Question 1

If there are K weeks and K additional days, how many total days is that?

Accepted Answer

Introduction / Context:This is a simple arithmetic translation problem. The phrase “K weeks and K days” means we convert weeks to days and add the extra days. Given Data / Assumptions: 1 week = 7 days. Total = (7 * K) + K days = 8K days. Options are numeric, implying a specific K. We align with the Recovery-First Policy and infer K = 12 to match a listed option. Concept / Approach:Compute 8K and choose the matching option. Step-by-Step Solution:Total days = 7K + K = 8K.For K = 12 → 8 * 12 = 96. Verification / Alternative check:Check each option against plausible K values: only 96 corresponds neatly to K = 12 with the 8K rule. Why Other Options Are Wrong:92 and 90 would imply non-integer K under 8K, or a misinterpretation of weeks → days. Common Pitfalls:Forgetting to multiply weeks by 7 or misreading the question as “K weeks or K days.” Final Answer:96

Question 2

How many days are there from the Kth day of the Kth week to the 2Kth day of the 2Kth week (counting by day positions)?

Accepted Answer

Introduction / Context:We map “Kth day of Kth week” and “2Kth day of 2Kth week” to absolute day positions and take the difference. This is a counting/indexing problem. Given Data / Assumptions: Each week has 7 days, days numbered 1..7. “Kth day of Kth week” corresponds to an absolute index. We interpret “between” as the difference in indices (exclusive counting is not requested explicitly; the standard approach is index difference). Concept / Approach:Absolute day index of the dth day of the wth week = 7*(w − 1) + d. Step-by-Step Solution:Index of Kth day of Kth week = 7*(K − 1) + K = 8K − 7.Index of 2Kth day of 2Kth week = 7*(2K − 1) + 2K = 16K − 7.Difference = (16K − 7) − (8K − 7) = 8K.With K = 12 (aligned to options via Recovery-First Policy), days = 8*12 = 96. Verification / Alternative check:Test with a small K (e.g., K = 2): indices 8*2−7 = 9 and 16*2−7 = 25 → difference 16 = 8K (K=2 ⇒ 16), pattern holds. Why Other Options Are Wrong:92 and 90 do not equal 8K for integer K when K is chosen to fit the list; 96 is the unique match. Common Pitfalls:Treating “2Kth day” as impossible because days per week are ≤7; here “day” is an absolute day index within a continuous sequence, not within a single week boundary. Final Answer:96

Question 3

Find the angle between the hour and minute hands at 6:30 a.m.

Accepted Answer

Introduction / Context:We compute the smaller angle between the hour and minute hands at a specified time. Given Data / Assumptions: Time: 6:30 a.m. Minute hand speed: 6 degrees per minute. Hour hand speed: 0.5 degrees per minute. Concept / Approach:The minute hand angle from 12 at t minutes is 6t. The hour hand angle at h:t is 30h + 0.5t. The absolute difference gives the angle; choose the acute one if two are possible. Step-by-Step Solution:At 6:30, minute angle = 6 * 30 = 180°.Hour angle = 30*6 + 0.5*30 = 180 + 15 = 195°.Difference = |195 − 180| = 15°. Verification / Alternative check:General formula for angle at h:m is |30h − 5.5m|; substituting 6:30 → |180 − 165| = 15°, matches. Why Other Options Are Wrong:22.5° is the angle between hour positions at quarter intervals in some cases, not here. 25° is arbitrary. “None of these” is unnecessary since 15° fits. Common Pitfalls:Forgetting the hour hand advances between hours (the 0.5m term). Final Answer:15°

Question 4

The last day of a century year (e.g., 1800, 1900, 2000) cannot be which weekdays?

Accepted Answer

Introduction / Context:Century years (multiples of 100) have special day-of-week behavior due to the number of “odd days” accumulated over 100 years in the Gregorian calendar. Given Data / Assumptions: Options list: I. Tuesday, II. Thursday, III. Saturday, IV. Sunday. We refer to the last day (December 31) of a century year. Concept / Approach:In the Gregorian calendar, the last day of a century year can only fall on Monday, Wednesday, Friday, or Sunday. Therefore, it cannot be Tuesday, Thursday, or Saturday. Step-by-Step Solution:Known results from odd-day analysis across centuries show the last day of century years excludes Tue/Thu/Sat.Thus the set of impossible weekdays = I, II, III. Verification / Alternative check:Check examples: 1900-12-31 was Monday; 2000-12-31 was Sunday. These align with the allowed set, not with Tue/Thu/Sat. Why Other Options Are Wrong:Any set including IV (Sunday) is incorrect because Sunday is possible (e.g., year 2000). Common Pitfalls:Confusing “first day” vs “last day,” and mixing Julian/Gregorian rules. Final Answer:I, II and III

Question 5

Mrs. Susheela celebrated her wedding anniversary on Tuesday, 30 September 1997. In which year will 30 September again fall on a Tuesday (next occurrence)?

Accepted Answer

Introduction / Context:We need the next year in which the date 30 September falls on the same weekday (Tuesday) as it did in 1997. Given Data / Assumptions: Reference: 30 Sep 1997 was Tuesday. Leap-year effect: moving from one year to the next changes the weekday by +1 in a common year and by +2 when crossing a leap year (for dates after February). Concept / Approach:Advance year by year from 1997 until 30 September falls on Tuesday again, accounting for leap-year shifts. Step-by-Step Solution:1998 (non-leap): Tuesday → Wednesday.1999 (non-leap): Wednesday → Thursday.2000 (leap year; date after Feb): +2 → Thursday → Saturday.2001 (non-leap): Saturday → Sunday.2002 (non-leap): Sunday → Monday.2003 (non-leap): Monday → Tuesday. Verification / Alternative check:Cross-check with a perpetual calendar or apply Zeller's congruence; 2003-09-30 is Tuesday. Why Other Options Are Wrong:2002 gives Monday; 2004 shifts beyond and is not the next. “30 October, 2003” changes the month entirely. Common Pitfalls:Forgetting the +2 shift for dates after February in leap years (e.g., across 2000). Final Answer:30th September, 2003

Question 6

If Republic day was celebrated in 1996 on Friday, on which day in 2000 Independence was celebrated?

Accepted Answer

26 January, 1996 to 15 August, 2000 days were = 1663 1663 having odd day = 4 ∴ In the year 2000, Independence day will on Tuesday.

Question 7

If 18th February, 1997 falls on Tuesday, then what will be the day on 18th February, 1999?

Accepted Answer

18th February, 1997 is Tuesday. So, 17th February, 1998 will also be Tuesday. Again 16th February, 1999 will also be Tuesday. Hence, 18th February, 1999 will be Thursday.

Question 8

I was born on August 11. Mohan is younger to me by 11 days. This year's independence day falls on Monday. The day on which Mohan's birthday will fall this year will be

Accepted Answer

Mohan was born on 31st July, August 15 falls on Monday.

Question 9

At what time, (in minutes) between 3 O'clock, and 4 O'clock, both the needles will coincide with each other?

Accepted Answer

At 3 O'clock, then minute hand is 15 min away from the hour hand. To coincide, it must gain 15 min as 55 min are gained in 60 min. 15 min are gained in (60/55) x 15 min = 164/11 min Hence, the hands are coincide at 164/11 min past 3.

Question 10

A clock is set right at 5 am. The clock loss 16 min in 24 h. What will be the right time when the clock indicates 10 Pm on the third day?

Accepted Answer

Time from 5 am of a particular day to 10 pm on the 4th day is 89 h. Now, the clock loses 16 min in 24 h or on other words, we can say that 23 h 44 min of this clock is equal to 24 h of the correct clock. or (23 + 44/60) ⇒ 356 h of this clock = 24 h of the correct clock ∴ 89 h of this clock = (24 x 15/356 x 89) h of correct clock = 90 h of the correct clock

Question 11

The Republic day of India was celebrated on Friday in year 2008. On which day, it will be celebrated in 2012?

Accepted Answer

26th January (Republic day) 2008 → Friday Clary, it will be on Wednesday in 2012.

Question 12

If 8th February, 2005 was a Tuesday, what was the day the 8th February,2004?

Accepted Answer

The year 2004 is a leap year. It has 2 odd days. The day on 8th February, 2004 is 2 days before the day on 8th February, 2005. Hence, this day is Sunday

Question 13

What was the day the 2nd July, 1984?

Accepted Answer

2nd July, 1984 means (1983 years 6 months and 2 days) 1900 years have 1 odd day 83 years have 20 leap years and 63 ordinary years = (40 + 63) odd days = 103 = 5 odd days 6 months and 2 days Jan 31 Feb 29 Mar 31 Apr 30 May 31 June 30 July 02 = 184 days = 2 odd days Total number of days = (1 + 5 + 2) = 8 odd days = 1 odd day Hence, it was Monday on 2nd July,1984.

Question 14

The year next to 1990 will have the same calender as that of the year 1990.

Accepted Answer

The year 1990 has 356 days, i.e., 1 odd day, year 1991 has 356 days, v 1 odd day, year 1992 has 366 days, i.e., 2 odd days. Likewise year 1993, 1994, 1995 have 1 odd days, so calculated from years 1990-95 = (1 + 1 + 2 + 1 + 1 + 1) = 7 odd days = 0 odd day Hence, the year 1996 will have the same calendar as that of the year 1990.

Question 15

If 27 March, 1995 was a Monday, then what day of the week was 1 November, 1994?

Accepted Answer

Here, 27th March 1995 was Monday. Now for calculating total number of odd days. First, we calculate total number of days till 1 Number 1994 ∴ Number of days in March 1995 = 27 Number of days in February 1995 = 28 Number of days in January 1995 = 31 Number of days in December 1994 = 31 Number of days in November 1994 = 29/146 ∴ Number of odd days = 146/7 = 206/7 So, 6 odd days. ∴ On November 1, 1994 = Monday - 6 = Tuesday

Question 16

If it was Saturday on December 17, 1899, what will be the day on December 22,1901?

Accepted Answer

Total number of odd days from December 17, 1899 to December 22, 1901 14 + 356 + 356 = 735 or, 735/7 = 105 = 0 odd days it was Saturday on December 17, 1899. So, it will be Saturday on December 22, 1901.

Question 17

In a month of 31 days, third Thursday falls on 16th. What will be the last day of the month?

Accepted Answer

Friday will fall on 3, 10, 17, 24, 31 So, it will be 5th Friday on 31 st

Question 18

How many times in a day (24 Hrs), are the hands of a clock in straight line but opposite in direction?

Accepted Answer

The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite direction at 6 'o clock only). So, in a day, the hands point in the opposite direction 22 times.

Question 19

On what dates of April, 2001 did Friday fall?

Accepted Answer

Lets find out 1st April from Zeller's Formula: f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C. In this case k = 1 (since 1st April) Month m = 2 (As march = 1, April = 2) D is the last two digit of year here D = 01 (As year is 2001) C is 1st two digit of century here C = 20 (As year is 2001) f = 1 + [13 x 2 - 1 / 5] + 01 +[01/4] + [20/4] -2 x 20 f = 1 + [25/5] + 01 + [0.5] + [5] - 40. f = 1 + 5 + 01 + 0 + 5 - 40 = -28 This - ve value of f can be made positive by adding multiple of 7 So f = - 28 + 28 = 0 So number of odd days is 0, So 1st April 2001 is Sunday, So 1st Friday is on 6th April , so next Fridays is 13th, 20th, 27th April.

Question 20

How many Sundays will be in a period of 100 years.

Accepted Answer

In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap year, as 1900 was not a leap year) Number of days in a period of 100 years is 365 x 100 + 23 or 365 x 100 + 24. If century year is not a leap year then number of days = 365 x 100 + 23, and number of weeks is 5217 and 4 odd and for leap year it will be 5217 weeks and 5 odd days, hence number of Sundays is either 5217 or 5218.

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