Question 1

A clock is set right at 5:00 am and loses 16 minutes in every 24 hours. What is the true time when this slow clock indicates 10:00 pm on the 4th day?

Accepted Answer

Introduction / Context:A slow clock shows less indicated time than true. Convert indicated elapsed time to true elapsed using the rate factor for loss. Given Data / Assumptions: Loss: 16 min per 24 h → rate r = indicated/true = (1440 − 16)/1440 = 1424/1440. Set right: 5:00 am (Day 1). Indicated: 10:00 pm on Day 4. Concept / Approach:Indicated elapsed from Day 1 5:00 am to Day 4 10:00 pm = 3 days 17 h = 89 h. True elapsed = 89 * (1440/1424) = 90 h. Step-by-Step Solution:Add 90 h to 5:00 am (Day 1) → Day 4 11:00 pm. Verification / Alternative check:Since the clock is slow, true time should be later than indicated; 11:00 pm is 1 hour later than 10:00 pm, consistent with the computed factor. Why Other Options Are Wrong:9:00 am/pm and 11:00 am violate direction (they imply earlier time), not later. Common Pitfalls:Using inverse factor (making the time earlier instead of later). Final Answer:11 : 00 pm

Question 2

If a clock takes 22 seconds to strike 12 times (uniform intervals), how long will it take to strike 6 times?

Accepted Answer

Introduction / Context:For striking clocks, the count of intervals equals (strikes − 1). Time scales with intervals if the interval length is constant. Given Data / Assumptions: 12 strikes → 11 intervals → 22 s total. Concept / Approach:Interval = 22/11 = 2 s. For 6 strikes → 5 intervals. Step-by-Step Solution:Time = 5 * 2 = 10 seconds. Verification / Alternative check:Proportional reasoning is straightforward. Why Other Options Are Wrong:12 s and 14 s assume wrong interval counts; 9 s ignores uniformity. Common Pitfalls:Using 6 instead of 5 intervals for 6 strikes. Final Answer:10 seconds

Question 3

March 5, 1999 was a Friday. What day of the week was March 5, 2000?

Accepted Answer

Introduction / Context:From a known weekday, advance one year while accounting for leap years to find the weekday on the same date next year. Given Data / Assumptions: Start: Friday (05-03-1999). Year 2000 is a leap year. Concept / Approach:From Mar 5, 1999 to Mar 5, 2000 includes Feb 29, 2000, so shift = +2 days (common year gives +1; leap year with Feb 29 between dates adds an extra +1). Step-by-Step Solution:Friday + 2 days → Sunday. Verification / Alternative check:Calendar lookup confirms Mar 5, 2000 was Sunday. Why Other Options Are Wrong:Monday/Friday/Tuesday correspond to miscounting leap-day effect. Common Pitfalls:Adding only +1 for a leap year even though Feb 29 lies between the two dates. Final Answer:Sunday

Question 4

India’s first Republic Day was celebrated on January 26, 1950. What was the day of the week on that date?

Accepted Answer

Introduction / Context:Historic weekday identification for 26-01-1950 (first Republic Day of India). We use standard day-of-week calculations or verified references. Given Data / Assumptions: Date: 26-01-1950. Concept / Approach:Apply a day-of-week algorithm (e.g., month codes/century codes) with leap-year handling for 1900s, or verify via authoritative calendar tables. Step-by-Step Solution (sketch):Combine century code (1900s), year code (50), month code (January with leap considerations after 1900), and date 26.Compute modulo 7 to arrive at Thursday. Verification / Alternative check:Public records and calendar archives list 26 January 1950 as Thursday. Why Other Options Are Wrong:They do not match the established weekday for that date. Common Pitfalls:For January/February, leap-year offsets are often mishandled for 1900-series years. Final Answer:Thursday

Question 5

If the day after tomorrow is Sunday, then what day was “tomorrow’s day before yesterday”? (Assume standard consecutive days.)

Accepted Answer

Introduction / Context:This is a verbal logic/calendar question testing careful parsing of relative-day phrases. Given Data / Assumptions: “Day after tomorrow” = Sunday. Concept / Approach:Let today be T. If T + 2 = Sunday, then T = Friday. Step-by-Step Solution:Tomorrow = T + 1 = Saturday.“Tomorrow’s day before yesterday” = (T + 1) − 2 = T − 1 = Thursday. Verification / Alternative check:Enumerate: Thu (T − 1), Fri (T), Sat (T + 1), Sun (T + 2). Works. Why Other Options Are Wrong:Friday/Monday/Tuesday result from offset mistakes (±1 error). Common Pitfalls:Reversing “tomorrow’s day before yesterday” as “day before tomorrow” (which would be today). Final Answer:Thursday

Question 6

A watch gains 3 minutes in every 30 minutes (i.e., runs fast). It is set correctly at 5:00 am. After 6 true hours, what time will this watch show?

Accepted Answer

Introduction / Context:A consistently fast watch accumulates extra indicated minutes. Convert a true-time interval to indicated time by adding the total gain. Given Data / Assumptions: Gain = 3 minutes per 30 minutes of true time. True interval considered = 6 hours = 12 halves of an hour. Concept / Approach:Total gain over 6 hours = 12 * 3 = 36 minutes. Step-by-Step Solution:True 6 hours after 5:00 am → 11:00 am.Indicated = 11:00 am + 36 minutes = 11:36 am. Verification / Alternative check:Rate form: in 60 true minutes, gain = 6 minutes; in 360 minutes, gain = 36; same result. Why Other Options Are Wrong:10:54 am implies a losing watch; 11:30 or 11:42 are incorrect gains. Common Pitfalls:Multiplying by 6 instead of by 12 half-hours. Final Answer:11 : 36 am

Question 7

Scheduled meeting time from relative earliness/lateness — Anuj arrived on Tuesday at 8:15 h (15 minutes before 8:30 h) and found himself 30 minutes earlier than a man who was 40 minutes late. What was the scheduled meeting time?

Accepted Answer

Introduction / Context:Time-of-appointment questions often tie together one person’s early arrival with another’s lateness. The trick is to express both observations relative to the same unknown “scheduled time,” then reconcile the two facts consistently. Given Data / Assumptions: Anuj reached at 8:15 h (since he was 15 minutes before 8:30 h). At 8:15 h, Anuj was 30 minutes earlier than a man who was 40 minutes late. “Late” and “early” are measured relative to the unknown scheduled time T. Concept / Approach:If the other man arrived at time A and was 40 minutes late, then T = A − 40 minutes. Anuj at 8:15 h was 30 minutes earlier than that man, so A = 8:15 h + 30 minutes = 8:45 h. Substitute to obtain T. Step-by-Step Solution:Other man’s arrival A = 8:15 + 0:30 = 8:45.He was 40 minutes late ⇒ scheduled T = 8:45 − 0:40 = 8:05.Therefore, the planned meeting time was 8:05 h. Verification / Alternative check:At T = 8:05, the late man at 8:45 is +40 minutes; Anuj at 8:15 is −10 minutes. Relative to the late man, Anuj is earlier by 30 minutes (8:45 − 8:15), matching the statement. Why Other Options Are Wrong:8:00 and 8:15 fail the 40-minute lateness for the other person given the 8:15/8:45 pair; 8:45 is an arrival, not the scheduled time. Common Pitfalls:Confusing “earlier than the late man” with “early relative to the schedule,” or subtracting 40 from 8:15 instead of from the late man’s 8:45 arrival. Final Answer:8:05 h

Question 8

Total buzzes made by a striking clock in 24 hours — A clock buzzes 1 time at 1 O’clock, 2 times at 2 O’clock, …, 12 times at 12 O’clock. How many total buzzes occur in a full day (24 hours)?

Accepted Answer

Introduction / Context:A striking clock repeats a daily pattern: during each 12-hour cycle it chimes 1 + 2 + … + 12 times at the top of each hour. Over 24 hours, this exact 12-hour sequence happens twice, so the total in a day is just double the 12-hour sum. Given Data / Assumptions: Within any 12-hour span, chimes follow the sequence 1, 2, 3, …, 12. The day contains two such spans: 0–12 and 12–24 (or a.m. and p.m.). We assume no extra chimes beyond the hour counts. Concept / Approach:Use the arithmetic series formula for 1 through 12, then double. Sum of first n integers is n(n + 1)/2. Here n = 12. Step-by-Step Solution:Sum(1..12) = 12 × 13 / 2 = 78.Daily total = 2 × 78 = 156. Verification / Alternative check:Spot-check: 1 + 2 + 3 + 4 = 10; 5 + 6 + 7 + 8 = 26; 9 + 10 + 11 + 12 = 42; total 10 + 26 + 42 = 78; doubling gives 156. Why Other Options Are Wrong:100 and 150 undercount; “None of these” is unnecessary because 156 precisely matches the calculation. Common Pitfalls:Accidentally adding 12 twice or forgetting to double for 24 hours. Another mistake is assuming 0 chimes at 12, but standard striking conventions use 12 chimes at 12 O’clock. Final Answer:156

Question 9

Right angles of clock hands in 24 hours — How many times are a clock’s hour and minute hands at a right angle (90°) in one full day?

Accepted Answer

Introduction / Context:Clock-hand problems rely on relative angular speed. The minute hand moves 6° per minute; the hour hand moves 0.5° per minute. The hands are at right angles (90° apart) twice in most hours, but not exactly in every single hour segment across 12 hours because of boundary effects. Given Data / Assumptions: Relative speed = 5.5° per minute (6 − 0.5). Right-angle separations occur when the angular difference is 90° or 270° (which is also 90° the other way around). We count distinct instants, not durations. Concept / Approach:In 12 hours, there are 22 right-angle positions (standard result). Intuitively, because 11 coincidences occur in 12 hours, and between each overlap there are usually two 90° separations. However, one of the potential right-angle events near the 2→3 or 8→9 boundary is missed, netting 22 rather than 24 in 12 hours. Step-by-Step Solution:Right angles in 12 hours = 22.Therefore, in 24 hours = 2 × 22 = 44. Verification / Alternative check:A known set of results: Overlaps per 12 h = 11; right angles per 12 h = 22; opposite positions per 12 h = 11. Doubling for 24 h gives 22, 44, 22 respectively. Why Other Options Are Wrong:22 and 24 are 12-hour counts misapplied; 48 assumes two right angles in every hour of 24, ignoring the misses near boundaries. Common Pitfalls:Assuming exactly two right angles every hour without exception; the pattern slips due to the continuous drift of the hands. Final Answer:44

Question 10

Coincidences of clock hands in 24 hours — How many times do the hour and minute hands coincide (overlap) in a day?

Accepted Answer

Introduction / Context:“Coincide” means the two hands line up exactly. Because the minute hand moves faster, it laps the hour hand slightly less than once per hour; the standard count for 12 hours is 11 overlaps. Given Data / Assumptions: Relative speed between hands: 5.5° per minute. Between two successive overlaps, the minute hand must gain 360° on the hour hand. We consider a continuous 24-hour period (not just daytime). Concept / Approach:Overlaps in 12 hours are 11 (not 12) due to the moving reference frame; one “potential” overlap shifts past the 12-hour boundary. Therefore, daily overlaps are exactly double that count. Step-by-Step Solution:Overlaps in 12 hours = 11.Overlaps in 24 hours = 2 × 11 = 22. Verification / Alternative check:Explicit formula: times (in minutes after H:00) are t = (60H)/11 for H = 0..10 within each 12-hour window, giving 11 solutions; repeat for the second 12 hours. Why Other Options Are Wrong:24 assumes one per hour; 21 or 20 undercount; 22 is the established standard result. Common Pitfalls:Forgetting the boundary effect that drops one overlap per 12-hour span. Final Answer:22

Question 11

Opposite directions (“pointing towards each other”) in 24 hours — How many times do the hour and minute hands face each other (180° apart) in a day?

Accepted Answer

Introduction / Context:“Pointing towards each other” is typically interpreted as the hands being in opposite directions, i.e., separated by 180°. Like overlaps, these opposite positions follow a repeating pattern tied to the relative speed of the hands. Given Data / Assumptions: Relative speed = 5.5° per minute. Opposite positions correspond to an angular difference of 180° (or equivalently −180°). The pattern over any 12-hour span repeats in the next 12 hours. Concept / Approach:In 12 hours, opposite positions occur 11 times (analogous to 11 overlaps). Therefore, in 24 hours, the count doubles. Step-by-Step Solution:Opposites in 12 hours = 11.Opposites in 24 hours = 2 × 11 = 22. Verification / Alternative check:General formula within an hour interval: t (minutes after H:00) satisfies |30H − 5.5t| = 180. Solving across H = 0..10 yields 11 solutions per 12-hour cycle (excluding boundary duplication). Why Other Options Are Wrong:24 assumes two per hour, which is incorrect; 20 and 12 undercount the standard result. Common Pitfalls:Mixing up counts for right angles (22 per 12 h) with opposites (11 per 12 h). Final Answer:22

Question 12

Day-of-week arithmetic — If today is Thursday, what day will it be after 363 days?

Accepted Answer

Introduction / Context:Days-of-week questions reduce to modular arithmetic modulo 7. Each span of 7 days returns the calendar to the same weekday. We only need the remainder of the day count when divided by 7 to shift forward from the given day. Given Data / Assumptions: Today is Thursday. We need the day after 363 days. A week is 7 days; weekdays cycle every 7. Concept / Approach:Compute 363 mod 7 to get the net weekday shift. 7 × 51 = 357; 363 − 357 = 6. Therefore, we move 6 days ahead from Thursday. Step-by-Step Solution:Remainder = 363 mod 7 = 6.Starting Thursday, +6 days: Friday (1), Saturday (2), Sunday (3), Monday (4), Tuesday (5), Wednesday (6).Therefore, the target day is Wednesday. Verification / Alternative check:Because 364 is divisible by 7, 363 is one day short of a full-week multiple; being one short of a multiple of 7 means “previous day” relative to Thursday + 7k, i.e., Wednesday. Why Other Options Are Wrong:Sunday, Saturday, and Thursday do not correspond to a +6 shift from Thursday; they represent remainders 3, 2, and 0 respectively. Common Pitfalls:Accidentally counting inclusively (starting at 0) or going backward instead of forward by the remainder. Final Answer:None of these (the correct day is Wednesday)

Question 13

Day-of-week after a fixed span — Today is Monday. What day will it be after 61 days?

Accepted Answer

Introduction / Context:Weekday progression repeats every 7 days. To find the day after a given number of days, reduce the number modulo 7 and shift forward from the starting day by that remainder. Given Data / Assumptions: Start: Monday. Advance: 61 days. 7-day cycle modulo arithmetic applies. Concept / Approach:Compute 61 mod 7 for the net shift. 7 × 8 = 56; 61 − 56 = 5. So we move 5 days ahead from Monday. Step-by-Step Solution:Remainder = 61 mod 7 = 5.From Monday, +5 → Tuesday (1), Wednesday (2), Thursday (3), Friday (4), Saturday (5).Target day: Saturday. Verification / Alternative check:61 = 56 + 5 = 8 full weeks + 5 days; eight full weeks return to Monday; five more days reach Saturday. Why Other Options Are Wrong:Wednesday (+2), Tuesday (+1), Thursday (+3) do not match the remainder 5. Common Pitfalls:Starting count at 0 rather than 1, which would shift all results by one day. Final Answer:Saturday

Question 14

Exact weekday after 84 days — If today is Wednesday, what day will it be after 84 days?

Accepted Answer

Introduction / Context:Because 7 divides evenly into many day counts, some questions land on the same weekday after the span. Here we check whether 84 is a multiple of 7 to determine if the weekday repeats exactly. Given Data / Assumptions: Today: Wednesday. Advance: 84 days. Week repeats every 7 days. Concept / Approach:Compute 84 mod 7. If the remainder is 0, the weekday remains unchanged after the span. Step-by-Step Solution:84 ÷ 7 = 12 exactly ⇒ remainder 0.Remainder 0 ⇒ same weekday as today.Therefore the day after 84 days is Wednesday. Verification / Alternative check:Each full week returns to the same weekday; 12 full weeks later remains Wednesday. Why Other Options Are Wrong:Monday, Tuesday, Sunday imply nonzero remainders (−2, −1, or +4), none of which match remainder 0. Common Pitfalls:Adding 1 because of inclusive counting; the standard method is remainder arithmetic relative to today. Final Answer:Wednesday

Question 15

Weekday after 59 days — Today is Thursday; what day will it be 59 days from today?

Accepted Answer

Introduction / Context:Again, reduce the day count modulo 7 to find the net shift from the given weekday. This is a straightforward application of periodicity. Given Data / Assumptions: Start day: Thursday. Advance: 59 days. Concept / Approach:Compute 59 mod 7. Since 7 × 8 = 56, the remainder is 3. Move forward three days from Thursday to reach the target weekday. Step-by-Step Solution:Remainder = 59 − 56 = 3.From Thursday: Friday (1), Saturday (2), Sunday (3).Therefore, the target day is Sunday. Verification / Alternative check:59 = 8 weeks + 3 days; eight weeks preserve the weekday; +3 shifts to Sunday. Why Other Options Are Wrong:Monday (+4), Tuesday (+5), Wednesday (+6) do not match remainder 3. Common Pitfalls:Subtracting 3 instead of adding 3, or accidentally using 60 instead of 59. Final Answer:Sunday

Question 16

Leap-year identification (Gregorian rules) — Which of the following listed years is not a leap year?

Accepted Answer

Introduction / Context:Under Gregorian rules, a year is a leap year if it is divisible by 4, except that century years must be divisible by 400 to be leap years. This question mixes non-century and century values to test the special 400 rule. Given Data / Assumptions: 707 (non-century). 800, 1200, 2000 (all century years). Apply divisibility by 4 and, for centuries, by 400. Concept / Approach:First check simple divisibility by 4 for non-century years. For centuries (years ending with 00), being divisible by 4 is not sufficient; they must be divisible by 400 to be leap years. Step-by-Step Solution:707 ÷ 4 leaves a remainder ⇒ not a leap year.800 ÷ 400 = 2 ⇒ leap year.1200 ÷ 400 = 3 ⇒ leap year.2000 ÷ 400 = 5 ⇒ leap year. Verification / Alternative check:Centuries 1700, 1800, 1900 are not leap years (not divisible by 400), while 2000 is (divisible by 400). This aligns with the rule we used. Why Other Options Are Wrong:800, 1200, 2000 all satisfy the 400 rule and thus are leap years; only 707 fails the leap-year criterion. Common Pitfalls:Assuming any year divisible by 4 is leap, forgetting the century exception. Final Answer:707

Question 17

First weekday of March in a leap year — If January 1 is a Friday in a leap year, what is the day on March 1 of that year?

Accepted Answer

Introduction / Context:Determining the weekday for the first of a later month requires adding the exact number of days elapsed since the known reference day. In a leap year, February has 29 days, which shifts the weekday by one extra day beyond a non-leap scenario. Given Data / Assumptions: Leap year; thus January has 31 days and February has 29 days. January 1 is Friday. We want the weekday on March 1 (after Jan + Feb days). Concept / Approach:Total days elapsed from Jan 1 to Mar 1 = 31 + 29 = 60. Compute 60 mod 7 to get the weekday shift from Friday. Step-by-Step Solution:60 ÷ 7 = 8 remainder 4.Shift 4 days forward from Friday: Saturday (1), Sunday (2), Monday (3), Tuesday (4).Therefore, March 1 is Tuesday. Verification / Alternative check:In a non-leap year, Jan + Feb = 59 ⇒ remainder 3 ⇒ Monday; adding 1 more day for leap years moves it to Tuesday, confirming the logic. Why Other Options Are Wrong:They correspond to incorrect remainders (3, 2, or 0) rather than the correct remainder 4. Common Pitfalls:Assuming February has 28 days in leap years or forgetting to apply modulo 7. Final Answer:Tuesday

Question 18

Bus interval reasoning — Buses leave every 30 minutes. An inquiry clerk says, “The last bus left 10 minutes ago and the next bus leaves at 9:35 am.” At what time did the clerk give this information?

Accepted Answer

Introduction / Context:This problem combines fixed departure intervals with two reference statements: the last departure was 10 minutes ago, and the next is scheduled at a specific absolute time. These two facts let us pin down the current time without enumerating the entire schedule. Given Data / Assumptions: Buses depart every 30 minutes (e.g., at hh:05 and hh:35, or hh:00 and hh:30 depending on alignment). Next bus leaves at 9:35 am. Previous bus departed 10 minutes before the present moment. Concept / Approach:If the next bus is at 9:35 am and intervals are 30 minutes, the prior scheduled departure was at 9:05 am. If the last bus left 10 minutes ago, and that last bus is the 9:05 departure, then the current time is 10 minutes after 9:05 am. Step-by-Step Solution:Next bus: 9:35 ⇒ prior bus: 9:05 (30-minute cadence).“Left 10 minutes ago”: now = 9:05 + 0:10 = 9:15.Therefore, the clerk spoke at 9:15 am. Verification / Alternative check:From 9:15 am, next at 9:35 is 20 minutes ahead; previous at 9:05 is 10 minutes earlier—both statements match. Why Other Options Are Wrong:9:10 would imply a 9:00 or 8:40 schedule; 8:55 or 9:08 do not align with a 9:35 next departure and a 30-minute interval with the given “10 minutes ago” constraint. Common Pitfalls:Mistaking the previous departure for 9:30 or confusing “10 minutes ago” as relative to the next bus instead of the previous one. Final Answer:9 : 15 am

Question 19

Layered relative days — “The day before the day before yesterday is three days after Saturday.” What day is it today?

Accepted Answer

Introduction / Context:These puzzles require carefully translating layered phrases into offsets from today. “Yesterday” is T − 1; “day before yesterday” is T − 2; “the day before the day before yesterday” is T − 3. Equate this to the named weekday + offset, and solve for today T. Given Data / Assumptions: T denotes today. T − 3 equals “three days after Saturday.” Three days after Saturday is Tuesday (Sun = +1, Mon = +2, Tue = +3). Concept / Approach:Set up the identity T − 3 = Tuesday. Then add 3 days to both sides to find T. This keeps orientation consistent and avoids miscounting. Step-by-Step Solution:“Three days after Saturday” = Tuesday.T − 3 = Tuesday ⇒ T = Tuesday + 3 = Friday. Verification / Alternative check:Check the phrasing backwards from Friday: Day before day before yesterday (T − 3) is Tuesday; indeed three days after Saturday, as stated. Why Other Options Are Wrong:Tuesday, Wednesday, Thursday are off by −3, −2, or −1 relative to Friday when mapped through the statement. Common Pitfalls:Confusing “before” counts (moving backward) with “after” counts (moving forward) and swapping signs mid-calculation. Final Answer:Friday

Question 20

Exact angle between clock hands at 4:20 — Compute the smaller angle between the hour and minute hands at 4:20.

Accepted Answer

Introduction / Context:Angles on a clock are computed from the hands’ absolute angles. The hour hand moves 0.5° per minute plus 30° per hour; the minute hand moves 6° per minute. The smaller angle is the absolute difference of these two absolute angles, clipped to ≤ 180° if needed. Given Data / Assumptions: Time: 4:20. Hour-hand angle = 30 × H + 0.5 × M. Minute-hand angle = 6 × M. Concept / Approach:Compute each hand’s angular position from 12 O’clock and then take the absolute difference. If the result exceeds 180°, subtract from 360° to get the smaller angle. Here it will already be acute. Step-by-Step Solution:Hour-hand angle: 30 × 4 + 0.5 × 20 = 120 + 10 = 130°.Minute-hand angle: 6 × 20 = 120°.Difference: |130 − 120| = 10°. Verification / Alternative check:At 4:00 the angle is 120°; in 20 minutes, the hour hand moves 10° while the minute hand moves 120° from 0°, placing them 10° apart. Why Other Options Are Wrong:12.5° and 15° are common guesses but do not match the computed positions; “None of these” is unnecessary. Common Pitfalls:Forgetting the hour hand’s additional 0.5° per minute, or assuming it stays fixed at the hour mark. Final Answer:10°

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