Examkin

Clocks and Calendars

29 questions

All types29Multiple choice16Numeric5Short answer8
Question 1Question →

Clock A continuously loses 22 minutes every hour, while Clock B continuously gains 11 minute every hour. Both clocks are synchronized to the exact true time at 8:00 AM on Monday, February 26, 1900.

Which of the following represents the exact day of the week, calendar date, and the time displayed by Clock B at the earliest instance when the time displayed by Clock A and Clock B are exactly 1212 hours apart?

Show answer & explanation

Answer: Thursday, March 8, 1900, at 12:00 Noon

Answer

Thursday, March 8, 1900, at 12:00 Noon
The two clocks drift apart at a combined rate of 3 minutes per hour. To achieve a 12-hour (720-minute) difference, exactly 240 true hours (10 days) must pass. Since 1900 is not a leap year, February has 28 days. Advancing 10 days from Monday, February 26 brings the true date to Thursday, March 8. In those 240 hours, Clock B gains 240 minutes (4 hours). Since the true time is 8:00 AM, Clock B displays 12:00 Noon.

Step-by-Step Solution

1
Calculate the relative speed of separation between the two clocks.
Clock A loses 2 mins/hr and Clock B gains 1 min/hr. Relative separation = 2 + 1 = 3 minutes every hour.
Since they move in opposite directions relative to the true time, their rates are added to find how fast they drift apart.
2
Determine the true time required for the clocks to be 12 hours apart.
12 hours = 720 minutes. Time required = 720 / 3 = 240 hours. 240 hours = 10 full days.
Dividing the total required separation by the hourly separation rate yields the total elapsed true time.
3
Calculate the true calendar date and day of the week after 10 days.
The year 1900 is not a leap year, so February has 28 days. Starting from Feb 26, adding 2 days reaches Feb 28. The remaining 8 days land on March 8. Monday + 10 days (or Monday + 3 odd days) = Thursday.
Century years must be divisible by 400 to be leap years. 1900 does not meet this condition.
4
Determine the time displayed by Clock B after 240 hours.
Clock B gains 1 minute per hour. In 240 hours, it gains 240 minutes, which is exactly 4 hours. True end time is 8:00 AM. 8:00 AM + 4 hours = 12:00 Noon.
The question specifically asks for the time displayed by Clock B, not the true time.

Key Concept

Relative speed of faulty clocks combined with century leap year constraints.
Question 2Question →

An analog 1212-hour clock is set to the exact correct standard time at 12:00 NOON on Sunday, February 26, 1896. This particular clock consistently gains exactly 44 minutes every 2424 hours. Assuming standard time follows the Gregorian calendar without any daylight saving adjustments, what time will this faulty clock display, and what will be the true day of the week, when exactly 88 standard years have passed (i.e., on February 26, 1904 at 12:00 NOON standard time)?

Show answer & explanation

Answer: 2:44 PM, Tuesday

Answer

The faulty clock will display 2:44 PM, and the true day of the week will be Tuesday.
Between February 26, 1896, and February 26, 1904, there is exactly one leap day crossed (February 29, 1896). The year 1900 is not a leap year due to the century rule, and the period ends before February 29, 1904. This results in precisely 29212921 elapsed days. Adding 22 odd days (2921(mod7)2921 \pmod 7) to Sunday determines the true day is Tuesday. The clock gains 1168411684 minutes (2921×42921 \times 4), which translates to 1616 full 1212-hour cycles plus an extra 164164 minutes (22 hours and 4444 minutes). Advancing 12:00 NOON by this remainder yields exactly 2:44 PM.

Step-by-Step Solution

1
Calculate the total number of true days elapsed between February 26, 1896, and February 26, 1904.
Identify that the period covers exactly 8 years, but requires careful evaluation of leap days.
Calendar and clock drift problems require the exact number of 24-hour periods that have passed.
2
Determine how many leap days (February 29ths) fall inside this exact date range.
Exactly 1 leap day is included (February 29, 1896).
1896 is a leap year and its leap day occurs after Feb 26. 1900 is a century year not divisible by 400, so it is NOT a leap year. 1904 is a leap year, but the period ends on Feb 26, before its leap day occurs.
3
Compute the total elapsed days.
8 years×365 days+1 leap day=2921 days8 \text{ years} \times 365 \text{ days} + 1 \text{ leap day} = 2921 \text{ days}.
This establishes the precise duration for both the day-of-week shift and the total minutes gained by the clock.
4
Calculate the true day of the week.
Tuesday.
2921(mod7)=22921 \pmod 7 = 2 odd days. Adding 2 days to the starting day (Sunday) yields Tuesday.
5
Calculate the total time gained by the faulty clock.
2921 days×4 minutes/day=11684 minutes2921 \text{ days} \times 4 \text{ minutes/day} = 11684 \text{ minutes}.
The clock continuously gains 4 minutes every 24 hours over the exact elapsed duration.
6
Determine the time displayed on the 12-hour analog dial.
2:44 PM.
A 12-hour clock resets every 720 minutes. 11684(mod720)=16411684 \pmod{720} = 164 minutes. 164164 minutes is exactly 2 hours and 44 minutes. Adding this shift to 12:00 NOON gives 2:44 PM.

Key Concept

Integration of Calendar Leap Exceptions with Continuous Clock Drift
Estimated Time:4m 0s
Question 3Question →

A university establishes a new scholarship fund on September 15, 2195, which happens to be a Friday. The founding document states that a major administrative review of the fund will occur exactly 1212 years later on September 15, 2207. What day of the week will this review take place?

Show answer & explanation

Answer: Friday; friday; FRIDAY

Answer

Friday
The 1212-year period from 2195 to 2207 contains only 22 leap years (2196 and 2204). The year 2200 is a century year not divisible by 400400, meaning it is a common year with 365365 days. The total shift in days of the week is equal to the number of years (1212) plus the number of extra leap days (22), resulting in a total shift of 1414 days. Since 1414 is a perfect multiple of 77, there are 00 odd days. The day of the week shifts by exactly 00 days, meaning the review will also take place on a Friday.

Step-by-Step Solution

1
Identify the total number of years in the given interval.
From 2195 to 2207 is exactly 1212 years.
This sets the baseline shift, as every standard year shifts the calendar by at least 11 day.
2
Determine the number of leap years in this 12-year interval.
The years divisible by 44 in this range are 2196, 2200, and 2204. However, century years are only leap years if divisible by 400400. Since 2200 is not divisible by 400400, it is not a leap year. Thus, there are only 22 leap years: 2196 and 2204.
Each leap year adds an additional 11 day to the shift because it has 366366 days instead of 365365.
3
Calculate the total number of odd days (total shift in days of the week).
Each normal year contributes 11 odd day, and each leap year contributes 22 odd days. For 1212 years with 22 leap years, the easiest calculation is: 1212 (one for each year) +2+ 2 (extra day for each leap year) =14= 14 days.
This gives the absolute number of days the weekday will advance.
4
Determine the final day of the week.
A shift of 1414 days is exactly 22 full weeks (1414 divided by 77 leaves a remainder of 00). Therefore, the day of the week remains the same. Friday +0+ 0 days == Friday.
Days of the week repeat every 77 days, so we use modulo 77 arithmetic.

Key Concept

Calculating day of the week shifts across century boundaries, accounting for the rule that non-400 century years are not leap years.
Question 4Question →

A historian finds a perfectly preserved wall calendar for the year 1892. In which of the following subsequent years could this exact same calendar be used for the entire year without any discrepancies?

Show answer & explanation

Answer: 1904

Answer

The exact same calendar as 1892 can be reused in the year 1904.
For a calendar to completely match a leap year like 1892, the target year must start on the same day of the week and also be a leap year. While leap years typically repeat every 28 years, the inclusion of the non-leap century year 1900 disrupts this cycle. Counting the odd days from 1892 forward yields exactly 14 odd days by the end of 1903. This means 1904 starts on the same weekday, and because 1904 is a leap year itself, it shares the exact same calendar.

Step-by-Step Solution

1
Determine the properties required for the calendar to match.
The target year must begin on the same day of the week and must also be a leap year, just like 1892.
A leap year has 366 days. If the target year is not a leap year, the days will not match after February 28th.
2
Calculate the cumulative odd days year by year starting from 1892.
1892 (2 odd days), 1893 (1), 1894 (1), 1895 (1), 1896 (2), 1897 (1), 1898 (1), 1899 (1). The sum so far is 10 odd days.
Every standard year adds 1 odd day, while a leap year adds 2 odd days. We must find a point where the sum is a multiple of 7.
3
Evaluate the century year 1900 and continue the odd days calculation.
1900 is a standard year (1 odd day). Continuing: 1900 (1), 1901 (1), 1902 (1), 1903 (1). The total sum of odd days from 1892 to 1903 is exactly 14.
Century years are only leap years if divisible by 400. Since 1900 is not divisible by 400, it is a standard year, which breaks the usual 28-year repetition cycle.
4
Verify the candidate year 1904.
Because the odd days up to 1903 sum to 14 (a multiple of 7), 1904 starts on the same day. Since 1904 is divisible by 4 and not a century year, it is a leap year.
Satisfying both requirements (same starting day and leap year status) means the entire calendar for 1904 is identical to 1892.

Key Concept

Calendar Repetition Across Non-Leap Century Years
Question 5Question →

A commemorative time capsule was sealed by a city council on Tuesday, August 15, 1876. The municipal charter mandates that it must be opened exactly 50 years later, on August 15, 1926. Based on the standard Gregorian calendar, on what day of the week will the capsule be opened?

Show answer & explanation

Answer: Sunday

Answer

The capsule will be opened on Sunday.
The correct answer accurately calculates the shift by accounting for 50 base years and exactly 11 leap days. It properly excludes 1900 as a leap year (due to the century rule) and correctly skips the 1876 leap day because the interval starts in August. Adding the resulting 61 odd days to Tuesday (61 mod 7 = 5) shifts the day forward by 5 days to Sunday.

Step-by-Step Solution

1
Determine the total number of base years in the interval.
1926 - 1876 = 50 years.
Each normal year adds exactly 1 odd day to the calendar shift.
2
Count the number of leap days crossed between August 15, 1876, and August 15, 1926.
There are 11 leap days (1880, 1884, 1888, 1892, 1896, 1904, 1908, 1912, 1916, 1920, 1924).
We must add 1 extra odd day for every February 29th crossed. The year 1876 is skipped because the start date is in August. The year 1900 is skipped because century years not divisible by 400 are not leap years.
3
Calculate the total number of odd days.
50 (from base years) + 11 (from leap days) = 61 total odd days.
Combining the normal yearly shift with the extra leap day shifts provides the total day shift.
4
Find the remainder when dividing the total odd days by 7 and apply it to the starting day.
61 divided by 7 leaves a remainder of 5. Tuesday + 5 days = Sunday.
The days of the week repeat in cycles of 7. Adding the remainder to the starting day gives the final day of the week.

Key Concept

Calculating odd days across century boundaries using the Gregorian calendar leap year rules.
Question 6Question →

A global climate summit was inaugurated on Tuesday, March 15, 2016. Based on the standard Gregorian calendar, on what day of the week will the same date fall in the year 2027?

Show answer & explanation

Answer: Monday; monday; MONDAY

Answer

Monday
To find the day of the week, we track the odd days. Between 2016 and 2027, there are 11 years, providing a baseline of 11 odd days. The interval strictly crosses two leap days (in 2020 and 2024), adding 2 more odd days. Crucially, the 2016 leap day is not counted because the interval starts in March, after February 29th. The total is 13 odd days13 \text{ odd days}, and since 13(mod7)=613 \pmod{7} = 6, shifting 6 days forward from Tuesday gives Monday.

Step-by-Step Solution

1
Calculate the total number of years between the two dates.
2027−2016=11 years2027 - 2016 = 11 \text{ years}
Each standard year shifts the calendar forward by 1 odd day1 \text{ odd day} because 365(mod7)=1365 \pmod{7} = 1.
2
Identify the leap days (February 29th) that actually fall within this specific date range.
Only 22 leap days are crossed: February 29, 2020, and February 29, 2024.
Although 2016 is a leap year, its February 29th occurred before the start date of March 15, 2016, so it does not add an extra day to this interval.
3
Calculate the total number of odd days accumulated over the interval.
11 base days+2 leap days=13 total odd days11 \text{ base days} + 2 \text{ leap days} = 13 \text{ total odd days}
Combining the normal year shifts and the extra days from leap years gives the total day shift.
4
Determine the final day of the week by taking the total odd days modulo 7.
13(mod7)=6 odd days13 \pmod{7} = 6 \text{ odd days}. Tuesday +6 days=Monday+ 6 \text{ days} = \text{Monday}.
The days of the week operate on a 7-day repeating cycle.

Key Concept

The inclusion of a leap year's extra day in a calculation strictly depends on whether February 29th falls within the exact given date interval, not just whether the year itself is a leap year.

Alternative Method

Instead of grouping odd days at the end, you can calculate the day year-by-year: 2016 to 2017 (+1, Wednesday), 2017 to 2018 (+1, Thursday), 2018 to 2019 (+1, Friday), 2019 to 2020 (+2 due to Feb 2020, Sunday), and so on, arriving at Monday for 2027.
Estimated Time:1m 0s
Question 7Question →

A mechanical watch, which gains time at a continuous and uniform rate, is observed to be exactly 5 minutes slow at 8:00 AM on a Sunday. By 8:00 AM on the immediately following Tuesday, the same watch is exactly 7 minutes fast. How many hours after 8:00 AM on that Sunday did the watch display the true, correct time?

Show answer & explanation

Answer: 20

Answer

20
The correct answer is derived by mapping the total error shift over the total elapsed time. The watch gains a total of 12 minutes over a span of 48 hours. This establishes a constant gain rate of 1 minute every 4 hours. Because the watch started exactly 5 minutes slow, it requires 5×4=205 \times 4 = 20 hours of elapsed true time for it to catch up and display the correct, synchronized time.

Step-by-Step Solution

1
Calculate the total true time elapsed between the two observations.
48 hours
To find the rate at which the watch gains time, we first need to determine the total duration of the period. From Sunday 8:00 AM to Tuesday 8:00 AM is precisely two full days, which equals 48 hours.
2
Calculate the total amount of time the watch gained over this period.
12 minutes
The watch transitions from being 5 minutes slow (-5) to being 7 minutes fast (+7). The total change in its display relative to true time is 7−(−5)=127 - (-5) = 12 minutes.
3
Determine the uniform rate at which the watch gains time.
1 minute gained every 4 hours
By dividing the total elapsed time by the total minutes gained (48÷1248 \div 12), we discover that the watch gains exactly 1 minute for every 4 hours of true time that pass.
4
Calculate the hours needed to make up the initial 5-minute deficit.
20 hours
In order to display the correct time, the watch must gain precisely the 5 minutes it was initially lagging behind. At a steady rate of 1 minute per 4 hours, this process will take 5×4=205 \times 4 = 20 hours.

Key Concept

Uniform time gain and synchronization of faulty clocks
Question 8Question →

If 1st January 2004 was a Thursday, what day of the week was 1st January 2005?

Show answer & explanation

Answer: Saturday; saturday

Answer

Saturday
The correct answer is Saturday. The year 2004 is a leap year containing 366 days. Since the time period from 1st January 2004 to 1st January 2005 includes February 29th, there are 2 odd days. Advancing 2 days from Thursday results in Saturday.

Step-by-Step Solution

1
Determine if the year 2004 is a leap year.
2004 is divisible by 4, so it is a leap year containing 366 days.
Leap years include an extra day (29th February).
2
Calculate the number of odd days between 1st January 2004 and 1st January 2005.
366 days divided by 7 equals 52 weeks and 2 odd days (366(mod7)=2366 \pmod 7 = 2).
Full weeks do not change the day of the week, so only the remainder (odd days) shifts the day.
3
Advance the given day by the number of odd days.
Thursday + 2 days = Saturday.
Adding 2 odd days to Thursday moves the day forward to Saturday.

Key Concept

Calculating day of the week shifts across leap years using odd days
Question 9Question →

A mechanical clock is synchronized to the correct standard time at exactly 00:00 (midnight) on February 15th of a leap year. This particular clock is known to gain exactly 44 minutes every 2424 hours of true time.

What is the acute angle (in degrees) between the hour and minute hands of this faulty clock at exactly 12:00 noon (true time) on March 2nd of the same leap year?

Show answer & explanation

Answer: 3

Answer

The acute angle between the hands is 3 degrees.
By determining the exact time elapsed across the leap year boundary (16.516.5 days), the total time gained by the clock is 16.5×4=6616.5 \times 4 = 66 minutes. Adding this to the true time of 12:00 yields a faulty clock time of 13:06 (or 1:06). At 1:06, the minute hand is at 36∘36^\circ (6×6∘6 \times 6^\circ) and the hour hand has moved past the 1 o'clock mark (30∘30^\circ) by an additional 3∘3^\circ ((6/60)×30∘(6/60) \times 30^\circ). The difference between 36∘36^\circ and 33∘33^\circ is 3∘3^\circ.

Step-by-Step Solution

1
Calculate the total true time elapsed in days.
The total elapsed time is 16.5 days.
From February 15th 00:00 to March 1st 00:00 in a leap year (29 days in February) is exactly 15 days. From March 1st to March 2nd 00:00 is 1 day. From 00:00 to 12:00 noon is 0.5 days. Total: 15 + 1 + 0.5 = 16.5 days.
2
Calculate the total time gained by the faulty clock.
The clock gains a total of 66 minutes.
The clock gains 4 minutes per 24 hours (1 day). For 16.5 days, the total gain is 16.5 * 4 = 66 minutes.
3
Determine the time shown on the faulty clock.
The faulty clock shows 13:06 (or 1:06 PM).
The true time is 12:00 noon. Adding the gained 66 minutes (1 hour and 6 minutes) gives 13:06.
4
Calculate the precise position of the minute and hour hands at 1:06.
Minute hand is at 36 degrees; Hour hand is at 33 degrees.
The minute hand moves 6 degrees per minute: 6 * 6 = 36 degrees from the 12 o'clock position. The hour hand moves 30 degrees per hour plus 0.5 degrees per minute: (1 * 30) + (6 * 0.5) = 30 + 3 = 33 degrees.
5
Find the acute angle between the two hands.
The acute angle is 3 degrees.
The difference between the two positions is |36 - 33| = 3 degrees.

Key Concept

Calculating elapsed time across leap year month boundaries combined with continuous clock drift and geometric clock angle formulas.
Estimated Time:2m 30s
Question 10Question →

A city hall maintains an antique mechanical tower clock that has a constant drift, falling behind standard time by exactly 2.52.5 minutes every 2424 hours. The maintenance crew calibrates the clock to the precise standard time on January 14, 19001900, at exactly 12:00 PM (Noon). Calculate the total amount of time, in minutes, that the clock will have drifted behind standard time by exactly 12:00 PM (Noon) on March 15, 19001900.

Show answer & explanation

Answer: 150

Answer

The clock will have accumulated a total drift of 150 minutes.
The correct answer is derived by accurately counting the exact number of days between January 14, 1900, and March 15, 1900. Since 1900 is not a leap year (it is a century year not divisible by 400), February has exactly 28 days. Adding the 17 days in January, 28 days in February, and 15 days in March gives a total of 60 days. Multiplying the 60 days by the daily drift rate of 2.5 minutes yields exactly 150 minutes.

Step-by-Step Solution

1
Calculate the days elapsed in January.
31−14=1731 - 14 = 17 days.
To find the time accumulated during the first month.
2
Determine the number of days in February 19001900.
February has 2828 days.
The year 19001900 is a century year not divisible by 400400, making it a standard non-leap year.
3
Sum the total days from January 14 to March 15.
17+28+15=6017 + 28 + 15 = 60 days.
To find the total 2424-hour periods over which the drift occurred.
4
Calculate the total drift in minutes.
60×2.5=15060 \times 2.5 = 150 minutes.
The clock loses exactly 2.52.5 minutes for every full 2424-hour period elapsed.

Key Concept

Century Leap Year Rule and Time Accumulation
Estimated Time:1m 30s
Question 11Question →

A rare astronomical alignment was recorded on Thursday, May 17, 1888. According to a research institute, this specific alignment occurs exactly every 13 years. On what day of the week will the second occurrence after the original 1888 recording take place?

Show answer & explanation

Answer: Sunday

Answer

Sunday
The second occurrence takes place 26 years after 1888, in the year 1914. Between May 1888 and May 1914, there are 26 years, 5 of which are leap years (1892, 1896, 1904, 1908, 1912). The year 1900 is not a leap year. This gives 26 + 5 = 31 odd days. Since 31 divided by 7 leaves a remainder of 3, the event occurs 3 days after Thursday, which is Sunday.

Step-by-Step Solution

1
Determine the target year for the second occurrence.
The interval is 13 years, so the second occurrence happens 2 × 13 = 26 years after the original recording. Target year is 1888 + 26 = 1914.
The question asks for the second occurrence, which requires adding two full intervals to the base year.
2
Calculate the total number of years and identify all leap years in the interval from May 17, 1888, to May 17, 1914.
Total years: 26. Leap years passed: 1892, 1896, 1904, 1908, 1912 (5 leap years). Note that 1900 is a century year not divisible by 400, so it is not a leap year.
To calculate the shift in the days of the week, we need the total number of normal years and leap years. Leap years add an extra odd day.
3
Calculate the total number of odd days over the 26-year period.
Total odd days = 26 + 5 = 31 odd days.
Each normal year contributes 1 odd day, and each leap year contributes an additional 1 odd day. Adding the number of leap years to the total number of years gives the total odd days.
4
Find the final day of the week by adding the remainder of odd days to the starting day.
31 divided by 7 leaves a remainder of 3. Three days after Thursday is Sunday.
The days of the week repeat every 7 days. Dividing the total odd days by 7 gives the effective shift.

Key Concept

Calculating odd days across a time interval that includes a century boundary to determine the day of the week.
Question 12Question →

A standard analog clock indicates the time is exactly 4:10. Calculate the smaller angle, in degrees, formed between the hour hand and the minute hand.

Show answer & explanation

Answer: 65; 65°; 65 degrees

Answer

65 degrees
At exactly 4:10, the minute hand has moved 60∘60^\circ from the top of the clock, and the hour hand has moved 120∘120^\circ for the four hours plus an extra 5∘5^\circ for the ten minutes passed. The difference between their specific positions (125∘125^\circ and 60∘60^\circ) is exactly 65∘65^\circ.

Step-by-Step Solution

1
Calculate the angular position of the minute hand relative to the 12 o'clock mark.
The minute hand is at 10 minutes, which corresponds to an angle of 10×6∘=60∘10 \times 6^\circ = 60^\circ.
The minute hand moves 360∘360^\circ in 60 minutes, which equals a speed of 6∘6^\circ per minute.
2
Calculate the angular position of the hour hand relative to the 12 o'clock mark.
The hour hand has moved past the 4 o'clock mark. Its position is 4×30∘+10×0.5∘=120∘+5∘=125∘4 \times 30^\circ + 10 \times 0.5^\circ = 120^\circ + 5^\circ = 125^\circ.
The hour hand moves 30∘30^\circ per hour and an additional 0.5∘0.5^\circ per minute due to continuous drift.
3
Find the absolute difference between the two angular positions to determine the angle between the hands.
∣125∘−60∘∣=65∘|125^\circ - 60^\circ| = 65^\circ.
The angle between the two hands is simply the absolute difference of their individual positions from the 12 o'clock reference point.

Key Concept

Clock hand angular positioning and the calculation of hour-hand drift.
Question 13Question →

If January 10, 1896, was a Friday, what day of the week was January 10, 1908?

Show answer & explanation

Answer: Friday

Answer

Friday
The correct day is Friday because evaluating the 12-year interval from January 10, 1896, to January 10, 1908, yields exactly 14 odd days. This breaks down into 12 baseline days (one for each year) plus 2 extra days for the leap years explicitly crossed (1896 and 1904). Importantly, the year 1900 is not a leap year, and the leap day of 1908 is not reached. Because 14 is perfectly divisible by 7 (remainder 0), the day of the week does not shift.

Step-by-Step Solution

1
Calculate the total number of years between the two dates.
The total span from January 10, 1896, to January 10, 1908, is exactly 12 years.
Each standard year contributes 1 odd day (since 365 mod 7 = 1) to the day of the week progression.
2
Identify the number of leap years crossed within this specific time frame.
The leap days crossed belong to the years 1896 and 1904. Total = 2 additional leap days.
1896 is a leap year, and we cross its February 29th because we start on January 10th. The year 1900 is a century year not divisible by 400, so it is a standard year (no leap day). 1904 is a leap year. Although 1908 is a leap year, the period ends on January 10th, so its February 29th is not crossed.
3
Calculate the total number of odd days and determine the final day of the week.
12 standard odd days + 2 leap odd days = 14 odd days. 14 divided by 7 leaves a remainder of 0. Friday + 0 days = Friday.
Adding the accumulated odd days to the original day provides the target day of the week. A remainder of 0 indicates a complete cycle.

Key Concept

Century leap year rules and date boundary logic in calendars
Question 14Question →

A scientist was born on February 29, 2096, which happens to be a Wednesday. Because of the specific rules governing the Gregorian calendar, they only celebrate their birthday on the exact date of February 29. On what day of the week will they celebrate their first actual birthday?

Show answer & explanation

Answer: Friday

Answer

Friday
The next leap year after 2096 is 2104, making an 8-year gap because 2100 is an ordinary year. This 8-year interval consists of 7 ordinary years and 1 leap year. Calculating the odd days yields (7 × 1) + (1 × 2) = 9 odd days. Dividing 9 by 7 gives a remainder of 2. Adding 2 days to Wednesday results in Friday.

Step-by-Step Solution

1
Determine the year of the first actual birthday.
February 29, 2104.
The year 2100 is a century year not divisible by 400, making it a common year. The next leap year containing February 29 after 2096 is 2104.
2
Calculate the total number of years and identify leap years in the interval.
8 years total, consisting of 7 ordinary years and 1 leap year (2104).
The period from February 29, 2096, to February 29, 2104, spans exactly 8 years. Only 2104 contributes a February 29th at the end of its respective annual cycle within this specific gap.
3
Calculate the total number of odd days.
9 odd days.
The 7 ordinary years contribute 1 odd day each (7), and the 1 leap year contributes 2 odd days. Total = 7 + 2 = 9.
4
Determine the final day of the week.
Friday.
9 days divided by 7 leaves a remainder of 2 odd days. Adding 2 days to the starting day of Wednesday results in Friday.

Key Concept

Leap year identification across century boundaries and odd day calculation.
Question 15Question →

If January 15, 1894 fell on a Monday, what day of the week was January 15, 1904?

Show answer & explanation

Answer: Friday; friday; FRIDAY

Answer

Friday
The period from January 15, 1894 to January 15, 1904 covers exactly 10 years. Within this period, we must count how many times February 29th is crossed. The year 1896 contributes one extra day. The year 1900 is a century year not divisible by 400, so it is a common year and contributes no extra days. The year 1904 is a leap year, but because the end date is January 15, we do not cross its February 29th, so it also contributes no extra days to this interval. The total number of odd days is 10 (for the 10 years) + 1 (for 1896) = 11. Dividing 11 by 7 leaves a remainder of 4. Four days after Monday is Friday.

Step-by-Step Solution

1
Determine the total number of years between the two dates.
The interval from January 15, 1894 to January 15, 1904 is exactly 10 years.
This provides the base number of odd days, as each standard 365-day year shifts the day of the week by 1.
2
Identify the leap years that contribute an extra day (February 29) strictly within this interval.
The only leap year whose February 29th is crossed is 1896.
The year 1896 is a regular leap year. The year 1900 is a century year not divisible by 400, so it is NOT a leap year. The year 1904 is a leap year, but the target date is January 15, meaning its February 29th has not yet occurred.
3
Calculate the total number of odd days.
10 (base years) + 1 (extra day for crossing February 29, 1896) = 11 odd days.
We add one extra day for each leap year February actually experienced in the given time window.
4
Determine the remainder when dividing the total odd days by 7.
11 divided by 7 leaves a remainder of 4.
The days of the week repeat in a standard 7-day cycle, so we use modulo 7 arithmetic.
5
Add the remainder to the starting day of the week.
Monday + 4 days = Friday.
Counting forward four days from Monday gives Tuesday, Wednesday, Thursday, and finally Friday.

Key Concept

Calculating odd days across a century boundary and evaluating leap year inclusion based on the month.

Alternative Method

You can count the shift year by year: 1894 to 1895 (+1), '95 to '96 (+1), '96 to '97 (+2, crosses Feb '96), '97 to '98 (+1), '98 to '99 (+1), '99 to 1900 (+1), 1900 to '01 (+1), '01 to '02 (+1), '02 to '03 (+1), '03 to 1904 (+1, stops in Jan). The total is 11, giving a net shift of 4 days.
Estimated Time:1m 0s
Question 16Question →

A newly constructed high-speed railway network begins its first commercial operation on Sunday, January 15, 2096. The operating company schedules a comprehensive decade-review on the exact same date 10 years later, January 15, 2106. On what day of the week will this decade-review take place?

Show answer & explanation

Answer: Friday

Answer

Friday
The interval from January 15, 2096, to January 15, 2106, covers exactly 10 years. During this time, we cross February 29th in the years 2096 and 2104, making them leap years. Crucially, the year 2100 is a century year. Century years are only leap years if they are divisible by 400. Since 2100 is not divisible by 400, it is a normal year. This means we have 8 normal years and 2 leap years in this 10-year span. Calculating the odd days: (8 × 1) + (2 × 2) = 12 odd days. Dividing 12 by 7 gives a remainder of 5. Counting 5 days forward from Sunday lands exactly on Friday.

Step-by-Step Solution

1
Determine the total number of years in the given time frame.
The period from January 15, 2096, to January 15, 2106, is exactly 10 years.
This establishes the baseline period for calculating odd days.
2
Identify the leap years that fall within this 10-year period.
The leap years are 2096 (since February 29th is crossed) and 2104. The year 2100 is a century year not divisible by 400, so it is a normal year.
Century years must be divisible by 400 to be leap years, which is a critical rule for crossing the year 2100 boundary.
3
Calculate the total number of normal years and leap years.
Out of 10 years, there are 2 leap years and 8 normal years.
Leap years have 2 odd days and normal years have 1 odd day.
4
Calculate the total number of odd days for the 10-year period.
(8 normal years × 1 odd day) + (2 leap years × 2 odd days) = 8 + 4 = 12 odd days.
This translates the years into a continuous shift in the days of the week.
5
Find the net shift in the day of the week by finding the remainder when dividing by 7.
12 divided by 7 leaves a remainder of 5 odd days.
The days of the week repeat every 7 days.
6
Add the odd days to the starting day of the week.
Sunday + 5 days = Friday.
This determines the exact final day of the week for the scheduled event.

Key Concept

Calculation of odd days across century boundaries
Question 17Question →

To correctly calculate dates across decades, one must accurately identify which years contain 366366 days. According to the rules of the Gregorian calendar, which of the following groups of years contains ONLY leap years?

Show answer & explanation

Answer: 1896,1904,20001896, 1904, 2000

Answer

The group containing 1896,19041896, 1904, and 20002000
The correct answer contains years that all perfectly satisfy leap year conditions. Both 18961896 and 19041904 are non-century years that divide evenly by 44. The year 20002000 is a century year that divides evenly by 400400. Therefore, all three are leap years.

Step-by-Step Solution

1
Recall the rules for determining a leap year.
A standard year is a leap year if it is divisible by 44. However, century years (ending in 0000) are only leap years if they are exactly divisible by 400400.
To evaluate each individual year listed in the options.
2
Evaluate the group containing 1896,19041896, 1904, and 20002000.
18961896 and 19041904 are divisible by 44. The year 20002000 is a century year and is divisible by 400400.
This confirms all years in this specific group satisfy the leap year rules.
3
Evaluate the remaining groups to find their non-leap years.
The year 21002100 is not divisible by 400400, the year 19001900 is not divisible by 400400, and the year 20182018 is not divisible by 44.
To eliminate the incorrect choices based on the presence of standard years.

Key Concept

Century Leap Year Logic
Estimated Time:45s
Question 18Question →

If January 1, 1900, was a Monday, what day of the week was January 1, 1901?

Show answer & explanation

Answer: Tuesday

Answer

Tuesday
The correct answer is Tuesday because the year 1900 is an ordinary year with 365 days, which produces exactly 1 odd day. Adding 1 day to the starting day of Monday brings us to Tuesday.

Step-by-Step Solution

1
Determine if the starting year (1900) is a leap year or an ordinary year.
1900 is an ordinary year.
Although 1900 is divisible by 4, century years must be divisible by 400 to be considered leap years. Since 1900 is not divisible by 400, it is an ordinary year.
2
Calculate the number of odd days in an ordinary year.
1 odd day.
An ordinary year has 365 days. Dividing 365 by 7 yields 52 weeks and a remainder of 1 day. This remainder represents the odd day.
3
Add the number of odd days to the given day of the week.
Tuesday.
Since January 1, 1900, was a Monday, moving forward by 1 odd day results in Tuesday.

Key Concept

Century Leap Year Rules and Odd Days Calculation
Question 19Question →

In the Gregorian calendar, a standard year has 365365 days, whereas a leap year has 366366 days. Based on the rules for determining leap years, which of the following years had exactly 366366 days?

Show answer & explanation

Answer: 20002000

Answer

The year 20002000 had exactly 366366 days because it is a century year perfectly divisible by 400400.
The correct answer is 20002000. All the given options are century years (ending in 0000). According to the Gregorian calendar, a century year is a leap year with 366366 days only if it is exactly divisible by 400400. Since 2000÷400=52000 \div 400 = 5, it is a leap year.

Step-by-Step Solution

1
Recall the criteria for determining a leap year in the Gregorian calendar.
A general year is a leap year if it is divisible by 44. However, century years (ending in 0000) must be divisible by 400400 to be considered leap years.
This special rule for centuries corrects the slight overestimation of the solar year caused by adding a leap day every 44 years.
2
Evaluate each given option using the century year rule.
Since 17001700, 18001800, 19001900, and 20002000 all end in 0000, divide each by 400400. Only 20002000 is perfectly divisible by 400400 (2000÷400=52000 \div 400 = 5).
Applying the correct divisibility rule isolated the only year that contains an extra day in February.

Key Concept

Identifying leap years across century boundaries.
Question 20Question →

If today is Wednesday, what day of the week will it be after exactly 4545 days?

Show answer & explanation

Answer: Saturday; saturday

Answer

Saturday
Because a week consists of 77 days, the day of the week repeats every 77 days. To find the day after 4545 days, divide 4545 by 77, which yields a quotient of 66 and a remainder of 33. This means exactly 66 full weeks and 33 additional days will pass. The 66 full weeks return the day to Wednesday. Counting forward 33 extra days from Wednesday results in Saturday.

Step-by-Step Solution

1
Calculate the number of odd days by finding the remainder when 45 is divided by 7.
45÷7=645 \div 7 = 6 with a remainder of 33.
The days of the week repeat every 7 days, meaning full weeks do not change the day. We only need the remainder (odd days).
2
Count forward by the number of odd days from the given day.
Wednesday + 33 days = Saturday.
Moving 33 days forward from Wednesday sequentially gives Thursday, Friday, and Saturday.

Key Concept

Calculating future days of the week using odd days (modulo 7 arithmetic).
Page 1 / 2Next
Clocks and Calendars Practice Questions — State PSC Exam | Examkin