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 SundaysCase (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

In a calendar based aptitude question, what day of the week was 15 August 1947, the historic date of Indian independence?

Accepted Answer

Introduction / Context:This question tests your understanding of calendar calculations and day of the week determination for a given historical date. Dates such as 15 August 1947, the date of Indian independence, are often used in competitive exams to check whether you know how to work with the Gregorian calendar or whether you know the result as a standard fact. Given Data / Assumptions:- We are working with the standard Gregorian calendar.- The date of interest is 15 August 1947.- We must determine the day of the week on this date.- No additional reference date is explicitly given in the question, so you can either use a known reference or a standard counting method such as odd days or Zellers congruence. Concept / Approach:There are two main ways to handle this problem in an exam:1. Treat it as a fact: 15 August 1947 is widely remembered as a Friday in the Gregorian calendar.2. Use calendar arithmetic: pick a known reference date and count the total number of days between that date and 15 August 1947, reduce that count modulo 7 to get the number of odd days, and then shift the weekday accordingly. In most exams, either method is acceptable, but knowing the fact saves time. Step-by-Step Solution:Step 1: Recognise that 15 August 1947 is the Indian independence day and is commonly recorded as a Friday.Step 2: If you want to verify, choose a convenient reference date whose weekday you know exactly, such as 1 January 1900 or 1 January 2000, and use a reliable day counting method.Step 3: Count complete years between the reference year and 1947, separating leap years and ordinary years, and convert the excess days into odd days.Step 4: Add the days from the start of 1947 up to 15 August (including all months January to July and 15 days of August) and again reduce modulo 7.Step 5: Combine the odd days from the years and the days in 1947, and shift the weekday of the reference date accordingly.Step 6: This systematic calculation confirms that the weekday is Friday. Verification / Alternative check:Because this is a famous date, you can cross check it mentally by recalling that celebrations and historical records consistently mention that independence was declared shortly after midnight going into Friday. In many calendars and history books, 15 August 1947 is explicitly labelled as a Friday, so the computed result matches the established historical fact. Why Other Options Are Wrong:Saturday, Sunday, Thursday and Monday are all one or more days shifted away from the correct weekday. If you perform the odd day calculation carefully or consult an accurate calendar, none of these alternatives align with 15 August 1947. Only Friday is compatible with both historical records and proper calendar arithmetic. Common Pitfalls:Typical mistakes include miscounting leap years, forgetting that century years are leap years only if divisible by 400, or adding an extra day for February in a non leap year. Another common issue is starting from an incorrect reference date. To avoid such errors, always be systematic in counting years and days, and when in doubt, rely on well known historical dates like this one as memorised facts. Final Answer:15 August 1947 fell on a Friday.

Question 16

Today is Monday in a calendar based reasoning question. After exactly 61 days from today, which day of the week will it be?

Accepted Answer

Introduction / Context:This problem checks your ability to handle modular arithmetic in the context of calendars. Instead of working with specific dates, it uses a relative description: starting from Monday, you must determine the weekday after a certain number of days. Such questions appear frequently in aptitude tests and can be solved quickly using remainders when dividing by 7. Given Data / Assumptions:- Today is Monday.- We are asked for the weekday after 61 days.- A standard week has 7 days.- We assume no changes to the calendar system (no skipped days or special rules). Concept / Approach:The key idea is that weekdays repeat every 7 days. Therefore, only the remainder when the number of elapsed days is divided by 7 matters. If the remainder is r, then we move forward by r steps from the starting weekday. This allows us to avoid counting all 61 days one by one. Step-by-Step Solution:Step 1: Note that the length of a week is 7 days, so we compute 61 modulo 7.Step 2: Divide 61 by 7: 7 * 8 = 56, and 61 - 56 = 5.Step 3: Therefore, 61 days correspond to 8 complete weeks plus a remainder of 5 days.Step 4: Complete weeks bring us back to the same weekday, so only the remainder of 5 days affects the weekday name.Step 5: Starting from Monday, count forward 5 days: Monday (0), Tuesday (1), Wednesday (2), Thursday (3), Friday (4), Saturday (5).Step 6: After moving 5 days ahead from Monday, we arrive at Saturday. Verification / Alternative check:You can verify by shorter steps. After 7 days it is again Monday, after 14 days again Monday, and so on. After 56 days (which is 7 * 8) it is still Monday. Then you simply count 5 more days: Tuesday, Wednesday, Thursday, Friday, Saturday. This quick mental confirmation reinforces that Saturday is correct. Why Other Options Are Wrong:Sunday, Monday, Tuesday and Wednesday correspond to different remainders. For example, Sunday would match a remainder of 6 days, Tuesday would match a remainder of 1 day, and so on. None of these align with the calculated remainder of 5 when 61 is divided by 7. Common Pitfalls:Learners often forget to take the remainder and instead count full weeks incorrectly, or they miscount the steps forward from the starting day. Another error is to treat the starting day as step 1 instead of step 0. To avoid confusion, treat the starting day as position 0 and then count the remainder days forward carefully. Final Answer:After 61 days from Monday, the day of the week will be Saturday.

Question 17

Using standard Gregorian calendar calculations, what was the day of the week on 16 July 1776?

Accepted Answer

Introduction / Context:This question asks for the weekday corresponding to a specific historical date, 16 July 1776. Calendar based aptitude problems like this one test your understanding of how the Gregorian calendar works and your ability to apply systematic methods such as the odd days method or Zellers congruence to find the day of the week for any given date. Given Data / Assumptions:- We are using the standard Gregorian calendar.- The date is 16 July 1776.- We must find which weekday (Monday, Tuesday, and so on) corresponds to this date.- No intermediate reference date is provided in the question, so you can either rely on a known result or apply a formula based method. Concept / Approach:There are two broad approaches for such historical date calculations:1. Memorisation of a few anchor dates with known weekdays and using them to compute other dates.2. Application of a formula like Zellers congruence, or the odd days method, where you represent the date numerically and compute a value modulo 7 to obtain the weekday index.In exams, having some well practiced methods greatly speeds up these computations. Step-by-Step Solution:Step 1: Represent the date as day = 16, month = July, year = 1776.Step 2: Convert the month and year into the codes required by your chosen method (for example, Zellers congruence uses month codes and century codes).Step 3: Add the day, month code, year code and century adjustment, then account for leap years up to that date.Step 4: Reduce the final sum modulo 7 to obtain a number from 0 to 6, corresponding to days from Sunday through Saturday or Monday through Sunday depending on the convention used.Step 5: When you carry out this calculation accurately for 16 July 1776, the result corresponds to Tuesday.Step 6: Therefore, the day of the week on 16 July 1776 was Tuesday. Verification / Alternative check:As an alternative check, you can cross verify using a different anchor date near 1776 whose weekday is known, then count forward or backward by complete weeks and additional days. When done correctly, both approaches should agree on Tuesday as the weekday for 16 July 1776. This cross checking helps confirm that no mistakes were made in the modular arithmetic or in counting leap years. Why Other Options Are Wrong:Monday, Wednesday, Friday and Saturday are all one or more days away from Tuesday. Any small error in the calculation of leap years or in the modulo 7 operation can lead to one of these incorrect alternatives, which is why they are good distractors in a multiple choice setting. Common Pitfalls:Common issues include forgetting that century years are not leap years unless divisible by 400, misapplying month codes, or using the wrong starting convention for the weekday index. Using two independent methods or at least checking the arithmetic carefully is the best way to avoid these errors. Final Answer:The day of the week on 16 July 1776 was Tuesday.

Question 18

It was Sunday on 1 January 2006 in the Gregorian calendar. Using calendar reasoning, what was the day of the week on 1 January 2010?

Accepted Answer

Introduction / Context:This question tests your understanding of how days of the week shift from year to year in the Gregorian calendar. You are given the weekday for 1 January 2006 and asked to find the weekday for 1 January 2010. This requires knowledge of how ordinary years and leap years contribute extra days when counting forward across several years. Given Data / Assumptions:- 1 January 2006 was a Sunday.- We are working in the Gregorian calendar.- We need the weekday on 1 January 2010.- The years between 2006 and 2010 are: 2006, 2007, 2008, 2009.- Among these, 2008 is a leap year, while 2006, 2007 and 2009 are ordinary years. Concept / Approach:Between one 1 January and the next, an ordinary year causes the weekday to move forward by 1 day (since 365 = 52 weeks + 1 day), and a leap year causes a shift of 2 days (since 366 = 52 weeks + 2 days). By summing these shifts across the years 2006 to 2009 and then applying the result modulo 7, we can determine the weekday on 1 January 2010 starting from the known Sunday in 2006. Step-by-Step Solution:Step 1: List the years from 2006 up to (but not including) 2010: these are 2006, 2007, 2008 and 2009.Step 2: Identify leap years in this range. Only 2008 is a leap year because it is divisible by 4 and not a century year.Step 3: Each ordinary year shifts the weekday by 1 day, and each leap year shifts it by 2 days.Step 4: Total shift = 1 (for 2006) + 1 (for 2007) + 2 (for 2008) + 1 (for 2009) = 5 days.Step 5: Starting from Sunday on 1 January 2006, move forward 5 days: Sunday (0), Monday (1), Tuesday (2), Wednesday (3), Thursday (4), Friday (5).Step 6: Therefore, 1 January 2010 falls on a Friday. Verification / Alternative check:If you prefer, you can check year by year. 1 January 2007 is Monday, 1 January 2008 is Tuesday, 1 January 2009 is Thursday (because of the leap year 2008 adding an extra shift), and 1 January 2010 becomes Friday. This year by year confirmation matches the earlier cumulative shift calculation and helps to ensure no year was misclassified as leap or ordinary. Why Other Options Are Wrong:Sunday would imply no net shift at all, which is incorrect over four years with at least one leap year. Saturday and Wednesday correspond to total shifts of 6 or 3 days respectively, while Monday would mean a shift of 1 day. None of these match the correct cumulative shift of 5 days from the starting Sunday. Common Pitfalls:Common mistakes include forgetting to treat leap years differently, misidentifying which years are leap years, or accidentally including the final year in the shift count again. Always list the years carefully, mark leap years clearly, compute the total shift, and then apply it step by step from the known starting weekday. Final Answer:The day of the week on 1 January 2010 was Friday.

Question 19

According to the standard Gregorian calendar, what was the day of the week on 28 May 2006?

Accepted Answer

Introduction / Context:This problem is another example of a calendar based aptitude question that asks for the day of the week for a specific date, 28 May 2006. Such questions can be solved using a stepwise day counting technique or a formula like Zellers congruence, and they are common in competitive examinations where speed and accuracy matter. Given Data / Assumptions:- Calendar system: Gregorian calendar.- Date of interest: 28 May 2006.- Required: the day of the week (Sunday, Monday, and so on).- No additional reference date is supplied in the question, so you must either know a useful nearby anchor date or apply a general method. Concept / Approach:A practical approach is to use a nearby date whose weekday is easier to determine or already known, and then count forward or backward. Since 1 January 2006 is known to be a Sunday (this is a widely used reference in many exam problems), we can count the number of days from 1 January 2006 to 28 May 2006, compute the remainder when dividing by 7, and then shift that many days from Sunday to reach the weekday for 28 May 2006. Step-by-Step Solution:Step 1: Take 1 January 2006 as Sunday.Step 2: Count days in the months leading up to May: January 31 days, February 28 days, March 31 days, April 30 days.Step 3: Days from 1 January to 30 April = 31 + 28 + 31 + 30 = 120 days.Step 4: From 1 May to 28 May there are 28 more days, so total days from 1 January to 28 May inclusive are 120 + 28 = 148 days, but from 1 January to just before 28 May, you move forward 147 days.Step 5: Compute 147 modulo 7. Since 7 * 21 = 147, the remainder is 0.Step 6: A remainder of 0 means that the weekday on 28 May 2006 is the same as on 1 January 2006, namely Sunday. Verification / Alternative check:Another way is to check an intermediate date such as 1 May 2006 and then count up to the 28th. As long as the total number of steps from the known Sunday is a multiple of 7, the weekday will again be Sunday. Both approaches, if calculated carefully, point to Sunday as the correct day. Why Other Options Are Wrong:Thursday, Friday, Saturday and Monday are all different weekday positions relative to Sunday. They correspond to remainders of 4, 5, 6 or 1 when dividing the total number of days by 7. Since our computation gave a remainder of 0, none of these alternatives can be correct for 28 May 2006. Common Pitfalls:Errors often occur when counting days in each month, especially if February in a leap year is involved. In this case, 2006 is not a leap year, so February has 28 days, but some students mistakenly use 29. Others forget whether to count the starting day or the ending day when converting between inclusive and exclusive counts. Carefully setting up the sum and then reducing modulo 7 avoids these issues. Final Answer:The day of the week on 28 May 2006 was Sunday.

Question 20

In terms of the variable x, how many total days are there in a time period of x weeks and x additional days?

Accepted Answer

Introduction / Context:This is a basic algebra and calendar question that combines the concept of weeks and days. It tests whether you can express a verbal time description, namely x weeks and x days, as a single algebraic expression in terms of the total number of days. Such conversions between units are very common in time and work, calendar and speed problems. Given Data / Assumptions:- The duration is x weeks and x days.- One week consists of exactly 7 days.- We must write the total duration in days as a simplified algebraic expression in x.- x is assumed to be a positive integer for typical aptitude questions, although the algebra works for any real x. Concept / Approach:To convert weeks to days, we multiply the number of weeks by 7. Then we add the extra days given. Because both quantities depend on x, we will end up with an expression like 7x plus x. Finally, we simplify this expression to a single term by combining like terms. This is a straightforward application of distributive and additive properties in algebra. Step-by-Step Solution:Step 1: Start with the expression x weeks and x days.Step 2: Convert x weeks into days. Since 1 week = 7 days, x weeks = 7 * x days = 7x days.Step 3: The total number of days is then 7x days from the weeks plus x additional days.Step 4: Write this as an algebraic sum: total days = 7x + x.Step 5: Combine like terms: 7x + x = 8x.Step 6: Therefore, the total number of days in x weeks and x days is 8x days. Verification / Alternative check:To verify the expression, choose a specific value for x. For example, let x = 2. Then the duration is 2 weeks and 2 days. Two weeks is 14 days and plus 2 days gives 16 days. Our algebraic formula 8x gives 8 * 2 = 16 days, which matches the direct calculation. Trying a few different values such as x = 3 or x = 5 will give the same agreement, confirming that 8x is correct. Why Other Options Are Wrong:The option 7x ignores the extra x days and counts only the weeks. The option 14x is what you would get if you incorrectly multiplied both the weeks and the days by 7. The option 7 gives a fixed number of days independent of x, which cannot be correct because the time period clearly grows as x increases. The expression x + 7 treats the contributions additively but does not scale the weeks properly and is not dimensionally consistent for large x values. Common Pitfalls:Learners sometimes forget to convert weeks into days at all, or they mistakenly think 1 week equals 10 days. Another common error is failing to combine like terms and leaving the answer as 7x + x instead of simplifying to 8x. Always remember basic unit conversions and algebraic simplification rules to avoid these mistakes. Final Answer:The total number of days in x weeks and x days is 8x days.

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