30 \times 8 - 5.5 \times 18| = |240 - 99| = 141$°.

$|30 \times 2 - 5.5 \times 44| = |60 - 242| = 182$°; it is over 180°, so the angle is $360 - 182 = 178$°.

$|30 \times 11 - 5.5 \times 47| = |330 - 258.5| = 71.5$°.

$|30 \times 6 - 5.5 \times 12| = |180 - 66| = 114$°.

$|30 \times 9 - 5.5 \times 50| = |270 - 275| = 5$°.

The clock shows 2:20. $|30 \times 2 - 5.5 \times 20| = |60 - 110| = 50$°.

The clock shows 5:15. $|30 \times 5 - 5.5 \times 15| = |150 - 82.5| = 67.5$°.

The clock shows 10:40. $|30 \times 10 - 5.5 \times 40| = |300 - 220| = 80$°.

The clock shows 7:35. $|30 \times 7 - 5.5 \times 35| = |210 - 192.5| = 17.5$°.

The clock shows 1:25. $|30 \times 1 - 5.5 \times 25| = |30 - 137.5| = 107.5$°.

The clock shows 4:48. $|30 \times 4 - 5.5 \times 48| = |120 - 264| = 144$°.

$|30 \times 3 - 5.5 \times 25| = |90 - 137.5| = 47.5$°. Reflex angle $= 360 - 47.5 = 312.5$°.

$|30 \times 8 - 5.5 \times 10| = |240 - 55| = 185$°; it is over 180°, so the angle is $360 - 185 = 175$°. Reflex angle $= 360 - 175 = 185$°.

$|30 \times 12 - 5.5 \times 20| = |360 - 110| = 250$°; it is over 180°, so the angle is $360 - 250 = 110$°. Reflex angle $= 360 - 110 = 250$°.

At 3:00 the gap is 90°. Closing it at 5.5° a minute takes $\frac{60 \times 3}{11} = 16\frac{4}{11}$ minutes.

At 5:00 the gap is 150°. Closing it at 5.5° a minute takes $\frac{60 \times 5}{11} = 27\frac{3}{11}$ minutes.

At 7:00 the gap is 210°. Closing it at 5.5° a minute takes $\frac{60 \times 7}{11} = 38\frac{2}{11}$ minutes.

At 2:00 the gap is 60°. Closing it at 5.5° a minute takes $\frac{60 \times 2}{11} = 10\frac{10}{11}$ minutes.

At 8:00 the gap is 240°. Closing it at 5.5° a minute takes $\frac{60 \times 8}{11} = 43\frac{7}{11}$ minutes.

At 10:00 the gap is 300°. Closing it at 5.5° a minute takes $\frac{60 \times 10}{11} = 54\frac{6}{11}$ minutes.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 1, θ = 180: the value in the hour is $38\frac{2}{11}$. Only this value lies inside the hour, so the answer is $38\frac{2}{11}$ minutes past 1.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 4, θ = 180: the value in the hour is $54\frac{6}{11}$. Only this value lies inside the hour, so the answer is $54\frac{6}{11}$ minutes past 4.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 8, θ = 180: the value in the hour is $10\frac{10}{11}$. Only this value lies inside the hour, so the answer is $10\frac{10}{11}$ minutes past 8.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 10, θ = 180: the value in the hour is $21\frac{9}{11}$. Only this value lies inside the hour, so the answer is $21\frac{9}{11}$ minutes past 10.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 4, θ = 90: the values in the hour are $5\frac{5}{11}$, $38\frac{2}{11}$. The first is $5\frac{5}{11}$ minutes past 4.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 4, θ = 90: the values in the hour are $5\frac{5}{11}$, $38\frac{2}{11}$. The second is $38\frac{2}{11}$ minutes past 4.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 6, θ = 90: the values in the hour are $16\frac{4}{11}$, $49\frac{1}{11}$. The first is $16\frac{4}{11}$ minutes past 6.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 7, θ = 90: the values in the hour are $21\frac{9}{11}$, $54\frac{6}{11}$. The second is $54\frac{6}{11}$ minutes past 7.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 5, θ = 60: the values in the hour are $16\frac{4}{11}$, $38\frac{2}{11}$. The first is $16\frac{4}{11}$ minutes past 5.

Use $M = \frac{2}{11}(30H \pm \theta)$ with H = 3, θ = 30: the values in the hour are $10\frac{10}{11}$, $21\frac{9}{11}$. The second is $21\frac{9}{11}$ minutes past 3.

The position comes twice per lap of 65 5/11 minutes, not per hour. Solving $5.5t = 90 + 360k$ or $360k - 90$ inside the interval gives 14 moments, the first at 2:27 3/11 and the last at 9:32 8/11.

The position comes once per lap of 65 5/11 minutes, not per hour. Solving $5.5t = 0 + 360k$ inside the interval gives 10 moments, the first at 1:05 5/11 and the last at 10:54 6/11.

The position comes once per lap of 65 5/11 minutes, not per hour. Solving $5.5t = 180 + 360k$ inside the interval gives 5 moments, the first at 6:00 and the last at 10:21 9/11.

The position comes twice per lap of 65 5/11 minutes, not per hour. Solving $5.5t = 90 + 360k$ or $360k - 90$ inside the interval gives 5 moments, the first at 1:21 9/11 and the last at 3:32 8/11.

The position comes once per lap of 65 5/11 minutes, not per hour. Solving $5.5t = 0 + 360k$ inside the interval gives 4 moments, the first at 2:10 10/11 and the last at 5:27 3/11.

330 minutes × 0.5° = 165°.

340 minutes × 0.5° = 170°.

25 minutes × 6° = 150°.

55 minutes × 6° = 330°.

720 minutes × 0.5° = 360°.

A correct clock meets every 65 5/11 min. Each meeting here comes $1\frac{5}{11}$ min early, and there are $\frac{1440}{64}$ meetings a day: total $32\frac{8}{11}$ min, so the clock gains.

A correct clock meets every 65 5/11 min. Each meeting here comes $5\frac{5}{11}$ min early, and there are $\frac{1440}{60}$ meetings a day: total $130\frac{10}{11}$ min, so the clock gains.

A correct clock meets every 65 5/11 min. Each meeting here comes $6\frac{6}{11}$ min late, and there are $\frac{1440}{72}$ meetings a day: total $130\frac{10}{11}$ min, so the clock loses.

6 hours pass, so it gains 3 × 6 = 18 minutes and shows 2:18 pm.

9 hours pass, so it gains 5 × 9 = 45 minutes and shows 4:45 pm.

In one true hour it shows 64 minutes. It has shown 320 minutes, so true time passed = 320 × 60/64 = 300 minutes: 11:00 am.

In one true hour it shows 66 minutes. It has shown 264 minutes, so true time passed = 264 × 60/66 = 240 minutes: 1:00 pm.

The clock shows 1424 minutes for every 1440 true minutes. It has shown 89 hours, so true time passed = 89 × 1440/1424 = 90 hours.

The clock shows 1416 minutes for every 1440 true minutes. It has shown 59 hours, so true time passed = 59 × 1440/1416 = 60 hours.

Clock (2) shows 9:00, where |30H − 5.5M| = 90°. The others give 12:15 → 82.5°, 6:15 → 97.5°, 3:30 → 75°.

Clock (3) shows 6:00, where |30H − 5.5M| = 180°. The others give 12:30 → 165°, 5:30 → 15°, 6:30 → 15°.

Clock (4) shows 3:00, where |30H − 5.5M| = 90°. The others give 9:30 → 105°, 2:30 → 105°, 3:15 → 7.5°.

The clock shows 4:00, a gap of 120°. At 5.5° a minute: 120 ÷ 5.5 = $21\frac{9}{11}$ minutes.

The clock shows 9:00, a gap of 270°. At 5.5° a minute: 270 ÷ 5.5 = $49\frac{1}{11}$ minutes.

6° − 0.5° = 5.5° per minute.

360° ÷ 12 = 30°.

11 meetings every 12 hours, so 22 in a day; the meeting near 12 is shared.

Two right angles per lap, 22 laps a day: 44.

22 meetings + 22 opposite positions = 44.

The minute hand gains 360° at 5.5° a minute: 360 ÷ 5.5 = 720/11 = 65 5/11 minutes.

Once per lap and 11 laps in 12 hours; the opposite position between 5 and 7 is only at 6:00.

One hour space = 30°.

5 × 6 = 30 minutes, × 12/11 = 360/11 = 32 8/11 minutes past 6.

0.5° × 30 minutes = 15°.

Gap of 360° at 5.5° a minute: 720/11 = 65 5/11 minutes.

The minute hand is on 6 (180°), the hour hand at 90 + 15 = 105°, so the angle is 75°. R gives exactly why it is not 90°.

R is true, and it gives 1440 ÷ (720/11) = 22 meetings a day, so A is false.

Strictly between 3 and 4 the right angle comes only at $32\frac{8}{11}$ min past 3 (the other is at 3:00 itself). Between 5 and 6 the hands are opposite only at 6:00, which is outside, so statement 2 is false (0 times).

30° per 60 minutes is 0.5° per minute, and at 9:00 the gap is |270 − 0| = 270°, whose smaller angle is 90°. Both are true.

100 = 14 × 7 + 2, so 2 odd days.

45 = 6 × 7 + 3, so 3 odd days.

365 = 52 × 7 + 1, so 1 odd day.

250 = 35 × 7 + 5, so 5 odd days.

1000 = 142 × 7 + 6, so 6 odd days.

From the 5-3-1-0 table, 300 years give 1 odd day.

From the 5-3-1-0 table, 200 years give 3 odd days.

500 = 400 + 100 years. Every 400 years give 0 odd days, and 100 years give 5.

1100 = 800 + 300 years. Every 400 years give 0 odd days, and 300 years give 1.

800 = 2 × 400 years, and every 400 years give 0 odd days, so the answer is 0.

2000 passes the leap test (a century year divisible by 400). 1900: century not divisible by 400, 2100: century not divisible by 400, 1800: century not divisible by 400.

1996 passes the leap test (divisible by 4, not a century year). 1998: not divisible by 4, 2002: not divisible by 4, 1990: not divisible by 4.

2400 passes the leap test (a century year divisible by 400). 2200: century not divisible by 400, 2300: century not divisible by 400, 2500: century not divisible by 400.

1600 passes the leap test (a century year divisible by 400). 1700: century not divisible by 400, 1500: century not divisible by 400, 1900: century not divisible by 400.

2028 passes the leap test (divisible by 4, not a century year). 2026: not divisible by 4, 2030: not divisible by 4, 2100: century not divisible by 400.

Completed years 1946: 1600 years → 0; 300 years → 1; 46 years (11 leap) → 57 → 1; 1947 up to the date: 227 days → 3. Total 5 → 5, which is Friday.

Completed years 1949: 1600 years → 0; 300 years → 1; 49 years (12 leap) → 61 → 5; 1950 up to the date: 26 days → 5. Total 11 → 4, which is Thursday.

Completed years 1868: 1600 years → 0; 200 years → 3; 68 years (17 leap) → 85 → 1; 1869 up to the date: 275 days → 2. Total 6 → 6, which is Saturday.

Completed years 1999: 1600 years → 0; 300 years → 1; 99 years (24 leap) → 123 → 4; 2000 up to the date: 1 day → 1. Total 6 → 6, which is Saturday.

Completed years 1982: 1600 years → 0; 300 years → 1; 82 years (20 leap) → 102 → 4; 1983 up to the date: 176 days → 1. Total 6 → 6, which is Saturday.

Completed years 2010: 2000 years → 0; 10 years (2 leap) → 12 → 5; 2011 up to the date: 92 days → 1. Total 6 → 6, which is Saturday.

Completed years 1856: 1600 years → 0; 200 years → 3; 56 years (14 leap) → 70 → 0; 1857 up to the date: 130 days → 4. Total 7 → 0, which is Sunday.

Completed years 2022: 2000 years → 0; 22 years (5 leap) → 27 → 6; 2023 up to the date: 235 days → 4. Total 10 → 3, which is Wednesday.

Completed years 1899: 1600 years → 0; 200 years → 3; 99 years (24 leap) → 123 → 4; 1900 up to the date: 60 days → 4. Total 11 → 4, which is Thursday.

Completed years 2003: 2000 years → 0; 3 years (0 leap) → 3 → 3; 2004 up to the date: 60 days → 4. Total 7 → 0, which is Sunday.

25 December 2023 is 288 days after 12 March 2023. 288 = 41 × 7 + 1, so move 1 day forward: Monday.

1 March 2024 is 60 days after 1 January 2024. 60 = 8 × 7 + 4, so move 4 days forward: Friday.

2 October 2025 is 48 days after 15 August 2025. 48 = 6 × 7 + 6, so move 6 days forward: Thursday.

5 June 2022 is 162 days before 14 November 2022. 162 = 23 × 7 + 1, so move 1 day back: Sunday.

10 May 2022 is 365 days after 10 May 2021. 365 = 52 × 7 + 1, so move 1 day forward: Tuesday.

4 July 2020 is 366 days after 4 July 2019. 366 = 52 × 7 + 2, so move 2 days forward: Saturday.

29 February 2020 is 1461 days after 29 February 2016. 1461 = 208 × 7 + 5, so move 5 days forward: Saturday.

8 February 2004 is 366 days before 8 February 2005. 366 = 52 × 7 + 2, so move 2 days back: Sunday.

14 November 2026 is 292 days after 26 January 2026. 292 = 41 × 7 + 5, so move 5 days forward: Saturday.

14 April 2018 is 171 days before 2 October 2018. 171 = 24 × 7 + 3, so move 3 days back: Saturday.

61 = 8 × 7 + 5; move 5 forward from Monday: Saturday.

100 = 14 × 7 + 2; move 2 forward from Wednesday: Friday.

75 = 10 × 7 + 5; move 5 back from Friday: Sunday.

200 = 28 × 7 + 4; move 4 forward from Sunday: Thursday.

50 = 7 × 7 + 1; move 1 back from Tuesday: Monday.

2025 is an ordinary year. From 2025 to 2030 there are 5 ordinary and 1 leap years: 5 × 1 + 1 × 2 = 7 odd days, a multiple of 7, and 2031 is also ordinary. No earlier year gives a multiple of 7 with the same type.

2024 is a leap year. From 2024 to 2051 there are 21 ordinary and 7 leap years: 21 × 1 + 7 × 2 = 35 odd days, a multiple of 7, and 2052 is also leap. No earlier year gives a multiple of 7 with the same type.

2026 is an ordinary year. From 2026 to 2036 there are 8 ordinary and 3 leap years: 8 × 1 + 3 × 2 = 14 odd days, a multiple of 7, and 2037 is also ordinary. No earlier year gives a multiple of 7 with the same type.

2027 is an ordinary year. From 2027 to 2037 there are 8 ordinary and 3 leap years: 8 × 1 + 3 × 2 = 14 odd days, a multiple of 7, and 2038 is also ordinary. No earlier year gives a multiple of 7 with the same type.

2013 is an ordinary year. From 2013 to 2018 there are 5 ordinary and 1 leap years: 5 × 1 + 1 × 2 = 7 odd days, a multiple of 7, and 2019 is also ordinary. No earlier year gives a multiple of 7 with the same type.

2096 is a leap year. From 2096 to 2107 there are 10 ordinary and 2 leap years: 10 × 1 + 2 × 2 = 14 odd days, a multiple of 7, and 2108 is also leap. No earlier year gives a multiple of 7 with the same type.

2023 has 365 days = 52 weeks + 1. It starts on Sunday, so the extra day(s) are Sunday. Sunday occurs 53 times.

2024 has 366 days = 52 weeks + 2. It starts on Monday, so the extra day(s) are Monday and Tuesday. Wednesday occurs 52 times.

2025 has 365 days = 52 weeks + 1. It starts on Wednesday, so the extra day(s) are Wednesday. Friday occurs 52 times.

2032 has 366 days = 52 weeks + 2. It starts on Thursday, so the extra day(s) are Thursday and Friday. Thursday occurs 53 times.

31 days = 4 weeks + 3. The extra days are Friday, Saturday, Sunday, so Sunday comes 5 times.

30 days = 4 weeks + 2. The extra days are Saturday, Sunday, so Saturday comes 5 times.

31 days = 4 weeks + 3. The extra days are Tuesday, Wednesday, Thursday, so Friday comes 4 times.

29 days = 4 weeks + 1. The extra days are Monday, so Monday comes 5 times.

30 days = 4 weeks + 2. The extra days are Wednesday, Thursday, so Tuesday comes 4 times.

The first Monday is the 5; add 7 for each later week: 5 + 7 × 2 = 19.

The first Friday is the 6; add 7 for each later week: 6 + 7 × 1 = 13.

The first Wednesday is the 5; add 7 for each later week: 5 + 7 × 3 = 26.

The first Sunday is the 6; add 7 for each later week: 6 + 7 × 3 = 27, and 34 would pass 31.

1 March is a Wednesday on the page. 1 June 2023 is 92 days later; 92 = 13 × 7 + 1, so move 1 forward: Thursday.

1 July is a Tuesday on the page. 25 December 2025 is 177 days later; 177 = 25 × 7 + 2, so move 2 forward: Thursday.

1 February is a Thursday on the page. 15 March 2024 is 43 days later; 43 = 6 × 7 + 1, so move 1 forward: Friday.

The page ends on Wednesday 30, so 1 May is a Thursday. 31 days = 4 weeks + 3, so Thursday occurs 5 times.

The page ends on Tuesday 30, so 1 July is a Wednesday. 31 days = 4 weeks + 3, so Wednesday occurs 5 times.

The gap is 366 days (29 February falls in between), which is 2 odd days: Wednesday.

The gap is 365 days (no 29 February in between), which is 1 odd day: Monday.

The gap is 366 days (29 February falls in between), which is 2 odd days: Monday.

The gap is 365 days (no 29 February in between), which is 1 odd day: Saturday.

100, 200, 300, 400 years give 5, 3, 1, 0 odd days: Friday, Wednesday, Monday, Sunday. Tuesday never appears.

The next day after Friday, Wednesday, Monday, Sunday: Saturday, Thursday, Tuesday, Monday. Of the options only Thursday fits.

365 days: 31 December is 364 days after 1 January, a whole number of weeks, so the same day: Wednesday.

31 December is 365 days later: 1 odd day, so Saturday + 1 = Sunday.

366 = 52 × 7 + 2.

366 = 52 weeks + 2: the two extra days are the first two days, Friday and Saturday.

Each century adds 5 odd days: 5, 10 → 3, 15 → 1, and the 400th year is a leap year so 20 + 1 = 21 → 0.

Check each year: the next year after 2042 with the same first day and the same leap status is 2053 (gaps 6, 11, 11).

1900 ÷ 400 is not whole, so it was ordinary; R is exactly the rule that decides it.

2024 is a leap year (R true), but 2030 is ordinary, so it cannot share 2024's calendar; the next match is 2052. A is false.

400 years give 0; 300 years give 15 → 1 odd day, not 3 (3 is for 200 years).

31 December is 364 days after 1 January, so statement 1 holds; 31 = 4 weeks + 3 days, so three weekdays occur 5 times. Both true.