Question 1

Counting months that have a 29th day in 400 years: How many times does a 29th day of the month occur in 400 consecutive years (Gregorian calendar)?

Accepted Answer

Introduction / Context: Every month except February in a common year has at least 29 days. In the Gregorian 400-year cycle, there are a fixed number of leap years (when February also has a 29th day). Counting total months and subtracting the non-leap Februaries yields the answer. Given Data / Assumptions: Total months in 400 years = 400 × 12 = 4800. Leap years in 400-year cycle = 97 (divisible by 4, except centuries not by 400; among 100 century years, only the one divisible by 400 is leap). Hence non-leap years = 400 − 97 = 303. Only February of non-leap years lacks a 29th day. Concept / Approach: All months contribute a 29th except the 303 February months of non-leap years. Subtract those from total months. Step-by-Step Solution: Occurrences = 4800 − 303 = 4497. Verification / Alternative check: Count leap Februaries with 29th: 97; the remaining 11 months for all 400 years contribute 400 × 11 = 4400; 4400 + 97 = 4497. Why Other Options Are Wrong: 4800 counts all months (including non-leap Februaries) incorrectly; 4400 ignores leap Februaries. Common Pitfalls: Miscounting leap years (assuming 100 instead of 97), or forgetting that every non-February month always has a 29th. Final Answer: 4497

Question 2

Back-counting within the same year: If March 2, 1994 was Wednesday, what day of the week was January 25, 1994?

Accepted Answer

Introduction / Context: To move between two dates in the same year, count the number of days between them and reduce modulo 7. Here, we go backward from a known Wednesday (Mar 2) to a January date, so we subtract the appropriate offset. Given Data / Assumptions: Mar 2, 1994 = Wednesday. 1994 is not a leap year (no Feb 29). Concept / Approach: Compute days from Jan 25 to Mar 2 and reduce mod 7. Since the target (Jan 25) precedes the reference (Mar 2), move backward by that many days. Step-by-Step Solution: Jan 25 → Jan 31: 6 days.Full February (non-leap): 28 days.Mar 1 → Mar 2: 1 day (or include both days beyond Jan 31 as 30 total; either way total is 36).Total = 6 + 28 + 2 = 36 days ≡ 1 (mod 7).Therefore Jan 25 is one weekday earlier than Mar 2 ⇒ Tuesday. Verification / Alternative check: Wednesday minus 1 day = Tuesday; the arithmetic agrees with 36 ≡ 1. Why Other Options Are Wrong: Wednesday/Thursday/Monday represent 0/+2/−2 shifts, which do not match 36 ≡ 1. Common Pitfalls: Mistally of days in February or off-by-one when counting inclusive/exclusive endpoints; always reduce mod 7. Final Answer: Tuesday

Question 3

Striking-clock interval logic: If a clock strikes 6 times in 5 seconds, how many times will it strike in 10 seconds?

Accepted Answer

Introduction / Context: For striking clocks, the duration refers to the time between the first and last strike, i.e., to the number of intervals, which is strikes − 1. If 6 strikes take 5 seconds, then each interval lasts 1 second. Scale this to the requested time span. Given Data / Assumptions: 6 strikes in 5 seconds ⇒ 5 intervals in 5 s ⇒ 1 s per interval. We want the number of strikes that can occur over 10 seconds. Concept / Approach: Number of intervals in 10 seconds = 10/1 = 10. Strikes = intervals + 1 = 11. Step-by-Step Solution: Interval = 1 s.Intervals in 10 s = 10 ⇒ strikes = 10 + 1 = 11. Verification / Alternative check: Check proportionality: doubling duration doubles the number of intervals; the strike count grows by one beyond those intervals. Why Other Options Are Wrong: 10/9/8 are counts of intervals, not strikes, or undercounts the extra end strike. Common Pitfalls: Confusing “strikes” with “intervals between strikes.” Final Answer: 11

Question 4

Weekday by offset from a known Sunday: Suresh was born on October 4, 1999. Shashikanth was born 6 days earlier. If Independence Day (Aug 15, 1999) was Sunday, which day was Shashikanth born?

Accepted Answer

Introduction / Context: We are given a reference weekday (Aug 15, 1999 = Sunday) and need to find the weekday for a later September date (6 days before Oct 4). By counting the total days forward and reducing modulo 7, we can determine the resulting weekday step from Sunday. Given Data / Assumptions: Suresh: Oct 4, 1999 ⇒ Shashikanth = Sep 28, 1999 (6 days earlier). Aug 15, 1999 = Sunday. September has 30 days; August has 31. Concept / Approach: Compute the number of days from Aug 15 to Sep 28: days remaining in Aug plus days up to Sep 28; reduce modulo 7 and step the weekday forward from Sunday. Step-by-Step Solution: Aug 15 → Aug 31: 16 days.Sep 1 → Sep 28: 28 days.Total = 44 days ≡ 2 (mod 7).Sunday + 2 = Tuesday. Verification / Alternative check: Breaking 44 as 35 + 9 ⇒ +0 + 2 = +2; consistent. Why Other Options Are Wrong: Wednesday/Monday/Sunday correspond to +3/+1/+0 steps, not +2. Common Pitfalls: Mistakes in counting the days in each month; always reduce modulo 7. Final Answer: Tuesday

Question 5

When two inaccurate clocks next agree: A wall clock gains 2 minutes in 12 hours; a table clock loses 2 minutes in 36 hours. Both are correct at Tuesday 12 noon. When next will they show the same time again?

Accepted Answer

Introduction / Context: Two clocks drift at different constant rates. They will next agree (show the same reading) when the faster clock advances ahead of the slower one by an integer multiple of 12 hours (720 minutes), since clock faces wrap every 12 hours. The required real time equals “wrap amount” divided by their relative rate difference. Given Data / Assumptions: Fast clock rate: +2 min per 12 h ⇒ +1/6 min/h = +0.166… min/h. Slow clock rate: −2 min per 36 h ⇒ −1/18 min/h ≈ −0.0555… min/h. Relative rate = 0.166… − (−0.0555…) = 0.2222… min/h = 2/9 min/h. Concept / Approach: They coincide when relative lead = 720 minutes (12 hours). Time required = 720 ÷ (2/9) = 720 × 9/2 = 3240 hours = 135 days. Step-by-Step Solution: Relative drift per hour = 2/9 min.Real hours to wrap = 720 ÷ (2/9) = 3240 h.3240 h = 135 days after Tue 12 noon ⇒ still 12 noon. Verification / Alternative check: Because 135 is an integer number of days, the real-time clock shows 12 noon again; both faces coincide at 12 noon. Why Other Options Are Wrong: Other options give incorrect durations or wrong time-of-day when accounting for 3240 hours. Common Pitfalls: Using 24-hour wrap (1440 min) instead of 12-hour wrap (720 min) for analog faces. Final Answer: 12 noon, after 135 days

Question 6

Run-fast watch, every 30 min it gains 3 min: After setting correct time at 5:00 am, what time will the watch show after 6 hours?

Accepted Answer

Introduction / Context: “Gains 3 minutes every 30 minutes” means the watch runs 10% fast relative to real time: in 30 real minutes, it shows 33 minutes. Scaling proportionally finds the shown reading after a given real-time interval. Given Data / Assumptions: Real elapsed time = 6 hours = 360 minutes. Watch runs 10% fast: shown time = real time × 1.1. Concept / Approach: Compute the shown minutes and add to the start time (5:00 am). Step-by-Step Solution: Shown minutes = 360 × 1.1 = 396 minutes = 6 h 36 min.5:00 am + 6:36 = 11:36 am. Verification / Alternative check: Chunking 6 hours as twelve 30-minute blocks: each gains 3 minutes ⇒ 12 × 3 = 36 minutes extra; 6:00 real becomes 6:36 shown. Why Other Options Are Wrong: 10:54/11:30/11:42 do not reflect the exact 10% gain over 6 hours. Common Pitfalls: Adding 3 minutes once instead of per every 30-minute block; mixing shown vs real intervals. Final Answer: 11:36 am

Question 7

Hands swapped (hour/minute) between given intervals: A man left between 3–4 and returned between 8–9. He found the hands had exactly interchanged positions. He returned at what time?

Accepted Answer

Introduction / Context: “Hands interchanged” means the minute hand's earlier minute-mark becomes the hour hand's later position and vice-versa. Using the standard positions (minute = M; hour = 5H + M/12, in minute-marks), we can write two linear equations for the unknown minute values at the two times and solve exactly. Given Data / Assumptions: First time: between 3 and 4 ⇒ hour = 3, minute m1. Second time: between 8 and 9 ⇒ hour = 8, minute m2. Interchange: m1 = 5×8 + m2/12 and m2 = 5×3 + m1/12. Concept / Approach: Write the two equations and solve simultaneously. Fractions with denominator 13 or 143 commonly emerge in such “swap” problems due to the 12-based hour-hand progress. Step-by-Step Solution: m2 = 15 + m1/12.m1 = 40 + m2/12 ⇒ 144 m1 = 5940 + m1 ⇒ 143 m1 = 5940 ⇒ m1 = 5940/143.m2 = 15 + (5940/143)/12 = 15 + 495/143 = 2640/143 = 18 66/143 ≈ 18.4615 min. Verification / Alternative check: 66/143 ≈ 0.4615 ≈ 6/13. Hence 8:18 6/13 is the standard exact form found in classic solutions. Why Other Options Are Wrong: Other minute values do not satisfy both interchange equations simultaneously. Common Pitfalls: Using degrees instead of minute-mark equations for swaps; rounding too early and missing the exact fractional form. Final Answer: 18 6/13 minutes past 8

Question 8

Right angle between 4:30 and 5:00 (repaired): At what time between 4:30 and 5:00 will the hands of a clock be at a right angle?

Accepted Answer

Introduction / Context: The original stem said “straight line,” but the options match known right-angle times. Applying the right-angle condition |30H − 5.5M| = 90 with H between 4 and 5 yields two solutions within the hour, one of which lies between 4:30 and 5:00. Given Data / Assumptions (Recovery Applied): Interpretation repaired to “right angle.” Hour H = 4; minutes M in (30, 60). Equation: |120 − 5.5M| = 90. Concept / Approach: Solving yields M = 21 9/11 and M = 54 6/11. Only 54 6/11 is between 4:30 and 5:00. Step-by-Step Solution: 120 − 5.5M = 90 ⇒ M = 21 9/11 (not in window).120 − 5.5M = −90 ⇒ 5.5M = 210 ⇒ M = 54 6/11 (in window). Verification / Alternative check: Substitute M = 54 6/11; angle equals 90° exactly. Why Other Options Are Wrong: 50, 42, 46 minutes are not exact right-angle solutions from the formula. Common Pitfalls: Forgetting that the hour hand moves (0.5 deg per minute), hence 5.5M term. Final Answer: 54 6/11 minutes past 4

Question 9

Right angles per day: How often are the hands of a clock at right angles in a 24-hour day?

Accepted Answer

Introduction / Context: In every 12-hour cycle, the hands form right angles 22 times (twice per roughly each hour, except around 2–4 hours where spacing shifts). Therefore, over 24 hours the count doubles to 44. Given Data / Assumptions: Standard 12-hour analog clock. Uniform motion. Concept / Approach: Use the known count per 12 h and multiply by 2 for a full day. Step-by-Step Solution: Right angles in 12 h: 22 ⇒ in 24 h: 44. Verification / Alternative check: Between consecutive hours there are typically two right-angle instants; summing carefully over 12 hours yields 22. Why Other Options Are Wrong: 38/40/48 miscount the standard total. Common Pitfalls: Assuming a fixed “two per hour” without accounting for boundary overlaps; the known total 22 per 12 h is robust. Final Answer: 44 times

Question 10

Hands pointing towards each other per day: How many times in a day do the hands of a clock point towards each other (coincide)?

Accepted Answer

Introduction / Context: The hands coincide 11 times in 12 hours (not 12, since the minute hand “laps” the hour hand slightly less than once per hour). Over a full day, this doubles to 22 coincidences. Given Data / Assumptions: Standard analog 12-hour cycle. Uniform hand motion. Concept / Approach: Use the standard count of overlaps per 12 h and multiply by 2. Step-by-Step Solution: Overlaps per 12 h: 11 ⇒ per 24 h: 22. Verification / Alternative check: Average interval between overlaps is about 65 5/11 minutes; 24 h / (65 5/11 min) ≈ 22. Why Other Options Are Wrong: 24 and 12 assume one per hour or one per 2 hours incorrectly; 20 undercounts. Common Pitfalls: Assuming exactly once per hour; the hour hand moves steadily, so the minute hand cannot lap exactly 12 times in 12 hours. Final Answer: 22

Question 11

Historical weekday (Gandhi Jayanti): Mahatma Gandhi was born on October 2, 1869. What day of the week was it?

Accepted Answer

Introduction / Context: Determining historical weekdays can be done by known references or algorithms (e.g., Zeller’s congruence). For 19th-century Gregorian dates in British India (post-adoption), October 2, 1869 for Mahatma Gandhi is historically recorded as a Saturday. Exam settings often expect recognition of this date or correct application of weekday arithmetic from a known anchor. Given Data / Assumptions: Date: 2-10-1869 (dd-mm-yyyy) in Gregorian calendar. We assume standard Gregorian weekday rules and no regional offset issues for the civil date. Concept / Approach: A quick Zeller-style computation or recalling the known commemorative weekday yields “Saturday.” While one may compute using century/month codes and leap adjustments, many aptitude keys list this as Saturday directly. Step-by-Step Solution (outline): Apply an algorithm (e.g., Zeller’s) to 2-10-1869.Account for month/century codes and leap-year corrections.Resulting code maps to Saturday. Verification / Alternative check: Independent almanacs/calendars confirm 02-Oct-1869 was Saturday. Why Other Options Are Wrong: Sunday/Monday/Friday do not match either algorithmic result or historical record. Common Pitfalls: Using Julian instead of Gregorian rules; misapplying month codes or leap-year adjustments for centuries. Final Answer: Saturday

Question 12

Calendar – Fridays in August 1988: On which date in August 1988 did Friday first occur? (List-style dates like 5, 12, 19, 26 correspond to Fridays in that month.)

Accepted Answer

Introduction / Context: Calendar questions often ask for specific weekdays on given dates. Here, we determine the dates in August 1988 that fall on Friday and identify the first such date. This uses weekday progression across a month once the weekday of the 1st is known. Given Data / Assumptions: Month: August 1988. We need the date of the first Friday in that month. Standard Gregorian calendar rules apply. Concept / Approach: The strategy is: (1) find the weekday of 1 August 1988; (2) move forward to the first Friday; (3) list the Friday pattern (adds of 7 days) to sanity-check. August has 31 days. Step-by-Step Solution: 1) Determine weekday of 1 Aug 1988. It is a Monday (1988 is a leap year; known anchor dates place 1988-08-01 as Monday).2) From Monday, the next Friday is 4 days later: Monday→Friday = +4 days.3) First Friday date = 1 + 4 = 5.4) Subsequent Fridays occur on 12, 19, 26; the first Friday is the 5th. Verification / Alternative check: You can verify with any perpetual calendar method (e.g., Zeller-like congruences or known 1988 references) to confirm 1 Aug 1988 is Monday, making 5 Aug 1988 a Friday. Why Other Options Are Wrong: 4 → Thursday, not Friday. 14 or 17 → do not correspond to the first Friday; 14 is a Sunday; 17 is a Wednesday in that month. Common Pitfalls: Assuming the first Friday is the 7th because Fridays “feel” like weekends, or miscounting from the 1st without confirming the weekday. Final Answer: 5

Question 13

Calendar – Same-year pattern: Which is the next year after 1991 that shares the same calendar as 1990?

Accepted Answer

Introduction / Context: Two years share the same calendar if all month layouts match (weekday of Jan 1 and leap/common status align so that all dates fall on the same weekdays). We want the next year after 1991 that matches the 1990 calendar. Given Data / Assumptions: Reference year: 1990 (a common year). Find the next year after 1991 with an identical calendar to 1990. Concept / Approach: For a common (non-leap) year, the same calendar generally repeats after 6 or 11 years depending on how leap years shift weekday starts. We test candidates after 1991, checking both leap status and weekday alignment. Step-by-Step Solution: 1) 1990 is a common year starting Monday; its exact month-by-month pattern is fixed.2) Check subsequent common years until all months align. The next match is 2001.3) Therefore 2001 is the first year after 1991 that shares 1990’s calendar. Verification / Alternative check: Compare month calendars for 1990 and 2001; each month layout (date vs weekday) matches. Many almanacs or calendar algorithms confirm this. Why Other Options Are Wrong: 1998, 2002, 2003 → their Jan 1 weekdays/leap offsets do not produce identical month layouts to 1990. Common Pitfalls: Assuming a constant 6-year repeat for all common years without checking intervening leap years. Final Answer: 2001

Question 14

Calendar – Day of week: Smt. Indira Gandhi died on 31 October 1984. What day of the week was it?

Accepted Answer

Introduction / Context: Determining weekdays for historic dates is a classic calendar problem solvable via congruence formulas or anchor-day references. The query asks for the weekday of 31 October 1984. Given Data / Assumptions: Date: 31 Oct 1984. Use the proleptic Gregorian calendar (modern standard). Concept / Approach: Methods include Zeller-like congruences, Doomsday algorithm, or verified anchor dates. By computing offsets for 1984 and stepping through month/day values, we identify the weekday. Step-by-Step Solution: 1) 1984 is a leap year.2) Using an anchor-day approach (Doomsday of 1984 is Wednesday), and known doomsday dates in leap years (e.g., 10/10, 8/8), we offset to 10/31.3) The offset yields Wednesday. Verification / Alternative check: Cross-check with a reliable perpetual calendar or mental Doomsday method: both confirm Wednesday. Why Other Options Are Wrong: Monday, Tuesday, Friday → do not match the computed weekday for 31-10-1984. Common Pitfalls: Forgetting 1984 is a leap year or misapplying month offsets can shift the result by ±1 day. Final Answer: Wednesday

Question 15

Calendar – Offset across years: Given that January 16, 1997 was a Thursday, what day of the week was January 4, 2000?

Accepted Answer

Introduction / Context: We have one fixed weekday (Jan 16, 1997 = Thursday) and must find the weekday for a later date (Jan 4, 2000). We step through whole years and partial months counting leap days to compute the weekday shift. Given Data / Assumptions: Jan 16, 1997 = Thursday. Find weekday for Jan 4, 2000. 1997 and 1999 are common years; 1998 is common; 2000 is a leap year. Concept / Approach: Weekday progression advances by +1 per common year and +2 per leap year when comparing same dates. We also adjust for partial-month offsets where needed. Step-by-Step Solution: 1) From Jan 16, 1997 (Thu) to Jan 16, 1998: +1 weekday (common year) → Friday.2) Jan 16, 1998 to Jan 16, 1999: +1 → Saturday.3) Jan 16, 1999 to Jan 16, 2000: +1 (1999 is common) → Sunday. 2000 is leap but Feb 29, 2000 lies after January, so no Jan-shift from leap day yet.4) Now move back from Jan 16, 2000 (Sunday) to Jan 4, 2000: 12 days earlier → Sunday − 12 ≡ Tuesday. Verification / Alternative check: Direct calculation via a congruence (or a calendar reference) confirms Jan 4, 2000 is Tuesday. Why Other Options Are Wrong: Thursday, Wednesday, Friday → do not match the computed weekday shift. Common Pitfalls: Miscounting leap-day effects in early months of a leap year (Feb 29 comes after January). Final Answer: Tuesday

Question 16

Calendar – Day of week: P. V. Narasimha Rao was elected party leader on 29 May 1991. What was the day of the week?

Accepted Answer

Introduction / Context: This is a routine day-of-week determination problem for a given historical date using standard calendar computations. Given Data / Assumptions: Date: 29 May 1991. Gregorian calendar rules. Concept / Approach: Compute the weekday using Doomsday/Zeller-like methods or verified anchor references for 1991, then apply month/day offsets to reach May 29. Step-by-Step Solution: 1) Establish 1991’s reference weekday (e.g., Doomsday falls on Thursday).2) Use May’s doomsday (in non-leap years, 5/9 is a doomsday) and offset to 5/29.3) The arithmetic yields Wednesday for 29 May 1991. Verification / Alternative check: Perpetual calendars confirm that 29-05-1991 was Wednesday. Why Other Options Are Wrong: Tuesday, Friday, Sunday → do not align with the computed weekday. Common Pitfalls: Misusing leap-year adjustments or month-specific reference days. Final Answer: Wednesday

Question 17

Clock – Fast watch (gains 10 min/day): A clock is set right at 10:00 am. It gains 10 minutes in 24 hours. What is the true time when the watch indicates 3:00 pm the next day?

Accepted Answer

Introduction / Context: Fast/slow clock problems compare indicated (watch) time to true time based on a constant daily gain or loss. Here, the watch gains 10 minutes per day. Given Data / Assumptions: Set correct at t=10:00 am (Day 1). Gain = +10 minutes per 24 hours. Indicated time later: 3:00 pm (Day 2), i.e., 29 hours indicated after 10:00 am previous day. Concept / Approach: Relate indicated time to true time via a rate factor. If a watch shows 24 h 10 m when 24 h have truly passed, the rate factor is (24 h + 10 m) / 24 h = 145/144. Step-by-Step Solution: 1) Indicated elapsed = 29 h (from 10:00 am Day 1 to 3:00 pm Day 2).2) True elapsed = indicated / (145/144) = 29 * 144/145 = 28.8 h = 28 h 48 m.3) Add to 10:00 am Day 1 → True time = Day 2 at 2:48 pm. Verification / Alternative check: Proportion method: gain 10 m per 24 h means ~0.4167 m per hour. Over 28 h 48 m true, watch gains about 12 m (rounding aligns to indicated 29 h). Why Other Options Are Wrong: 2:12 pm, 2:30 pm, 2:45 pm → do not match the computed 2:48 pm true time. Common Pitfalls: Subtracting 10 minutes directly from the indicated time or forgetting that gain accumulates proportionally with elapsed true time. Final Answer: 48 min past 2 pm

Question 18

Calendar – Same calendar as 2008: Which of the following years has the same calendar as 2008?

Accepted Answer

Introduction / Context: Identical calendars require identical leap/common status and matching weekday alignment for every month. 2008 is a leap year starting on Tuesday. We must find a future year whose entire calendar matches. Given Data / Assumptions: Reference: 2008 (leap year). We test candidates to find identical month layouts. Concept / Approach: In the Gregorian system, the full calendar pattern for a leap year typically repeats every 28 years unless century rules interfere. Thus 2008 + 28 = 2036 is the first match. Step-by-Step Solution: 1) Verify that 2036 is a leap year and that Jan 1 weekday alignment creates the same monthly layout as 2008.2) Month-by-month comparison confirms 2036 mirrors 2008. Verification / Alternative check: Comparing monthly calendars or using a repetition rule (28-year cycle for leap years not crossing century anomalies) confirms 2036. Why Other Options Are Wrong: 2016, 2020 → leap years but different starting weekdays; 2017/2019 → common years, cannot match a leap-year layout. Common Pitfalls: Assuming any leap year within 8 years matches; the correct full repeat is 28 years (barring special century cases). Final Answer: 2036

Question 19

Clock – Right angles between 5 and 6: At what times between 5:00 and 6:00 will the hour and minute hands be at right angles (90°)?

Accepted Answer

Introduction / Context: Right-angle (90°) problems require setting the angle between the hour and minute hands to 90°. Between 5:00 and 6:00 there are generally two instants where the condition holds. Given Data / Assumptions: Time window: 5:00 to 6:00. Minute-hand speed: 6° per minute. Hour-hand speed: 0.5° per minute; position at 5:00 is 150°. Concept / Approach: Angle formula: |30h − 5.5m| where h is the hour and m is minutes past the hour. For h=5, solve |150 − 5.5m| = 90. Step-by-Step Solution: 1) Case A: 150 − 5.5m = 90 → 5.5m = 60 → m = 10 10/11 minutes.2) Case B: 150 − 5.5m = −90 → 5.5m = 240 → m = 43 7/11 minutes.3) Both moments lie within (5:00, 6:00). Verification / Alternative check: Plug m back into angle formula to confirm 90° (within fractional precision). Why Other Options Are Wrong: Options involving times past 6:00 or mis-stated fractions do not satisfy the 5:00–6:00 window or the 90° condition. Common Pitfalls: Using 6°−0.5° = 5.5° difference correctly but mixing signs; forgetting there are two solutions. Final Answer: 10 10/11 min and 43 7/11 min past 5

Question 20

Clock – Uniform gain/lose period: A watch is 5 minutes slow at 7:00 am on Sunday and 5 minutes 48 seconds fast at 7:00 pm the following Sunday. On which day was it exactly correct?

Accepted Answer

Introduction / Context: When a watch gains uniformly, knowing two offsets at two times lets us find the instant when the error was zero. Here the watch goes from being 5 minutes slow to 5 minutes 48 seconds fast over one week plus 12 hours. Given Data / Assumptions: T1: Sunday 7:00 am, error = −5:00 (slow). T2: Next Sunday 7:00 pm, error = +5:48 (fast). Uniform gain across the interval. Concept / Approach: Total gain = 5:48 − (−5:00) = 10:48 = 648 s. Total real time = 7 days 12 hours = 180 hours = 648,000 s. Rate of gain = 648 / 648,000 = 0.001 s/s = 3.6 s per hour. Time from T1 to zero error = 300 seconds / 3.6 s per hour = 83 1/3 hours. Step-by-Step Solution: 1) Compute gain rate: 10 m 48 s over 180 h → 3.6 s/h.2) Time to correct from −5 m (−300 s): 300 / 3.6 = 83 1/3 h = 83 h 20 m.3) From Sunday 7:00 am + 83 h 20 m → Wednesday 6:20 pm. Verification / Alternative check: Check mid-interval symmetry: the correction time should lie roughly midweek; Wednesday evening fits. Why Other Options Are Wrong: They correspond to different weekdays; only Wednesday aligns with 83 h 20 m after the initial Sunday morning. Common Pitfalls: Confusing “fast” with “slow” sign conventions; mixing minutes and seconds during rate computation. Final Answer: Wednesday

Calendar Questions