Question 1

Calendar of which one of the following 2 years is similar?

Accepted Answer

Calendar of 2 years is similar if number of odd days between these two years is zero. Consider option one by one (a) Between 1940 to 1946 we have two leap years 1940 and 1944 so number of odd days is 2 + 1 + 1 + 1 + 2 + 1 = 8 or 1 hence calendar of these two years is not similar. (b) Between 1977 to 1982 we have one leap years 1980 so number of odd days is 1 + 1 + 1 + 1 + 2 + 1 = 6 calendar of these two years is not similar. (c) Between 1912 to 1916 we have one leap years 1912 so number of odd days is 2 + 1 + 1 + 1 = 5 hence calendar of these two years is not similar.

Question 2

Calendar of 2013 is related to calendar of 2015 same as calendar of 1977 to which one of the following year?

Accepted Answer

Number of odd days between 2013 and 2015 is 2, so we have to find the year which will have 2 odd days between 1977 and required year. Consider options one by one - (a) Number of odd days between 1977 and 1981 is 1 + 1 + 1 + 2 = 5 odd days (b) Number of odd days between 1977 and 1985 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 10 odd days or 3 odd days (c) Number of odd days between 1977 and 1990 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 = 16 odd days or 2 odd days, hence calendar of 1990 is the answer.

Question 3

Which of the following year has 53 Sundays?

Accepted Answer

In a year we have 53 Sundays only when year start with Sunday (For non leap year) and either with Saturday of Sunday (For leap year). From zeller's formula 1st January 2001 was Monday From number of odd days 1st January 2006 will start with Sunday so it had 53 Sunday Similar 1stJanuary 2009 will start with Thursday.

Question 4

If in a particular year 'X' there are 53 Sundays then how many Sundays will be there in a period of four years X to X + 3 year.

Accepted Answer

Let us take two cases Case (i) : When year start with Sunday then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays Case (ii) : When year start with Sunday and then we have 53 Sundays that means year is a leap year then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays.

Question 5

If in a particular month there are 5 Thursday then how many Fridays are there in next month.

Accepted Answer

If month ends with Thursday then next month will start with Friday and it may have 5 Friday otherwise it may have 4 Fridays.

Question 6

Calendar reasoning — Given that Monday falls on 20 March 1995, determine the day of the week on 3 November 1994 (use the given anchor date, not real-world memory).

Accepted Answer

Introduction / Context: Calendar questions often provide one “anchor” date and ask you to deduce the day on another date. The skill is to convert dates to an offset in days and then take that offset modulo 7 to map back to a weekday. Given Data / Assumptions: Anchor: 20 March 1995 is a Monday (given). Target: 3 November 1994 (earlier date). Weekdays repeat every 7 days. Concept / Approach: When moving from the anchor to another date, count the exact number of days between the dates, then shift the weekday by (difference mod 7). A negative shift just means “go backward.” Step-by-Step Solution: 1) Count days from 3 Nov 1994 to 20 Mar 1995.2) From 3 Nov 1994 to 30 Nov 1994 = 27 days.3) Dec 1994 = 31, Jan 1995 = 31, Feb 1995 = 28 (1995 is not leap), total so far 27+31+31+28 = 117.4) Add 1–20 Mar 1995 = 20; grand total = 137 days.5) 137 mod 7 = 4 (since 7×19 = 133).6) If 20 Mar 1995 is Monday, then 4 days earlier is Thursday. Verification / Alternative check: Working forward: 3 Nov 1994 → add 137 days → Monday implies the start was Monday − 4 = Thursday. Same result. Why Other Options Are Wrong: Tuesday/Sunday: These correspond to different remainders mod 7 and do not match the computed offset. None of these: A valid weekday (Thursday) matches. Common Pitfalls: Forgetting that 1995 is not a leap year. Counting 3 Nov inclusively or exclusively inconsistently; keep a consistent convention. Final Answer: Thursday

Question 7

Day-of-week cycles — If 94 days ago it was Monday, after how many days from today will it next be Sunday?

Accepted Answer

Introduction / Context: This problem checks understanding of 7-day modular cycles. If we know the weekday some fixed number of days ago, we can deduce today’s weekday and then count forward to the desired day. Given Data / Assumptions: 94 days ago: Monday. Week repeats every 7 days. We want the number of days from today to the next Sunday. Concept / Approach: Compute today’s day first. Since 94 mod 7 = 3, moving forward 94 days shifts the weekday by +3. Step-by-Step Solution: 1) 94 mod 7 = 3.2) 94 days after Monday → Monday + 3 = Thursday (today).3) From Thursday to Sunday is 3 days (Fri=1, Sat=2, Sun=3). Verification / Alternative check: Count directly on fingers or a 7-step cycle: Thu→Fri(1), Sat(2), Sun(3). Why Other Options Are Wrong: 5, 24, 27 days do not match the computed shortest forward gap (3). Since 3 is not listed, the correct choice per the options is “None of these.” Common Pitfalls: Taking 94 mod 7 incorrectly. Accidentally counting the current day as day 1 instead of day 0. Final Answer: None of these (actual gap is 3 days).

Question 8

Century-year endings — Which weekday can be the last day (31 December) of a century year?

Accepted Answer

Introduction / Context: Century years (…1700, 1800, 1900, 2000, …) have distinctive leap-year behavior: only years divisible by 400 are leap years. This shapes which weekdays can occur on year-end dates. Given Data / Assumptions: We consider Gregorian calendar rules. “Last day of a century” means 31 December in a year ending with 00. Concept / Approach: Across centuries, weekday patterns cycle. For example: 1700-12-31, 1800-12-31, 1900-12-31, 2000-12-31 are known benchmarks generating allowable weekdays. Non-400 centuries shift differently than 400-multiples. Step-by-Step Solution: 1) Check known samples under Gregorian rules:— 1700-12-31: Friday.— 1800-12-31: Wednesday.— 1900-12-31: Monday.— 2000-12-31: Sunday (400-multiple leap century).2) Hence possible last-day weekdays include Monday, Wednesday, Friday, Sunday. Verification / Alternative check: Day-counting modulo 7 over 100-year spans shows non-400 centuries shift by +5 weekdays; 400-multiples net a +6 shift within that block, yielding the observed set. Why Other Options Are Wrong: Tuesday/Thursday/Saturday do not occur as the last day of century years under Gregorian patterns illustrated above. Common Pitfalls: Assuming every 100th year is a leap year (not true unless divisible by 400). Final Answer: Friday

Question 9

Weekend-day counts — If December of a certain year has exactly 9 weekend days (Saturdays and Sundays), how many weekend days did January of that same year have?

Accepted Answer

Introduction / Context: Counting weekend days (all Saturdays and Sundays) across months requires careful attention to the calendar layout. The question links December's weekend count to January of the same year. Given Data / Assumptions: December has exactly 9 weekend days. “Weekend” means Saturdays + Sundays (individual days, not pairs). We want January's weekend-day count of the same year. Concept / Approach: January and December are 11 months apart. Depending on where weekdays fall and whether the year is leap or common, January can have either 8 or 10 weekend days while December has 9. There is no unique single value constrained by “December = 9.” Step-by-Step Solution: 1) Enumerate possibilities (conceptually): month lengths and start weekdays shift the placement of Saturdays/Sundays.2) Real calendar enumerations produce two valid January counts when December has 9 weekend days: sometimes 8, sometimes 10.3) Because both 8 and 10 are feasible, there is no single forced answer. Verification / Alternative check: Test multiple Gregorian years: some with December=9 yield January=8; others yield January=10. Hence non-unique. Why Other Options Are Wrong: 8 and 10 can both be correct depending on the year; since the item expects exactly one correct option, selecting either alone would be misleading. 9: January having 9 with December=9 is not guaranteed and is not the only outcome. Common Pitfalls: Confusing “weekend days” with “weekends as pairs.” Assuming a unique answer without checking calendar variants. Final Answer: None of these (January can be 8 or 10 weekend days).

Question 10

Year with 53 Saturdays and 53 Sundays — How many weekend days does its January contain?

Accepted Answer

Introduction / Context: A year having 53 Saturdays and 53 Sundays implies specific start-day and leap-year structure. We use that structure to infer January's weekend-day count. Given Data / Assumptions: Total Saturdays = 53 and total Sundays = 53 in the year. Gregorian rule: such years occur when the year is a leap year starting on a Saturday (e.g., 2000). Concept / Approach: In a leap year that starts on Saturday, January contains five Saturdays and five Sundays (because Jan 1 is Saturday and the month has 31 days), yielding 10 weekend days in January. Step-by-Step Solution: 1) For 53 Saturdays and 53 Sundays, the configuration must allow both weekend days to occur 53 times; this aligns with a leap year starting on Saturday.2) January of such a year starts on Saturday and runs 31 days → Saturdays on 1, 8, 15, 22, 29 and Sundays on 2, 9, 16, 23, 30.3) Count = 10 weekend days. Verification / Alternative check: Check a known example (Year 2000): Jan 2000 indeed had 10 weekend days. Why Other Options Are Wrong: 8 or 9: too few given the start-day pattern in this special case. None of these: 10 is a valid, determinable value. Common Pitfalls: Interpreting “53 weekends” as 53 week-pairs instead of separate Saturday and Sunday counts. Final Answer: 10

Question 11

Century boundaries — If the 1st day of a century is Thursday, the 1st day of the next century must be which weekday?

Accepted Answer

Introduction / Context: Weekday shifts over 100 years differ depending on whether the century year at the end of the block is a leap year (divisible by 400) or not. This dictates what the next century's first day can be. Given Data / Assumptions: Current century starts on Thursday. Next century begins 100 years later. Only years divisible by 400 are leap years among the centuries. Concept / Approach: Over a 100-year span, the weekday advances by +5 if the closing century is not a 400-multiple; if it is a 400-multiple, the net shift is +6 (one extra leap day). Starting from Thursday, +5 → Tuesday; +6 → Wednesday. Step-by-Step Solution: 1) Compute weekday shift across 100 years.2) Non-400 century: +5 → Thursday + 5 = Tuesday.3) 400-multiple century: +6 → Thursday + 6 = Wednesday. Verification / Alternative check: Historical patterns of 1700→1800 and 1600→1700 illustrate the +5 vs +6 behavior. Why Other Options Are Wrong: Other pairs do not correspond to the possible +5/+6 shifts from Thursday. Common Pitfalls: Treating every century as leap; only 400-multiples are. Final Answer: Tuesday or Wednesday

Question 12

Hundred-year shift — If 1 January of year x is Monday, what is the first day of year x + 100?

Accepted Answer

Introduction / Context: Advancing exactly 100 years changes the weekday by +5 (ordinary block) or +6 (if the terminal century year is a 400-multiple). We apply this to a Monday start. Given Data / Assumptions: Year x starts on Monday (1 Jan). We seek the start day of year x+100. Concept / Approach: From Monday, add +5 → Saturday (non-400 century) or +6 → Sunday (400-multiple). Thus two possibilities exist. Step-by-Step Solution: 1) Determine whether the 100-year block ends at a 400-multiple year or not.2) Map the weekday shift accordingly: +5 or +6.3) Monday + 5 = Saturday; Monday + 6 = Sunday. Verification / Alternative check: Cross-check with known sequences around 1900/2000 to see Saturday/Sunday outcomes. Why Other Options Are Wrong: Friday/Friday or Saturday do not match the +5/+6 mapping from Monday. Common Pitfalls: Assuming a single unique weekday when two are possible. Final Answer: Saturday or Sunday

Question 13

Century starts — Which weekdays can a century year (…00) NOT start on?

Accepted Answer

Introduction / Context: Not all weekdays are possible for 1 January of a century year. Under Gregorian rules, century starts occur only on a subset of weekdays. Given Data / Assumptions: Consider years …, 1700, 1800, 1900, 2000, 2100, … We examine which weekdays are attainable on 1 January of those years. Concept / Approach: Empirical (and theoretical) analysis shows that century years start only on Monday, Wednesday, Friday, or Saturday. Step-by-Step Solution: 1) Known examples: 1900 began on Monday; 2000 on Saturday; 2100 on Friday; 1800 on Wednesday.2) Therefore, the possible set is {Monday, Wednesday, Friday, Saturday}.3) The complements (Tuesday, Thursday, Sunday) are not possible. Verification / Alternative check: Apply 100-year weekday shifts (+5, +6) and cycle through samples to see the attainable set. Why Other Options Are Wrong: Options A–C include weekdays that actually can occur (e.g., Wednesday, Friday, Saturday). Hence none of those groupings are correct. Common Pitfalls: Mistyping “Sunday” as “Sunady.” Assuming any weekday is possible. Final Answer: None of these

Question 14

Month-to-month calendar reuse — The calendar layout of July this year matches which month of the previous year?

Accepted Answer

Introduction / Context: Some months share the same weekday layout as months in another year. The problem asks which month in the preceding year matches this year’s July. Given Data / Assumptions: Target: July of the current year (31 days). We search the previous year's month with identical day-of-week alignment and length. Concept / Approach: July (31 days) most commonly matches May of the previous year, and in other cases January/October depending on leap structure. Among the options provided, only May is feasible. Step-by-Step Solution: 1) Candidate months in prior year must have 31 days.2) Checking weekday starts across years shows that May (previous year) frequently aligns with July (current year) in layout.3) April has 30 days; September has 30 days; they cannot match a 31-day layout including weekday positions. Verification / Alternative check: Try multiple actual year pairs; you will find repeated May↔July alignments (previous-year May with current-year July). Why Other Options Are Wrong: September/April are 30-day months and cannot match a 31-day July layout. None of these is unnecessary since May works. Common Pitfalls: Ignoring month length when matching layouts. Final Answer: May

Question 15

What was the day on 15th august 1947 ?

Accepted Answer

To find the day of the week for any given date, we can use a formula called Zeller's Congruence. Here are the steps: Write down the date: 15th August 1947. Day (q) = 15 Month (m) = 8 (August) Year = 1947, so we get: K = 47 (last two digits of the year) J = 19 (first two digits of the year) Apply the formula: The formula for Zeller's Congruence is: h = (q + floor(13(m + 1) / 5) + K + floor(K / 4) + floor(J / 4) - 2J) mod 7 Plug the values into the formula: h = (15 + floor(13(8 + 1) / 5) + 47 + floor(47 / 4) + floor(19 / 4) - 2(19)) mod 7 Calculate the terms: floor(13(9) / 5) = floor(117 / 5) = 23 floor(47 / 4) = 11 floor(19 / 4) = 4 Final calculation: h = (15 + 23 + 47 + 11 + 4 - 38) mod 7 Simplified: h = 62 mod 7 = 6 Interpret the result: h = 6, which corresponds to Friday (where 0 = Saturday, 1 = Sunday, ..., 6 = Friday). Thus, 15th August 1947 was a Friday.

Question 16

Today is Monday. After 61 days, it will be :

Accepted Answer

Each day of the week is repeated after 7 days. So, after 63 days, it will be Monday. After 61 days, it will be Saturday.

Question 17

What was the day of the week on, 16th July, 1776?

Accepted Answer

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776) Counting of odd days : 1600 years have 0 odd day. 100 years have 5 odd days. 75 years = (18 leap years + 57 ordinary years) = [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days 1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. Jan Feb Mar Apr May Jun Jul 31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) Total number of odd days = (0 + 2) = 2. Required day was 'Tuesday'.

Question 18

It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

Accepted Answer

On 31st December, 2005 it was Saturday. Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. therfore, on 31st December 2009, it was Thursday. Thus, on 1st Jan, 2010 it is Friday

Question 19

What was the day of the week on 28th May, 2006?

Accepted Answer

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006) Odd days in 1600 years = 0 Odd days in 400 years = 0 5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days 148 days = (21 weeks + 1 day) 1 odd day. Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day. Given day is Sunday.

Question 20

How many days are there in x weeks x days?

Accepted Answer

x weeks x days = (7x + x) days = 8x days.

Calendar Questions