Q: Opposite directions between 2 and 3 o’clock: At what time between 2:00 and 3:00 will the clock hands be in opposite directions (180° apart)?

Introduction / Context: We again use relative angular speed to find when the hands are 180° apart, now between 2 and 3. Given Data / Assumptions: At 2:00, hour hand = 60°. Hour hand speed = 0.5°/min; minute hand speed = 6°/min. Concept / Approach: Angle difference after t minutes: |(60 + 0.5t) − 6t| = |60 − 5.5t|. Set equal to 180 and solve in 0 < t < 60. Step-by-Step Solution: 1) |60 − 5.5t| = 180.2) 60 − 5.5t = −180 ⇒ 5.5t = 240 ⇒ t = 240/5.5 = 480/11 minutes.3) The other case gives negative t; discard. Verification / Alternative check: t ≈ 43 7/11 min; substituting yields a 180° separation. Why Other Options Are Wrong: Other fractions do not satisfy |60 − 5.5t| = 180 within the hour. Common Pitfalls: Mixing 90° vs 180°; ignoring the hour-hand movement. Final Answer: 480/11 min past 2

Q: Slow clock — loses 16 minutes/day: A clock is set right at 6:00 a.m. It loses 16 minutes in every 24 hours. If it shows 11:00 p.m. on the 4th day, what is the true time?

Introduction / Context: We convert a shown time from a slow clock into true time using a rate ratio. Given Data / Assumptions: Clock correct at 6:00 a.m. (Day 1). Loss = 16 minutes per 24 true hours. Shown time: 11:00 p.m. on the 4th day. Concept / Approach: In 24 true hours, the clock shows 24 − 16/60 = 23 44/60 = 23 11/15 hours. Thus indicated/true = (23 11/15)/24 = (356/15)/24 = 89/90. Hence true elapsed = indicated × (90/89). Step-by-Step Solution: 1) Indicated elapsed from Day 1, 6:00 a.m. to Day 4, 11:00 p.m.: Day1 6→Day4 6 = 72 h; plus 6→11 p.m. = 17 h; total indicated = 89 h.2) True elapsed = 89 × (90/89) = 90 h.3) True time = Day 1, 6:00 a.m. + 90 h = Day 5, 12:00 a.m. (i.e., midnight as Day 4 ends). Verification / Alternative check: Loses 16 min/day ⇒ in 3 days 18 h (90 h) it would have lost 60 minutes, so it would read one hour behind: 11 p.m. shown while true is 12 o’clock. Why Other Options Are Wrong: 11 p.m. ignores the loss; 2 p.m./9 p.m. are unrelated. Common Pitfalls: Using 16/24 hours directly; forgetting to compute indicated elapsed first. Final Answer: 12 o' clock

Q: Fast clock — gains 10 minutes/day: A clock is correct at 9:00 p.m. It gains 10 minutes in 24 hours. What is the true time when this fast clock indicates 2:00 p.m. on the following day?

Introduction / Context: Here the clock runs fast, so its indicated interval exceeds the true interval. We must convert the shown time back to true time. Given Data / Assumptions: Clock correct at 9:00 p.m. (start). Gain = 10 minutes per 24 true hours. Shown (indicated) time later: 2:00 p.m. next day. Concept / Approach: In 24 true hours, indicated = 24 h + 10 min = 24 + 1/6 = 145/6 h. Thus indicated/true = (145/6)/24 = 145/144. Therefore true elapsed = indicated × (144/145). Step-by-Step Solution: 1) Indicated elapsed from 9:00 p.m. to next-day 2:00 p.m. = 17 h.2) True elapsed = 17 × (144/145) = 2448/145 h = 16 h + (128/145) h.3) Convert fractional hour: (128/145) × 60 ≈ 52.97 min ≈ 53 min.4) True time ≈ 9:00 p.m. + 16 h 53 m = next day about 1:53 p.m. Verification / Alternative check: Check proportion: a rate of +10 min/day ≈ +0.694% fast. Over ~17 h, that is ~7.07 minutes of fastness; 2:00 p.m. minus ~7 min ≈ 1:53 p.m., consistent. Why Other Options Are Wrong: “48 past …” choices assume ~+48 min/day or misuse the ratio; 2:48 or 1:48 do not match the precise ratio 145/144. Common Pitfalls: Subtracting 10 minutes instead of scaling by the correct factor; mixing “true” and “indicated” intervals. Final Answer: About 1 : 53 pm

Q: Minute hand overtakes hour hand every 70 minutes (true time): How much does the clock gain or lose in a day?

Introduction / Context: On a perfect clock, consecutive coincidences (minute hand overtakes hour hand) occur every 720/11 minutes ≈ 65 5/11 min. If the observed interval (by true time) is 70 minutes, the clock’s mechanism runs slow by a constant factor. Given Data / Assumptions: Perfect interval T₀ = 720/11 min. Observed true-time interval T = 70 min. Uniform scale factor r on hand speeds (same for both hands). Concept / Approach: Relative speed scales by r, so T = T₀ / r ⇒ r = T₀ / T = (720/11) / 70 = 72/77. Hence the clock runs at 72/77 of real time (slow). In 24 true hours, indicated time = 24 × r hours; daily loss = 24 − 24r. Step-by-Step Solution: 1) r = 72/77.2) Indicated in 24 h true = 24 × 72/77 = 1728/77 h.3) Daily loss = 24 − 1728/77 = (1848 − 1728)/77 = 120/77 h = (120 × 60)/77 = 7200/77 minutes. Verification / Alternative check: Since T > T₀, the clock must be slow ⇒ a loss (not a gain). The magnitude matches 7200/77 ≈ 93.5 min/day. Why Other Options Are Wrong: “Gain” contradicts T > T₀; 7300/77 and 7300/78 use wrong bases. Common Pitfalls: Using (T − T₀) proportion instead of scaling the rate; forgetting both hands scale equally. Final Answer: 7200/77 min loss

Question 1

If a month in an year start with Monday, then the date of the fourth day after the second Saturday in that month, will be?

Accepted Answer

1st day of month is Monday. 6th day of month is Saturday. 13th day of month is second Saturday. 17th day is the fourth day after second Saturday.

Question 2

In a clock, the angle between the hour hand and minute hand at 5 h 10 min, is?

Accepted Answer

The angle between hour hand and minute hand at 5 O'clock = 5 x 30° = 150° Minute hand move in 10 min = 2 x 30° = 60° Hour hand move in 10 min = 30°/6 = 5° ∴ Angle between hour hand and minute hand at 5 h 10 min = 150° - 60° + 5deg; = 95°

Question 3

The director came to the office to attend a meeting at 15 min past 12 and he came 25 min before his steno who in turn was late by 30 min to the meeting. At what time was the meeting supposed to start?

Accepted Answer

Director came at 12 : 15. Steno came at 12 : 40. Since, steno was late by 30 min to the meeting. ∴ Meeting time 12 : 10

Question 4

When the time in the wall clock in 3:25 pm, the acute angle between the hour hand and minute hand is?

Accepted Answer

Since, angle between two number is 360°/12 = 30° When the time is 3:25 pm, the minute hand will be at 5 and will have moved 60° from 3 and hour hand would be between 3 and 4 it will move 30° in 60 min and so in 25 min it would have moved 30 x 25/60 = 12.5° So, angle between two hand will be 60 - 12.5 = 47.5°

Question 5

Slow clock — from Monday 7:00 a.m. to “Friday 6:00 a.m.” (shown): A clock is correct at 7:00 a.m. on Monday but loses 15 minutes every 24 hours. When this slow clock shows 6:00 a.m. on the following Friday, what is the true time and weekday?

Accepted Answer

Introduction / Context: This is a “gain/loss” clock problem. The clock loses 15 minutes per 24 hours, so its indicated time runs slow relative to true time. We must map a shown (indicated) time back to the true time. Given Data / Assumptions: Clock correct at Monday 7:00 a.m. (start anchor). Loss rate = 15 minutes per 24 true hours. Clock shows Friday 6:00 a.m.; find true time then. Concept / Approach: If a clock loses 15 minutes in 24 hours, then in 24 true hours it shows only 23 h 45 m = 23.75 h. Thus indicated/true time ratio = 23.75/24 = 95/96. Equivalently, true time = indicated × (96/95). We first compute the indicated elapsed time from the anchor, then scale. Step-by-Step Solution: 1) Indicated elapsed from Mon 7:00 a.m. to Fri 6:00 a.m.: 3 full days (Tue–Thu) = 72 h, plus Mon 7→Tue 7 = 24 h (now 96 h), but we need Fri 6 which is 1 hour earlier than Fri 7, so total indicated = 95 h.2) Convert to true elapsed: 95 × (96/95) = 96 h.3) True time = Monday 7:00 a.m. + 96 h = Friday 7:00 a.m. Verification / Alternative check: Over 4 days, the slow clock loses 4 × 15 = 60 minutes, i.e., it reads one hour behind. So when it reads 6:00 a.m., true is 7:00 a.m. Why Other Options Are Wrong: 6:15 or 6:30 a.m. under-correct the one-hour deficit; 7:15 a.m. over-corrects. Common Pitfalls: Using 15/24 as hours directly without forming the ratio; forgetting to compute indicated elapsed time first. Final Answer: 7 : 00 am Friday

Question 6

Straight-line (180°) configuration between 9 and 10 o’clock: At what time between 9:00 and 10:00 are the hands in the same straight line but not together (i.e., 180° apart)?

Accepted Answer

Introduction / Context: We seek the first time after 9:00 when the hands are 180° apart (a straight line). This is a standard relative-speed problem for clock hands. Given Data / Assumptions: Hour hand speed = 0.5° per minute. Minute hand speed = 6° per minute. At 9:00, hour hand angle = 270°, minute hand angle = 0°. Concept / Approach: Angle difference at t minutes after 9:00 is |(270 + 0.5t) − 6t| = |270 − 5.5t|. For a straight line, set this to 180° and solve for t within 0 < t < 60. Step-by-Step Solution: 1) |270 − 5.5t| = 180.2) 270 − 5.5t = 180 ⇒ 5.5t = 90 ⇒ t = 90/5.5 = 180/11 minutes.3) The other case 270 − 5.5t = −180 ⇒ t = 900/11 ≈ 81.82 min exceeds the hour; reject. Verification / Alternative check: At t = 180/11, the computed difference is exactly 180°. Why Other Options Are Wrong: 162/11 and 164/11 are distractors that come from mixing the 90° case or rounding; 162/11 past 10 is outside the 9–10 window. Common Pitfalls: Confusing 90° (right angle) with 180°; forgetting the hour hand moves. Final Answer: 180/11 min past 9

Question 7

Opposite directions between 2 and 3 o’clock: At what time between 2:00 and 3:00 will the clock hands be in opposite directions (180° apart)?

Accepted Answer

Introduction / Context: We again use relative angular speed to find when the hands are 180° apart, now between 2 and 3. Given Data / Assumptions: At 2:00, hour hand = 60°. Hour hand speed = 0.5°/min; minute hand speed = 6°/min. Concept / Approach: Angle difference after t minutes: |(60 + 0.5t) − 6t| = |60 − 5.5t|. Set equal to 180 and solve in 0 < t < 60. Step-by-Step Solution: 1) |60 − 5.5t| = 180.2) 60 − 5.5t = −180 ⇒ 5.5t = 240 ⇒ t = 240/5.5 = 480/11 minutes.3) The other case gives negative t; discard. Verification / Alternative check: t ≈ 43 7/11 min; substituting yields a 180° separation. Why Other Options Are Wrong: Other fractions do not satisfy |60 − 5.5t| = 180 within the hour. Common Pitfalls: Mixing 90° vs 180°; ignoring the hour-hand movement. Final Answer: 480/11 min past 2

Question 8

Slow clock — loses 16 minutes/day: A clock is set right at 6:00 a.m. It loses 16 minutes in every 24 hours. If it shows 11:00 p.m. on the 4th day, what is the true time?

Accepted Answer

Introduction / Context: We convert a shown time from a slow clock into true time using a rate ratio. Given Data / Assumptions: Clock correct at 6:00 a.m. (Day 1). Loss = 16 minutes per 24 true hours. Shown time: 11:00 p.m. on the 4th day. Concept / Approach: In 24 true hours, the clock shows 24 − 16/60 = 23 44/60 = 23 11/15 hours. Thus indicated/true = (23 11/15)/24 = (356/15)/24 = 89/90. Hence true elapsed = indicated × (90/89). Step-by-Step Solution: 1) Indicated elapsed from Day 1, 6:00 a.m. to Day 4, 11:00 p.m.: Day1 6→Day4 6 = 72 h; plus 6→11 p.m. = 17 h; total indicated = 89 h.2) True elapsed = 89 × (90/89) = 90 h.3) True time = Day 1, 6:00 a.m. + 90 h = Day 5, 12:00 a.m. (i.e., midnight as Day 4 ends). Verification / Alternative check: Loses 16 min/day ⇒ in 3 days 18 h (90 h) it would have lost 60 minutes, so it would read one hour behind: 11 p.m. shown while true is 12 o’clock. Why Other Options Are Wrong: 11 p.m. ignores the loss; 2 p.m./9 p.m. are unrelated. Common Pitfalls: Using 16/24 hours directly; forgetting to compute indicated elapsed first. Final Answer: 12 o' clock

Question 9

Fast clock — gains 10 minutes/day: A clock is correct at 9:00 p.m. It gains 10 minutes in 24 hours. What is the true time when this fast clock indicates 2:00 p.m. on the following day?

Accepted Answer

Introduction / Context: Here the clock runs fast, so its indicated interval exceeds the true interval. We must convert the shown time back to true time. Given Data / Assumptions: Clock correct at 9:00 p.m. (start). Gain = 10 minutes per 24 true hours. Shown (indicated) time later: 2:00 p.m. next day. Concept / Approach: In 24 true hours, indicated = 24 h + 10 min = 24 + 1/6 = 145/6 h. Thus indicated/true = (145/6)/24 = 145/144. Therefore true elapsed = indicated × (144/145). Step-by-Step Solution: 1) Indicated elapsed from 9:00 p.m. to next-day 2:00 p.m. = 17 h.2) True elapsed = 17 × (144/145) = 2448/145 h = 16 h + (128/145) h.3) Convert fractional hour: (128/145) × 60 ≈ 52.97 min ≈ 53 min.4) True time ≈ 9:00 p.m. + 16 h 53 m = next day about 1:53 p.m. Verification / Alternative check: Check proportion: a rate of +10 min/day ≈ +0.694% fast. Over ~17 h, that is ~7.07 minutes of fastness; 2:00 p.m. minus ~7 min ≈ 1:53 p.m., consistent. Why Other Options Are Wrong: “48 past …” choices assume ~+48 min/day or misuse the ratio; 2:48 or 1:48 do not match the precise ratio 145/144. Common Pitfalls: Subtracting 10 minutes instead of scaling by the correct factor; mixing “true” and “indicated” intervals. Final Answer: About 1 : 53 pm

Question 10

Minute hand overtakes hour hand every 70 minutes (true time): How much does the clock gain or lose in a day?

Accepted Answer

Introduction / Context: On a perfect clock, consecutive coincidences (minute hand overtakes hour hand) occur every 720/11 minutes ≈ 65 5/11 min. If the observed interval (by true time) is 70 minutes, the clock’s mechanism runs slow by a constant factor. Given Data / Assumptions: Perfect interval T₀ = 720/11 min. Observed true-time interval T = 70 min. Uniform scale factor r on hand speeds (same for both hands). Concept / Approach: Relative speed scales by r, so T = T₀ / r ⇒ r = T₀ / T = (720/11) / 70 = 72/77. Hence the clock runs at 72/77 of real time (slow). In 24 true hours, indicated time = 24 × r hours; daily loss = 24 − 24r. Step-by-Step Solution: 1) r = 72/77.2) Indicated in 24 h true = 24 × 72/77 = 1728/77 h.3) Daily loss = 24 − 1728/77 = (1848 − 1728)/77 = 120/77 h = (120 × 60)/77 = 7200/77 minutes. Verification / Alternative check: Since T > T₀, the clock must be slow ⇒ a loss (not a gain). The magnitude matches 7200/77 ≈ 93.5 min/day. Why Other Options Are Wrong: “Gain” contradicts T > T₀; 7300/77 and 7300/78 use wrong bases. Common Pitfalls: Using (T − T₀) proportion instead of scaling the rate; forgetting both hands scale equally. Final Answer: 7200/77 min loss

Question 11

Hands coincide every 64 minutes: By how much does the watch gain (per day) if successive coincidences occur every 64 minutes?

Accepted Answer

Introduction / Context: On a perfect clock, the time between coincidences is T₀ = 720/11 minutes ≈ 65 5/11 min. If the observed interval (in true time) is shorter, 64 minutes, the watch runs fast by a constant scale factor. Given Data / Assumptions: Perfect coincidence interval T₀ = 720/11 min. Observed true-time interval T = 64 min. Uniform speedup factor r (applies to both hands equally). Concept / Approach: With scaling factor r, T = T₀ / r ⇒ r = T₀ / T = (720/11)/64 = 720/704 = 45/44 > 1 (fast). Daily gain = indicated − true over 24 hours = (r − 1) × 24 hours. Step-by-Step Solution: 1) r = 45/44.2) Daily gain = (45/44 − 1) × 24 h = (1/44) × 24 h = 24/44 h = 6/11 h.3) Convert to minutes: (6/11) × 60 = 360/11 min = 32 8/11 min. Verification / Alternative check: Because 64 < 65 5/11, the watch must gain time; 32 8/11 min ≈ 32.73 min/day. Why Other Options Are Wrong: 37 and 31 minutes are off; “None of the above” is incorrect since an exact value exists. Common Pitfalls: Subtracting intervals directly (T₀ − T) without converting to a rate; forgetting both hands scale equally. Final Answer: 32 8/11 min

Question 12

Weekday shift across a non-leap year boundary: If January 3, 2007 was a Wednesday, what weekday was January 3, 2008?

Accepted Answer

Introduction / Context: We are asked to find the weekday on the same calendar date one year later. The key is whether the intervening year is a leap year and whether Feb 29 lies within the interval affecting the shift. Given Data / Assumptions: Start: Jan 3, 2007 (Wednesday). End: Jan 3, 2008. 2007 is not a leap year (365 days). Concept / Approach: From a date D in year Y to the same date D in year Y+1, the weekday shifts by +1 for a normal year (365 ≡ 1 mod 7), and by +2 only when the span includes Feb 29 (i.e., when the starting date is on/after March in a leap-to-next-year span). Here, we traverse one normal year (no Feb 29 inside the span before the endpoint). Step-by-Step Solution: 1) Days elapsed: 365 (non-leap year 2007).2) 365 mod 7 = 1 ⇒ weekday advances by +1.3) Wednesday + 1 day = Thursday. Verification / Alternative check: Since the destination date (Jan 3, 2008) lies before Feb 29, even though 2008 is leap year, the extra day does not affect this span. Why Other Options Are Wrong: Tuesday is −1; Friday/Saturday imply +2/+3, which do not apply here. Common Pitfalls: Assuming any leap year nearby automatically adds +2 days; forgetting the mod-7 behavior. Final Answer: Thursday

Question 13

Same date next year (leap-year shift): If January 3, 2008 was a Friday, what day of the week fell on January 3, 2009?

Accepted Answer

Introduction / Context: When moving from a given calendar date to the same date in the next year, the weekday shift depends on whether the year in between is a leap year (366 days) or a common year (365 days). A 365-day span advances the weekday by 1; a 366-day span advances it by 2, because 365 ≡ 1 (mod 7) and 366 ≡ 2 (mod 7). Given Data / Assumptions: January 3, 2008 = Friday. We need the weekday on January 3, 2009. Year 2008 is a leap year (divisible by 4 and also by 400 rule for centuries; 2008 qualifies by divisibility by 4 and is not a century year). Concept / Approach: If the span from one year’s date to the next passes through a leap year, the weekday advances by 2. Otherwise, it advances by 1. Here we cross the body of 2008 (a leap year). Step-by-Step Solution: From Jan 3, 2008 to Jan 3, 2009 is 366 days.366 days ≡ 2 (mod 7) ⇒ weekday + 2.Friday + 2 days = Sunday. Verification / Alternative check: Enumerating two steps from Friday: Saturday (1), Sunday (2). Matches the modular reasoning. Why Other Options Are Wrong: Monday/Tuesday/Saturday correspond to +3/+4/ +1 steps, which do not match a leap-year span of +2. Common Pitfalls: Forgetting that leap years shift by two days for the same date next year, not one day. Final Answer: Sunday

Question 14

Weekday across a leap year boundary: If March 5, 1999 was Friday, what day of the week was March 9, 2000?

Accepted Answer

Introduction / Context: Weekday movement over a year depends on whether the interval covers a leap day (Feb 29). From a date in 1999 to the same date in 2000, we cross Feb 29, 2000 because 2000 is a leap year (divisible by 400). Hence the weekday advances by 2 for exactly one year, plus any extra days beyond that “same-date” point. Given Data / Assumptions: March 5, 1999 = Friday. Find weekday on March 9, 2000. 2000 is a leap year under Gregorian rules. Concept / Approach: First move to March 5, 2000 (same calendar date): advance by +2 (leap span). Then move four more days to March 9, 2000. Total shift = +2 + 4 = +6 weekdays. Step-by-Step Solution: Mar 5, 1999 → Mar 5, 2000: +2 days ⇒ Friday → Sunday.Mar 5 → Mar 9 adds 4 days ⇒ Sunday + 4 = Thursday. Verification / Alternative check: Counting forward: Sun (Mar 5), Mon (6), Tue (7), Wed (8), Thu (9). Why Other Options Are Wrong: Wednesday/Saturday/Friday correspond to +5/+1/0 moves, inconsistent with +6. Common Pitfalls: Applying a +1 shift for the “year later” step despite crossing a leap day. Final Answer: Thursday

Question 15

Relating known weekday to a prior year: If March 6, 2005 was a Monday, what was the day of the week on March 7, 2004?

Accepted Answer

Introduction / Context: Relating weekdays across years demands careful handling of leap days. Here, we are given a weekday in 2005 and asked for a weekday in 2004. The key idea is to use the same calendar date one year apart (March 7, 2004 ↔ March 7, 2005) and the fact that between those dates there are 365 days (not 366), since Feb 29, 2004 lies before March 7, 2004 and is therefore not traversed in that one-year span. Given Data / Assumptions: March 6, 2005 = Monday. We want March 7, 2004. From March 7, 2004 to March 7, 2005 is 365 days ⇒ +1 weekday. Concept / Approach: First find March 7, 2005 from March 6, 2005, then step back one year to March 7, 2004 by subtracting one weekday (the inverse of +1). Step-by-Step Solution: March 6, 2005 = Monday ⇒ March 7, 2005 = Tuesday.One year earlier (365-day span): Tuesday → Monday. Verification / Alternative check: Directly: Mar 7, 2004 + 365 days = Mar 7, 2005. So the weekday must advance by +1; reversing gives Monday. Why Other Options Are Wrong: Tuesday/Friday/Sunday correspond to different offsets (0, ±2, etc.) that do not match the 365-day relationship used here. Common Pitfalls: Assuming +2 because 2004 is a leap year; the extra day does not lie within Mar 7, 2004 → Mar 7, 2005. Final Answer: Monday

Question 16

Multi-year weekday shift from a known Jan 1: If it was Sunday on January 1, 2006, what was the day of the week on January 2, 2010?

Accepted Answer

Introduction / Context: Across several years, we sum yearly shifts: +1 for each common year and +2 for each leap year encountered. After reaching the target year’s same date, we add the remaining day offset to reach the exact target date. Given Data / Assumptions: Jan 1, 2006 = Sunday. We need Jan 2, 2010. Year-by-year: 2006 (+1), 2007 (+1), 2008 (leap, +2), 2009 (+1). Concept / Approach: Compute the net shift to Jan 1, 2010, then step one more day to Jan 2, 2010. Step-by-Step Solution: Total to Jan 1, 2010 = 1 + 1 + 2 + 1 = +5 ⇒ Sunday → Friday.Jan 2, 2010 = Friday + 1 = Saturday. Verification / Alternative check: Modulo arithmetic: 365 ≡ 1, 366 ≡ 2 (mod 7); accumulated = 5; then +1 day. Why Other Options Are Wrong: Thursday/Sunday/Friday reflect partial steps; only Saturday matches full accumulation. Common Pitfalls: Forgetting to add the extra one day from Jan 1 to Jan 2 after computing the multi-year shift. Final Answer: Saturday

Question 17

Right angle between 7 and 8 o’clock: At what time between 7 and 8 o’clock will the hands of a clock be at right angles?

Accepted Answer

Introduction / Context: For clock problems, the angle between hands at time H:M is given by |30H − 5.5M| degrees. Right angles occur when this magnitude equals 90°. Between any two consecutive hours, there are typically two right-angle instants. Given Data / Assumptions: Hour between 7 and 8 ⇒ H = 7. Right angle condition: |30H − 5.5M| = 90. Concept / Approach: Set up |210 − 5.5M| = 90 and solve the two linear cases to find the pair of minutes M in (0, 60). Convert fractional minutes to a mixed number of minutes and 11ths (since 5.5 = 11/2 commonly yields denominators of 11). Step-by-Step Solution: Case 1: 210 − 5.5M = 90 ⇒ 5.5M = 120 ⇒ M = 21 9/11.Case 2: 210 − 5.5M = −90 ⇒ 5.5M = 300 ⇒ M = 54 6/11. Verification / Alternative check: Substitute each M to get |210 − 5.5M| = 90 exactly. Why Other Options Are Wrong: 19 5/11 or 21 5/11 are not the correct solutions; 18 min is approximate and off. Since none match either 21 9/11 or 54 6/11, “None of these” is correct. Common Pitfalls: Dropping the negative case or rounding fractions prematurely; there are two answers within the hour. Final Answer: None of these

Question 18

Hands 3 minutes apart between 5 and 6: At what time between 5 and 6 o’clock are the hands of a clock exactly 3 minute-marks apart?

Accepted Answer

Introduction / Context: “3 minutes apart” means an angle of 3 × 6 = 18 degrees between hands. Using the standard angle formula |30H − 5.5M|, we can solve for M in the hour after 5 o’clock. Typically two instants satisfy a small separation within an hour. Given Data / Assumptions: Hour between 5 and 6 ⇒ H = 5. Angle condition: |30H − 5.5M| = 18. Concept / Approach: Write |150 − 5.5M| = 18 and solve both linear cases. Convert any fractional minute to a mixed fraction as needed. Step-by-Step Solution: 150 − 5.5M = 18 ⇒ 5.5M = 132 ⇒ M = 24.150 − 5.5M = −18 ⇒ 5.5M = 168 ⇒ M = 30 6/11. Verification / Alternative check: At 5:24, angle = |150 − 5.5 × 24| = |150 − 132| = 18°. The second instant (≈ 5:30 6/11) also works, but if only one option matches exactly, choose that. Why Other Options Are Wrong: 22 and 26 past 5 are not the exact solutions from the equation; “None” is false because 24 appears and is correct. Common Pitfalls: Confusing “minutes apart” with “minutes past” or using 5.0M instead of 5.5M for the minute contribution of the hour hand. Final Answer: 24 minutes past 5

Question 19

Straight line (0° or 180°) each day: How many times in a day are the hands of a clock in a straight line (either overlapping or opposite)?

Accepted Answer

Introduction / Context: A “straight line” means either 0° (overlap) or 180° (opposite). In 12 hours, the hands overlap 11 times and are opposite 11 times. Over a full day (24 hours), each of these occurs twice as often, totaling 44 straight-line occurrences. Given Data / Assumptions: Uniform hand motion; standard 12-hour analog clock. Co-linear events comprise both overlaps and oppositions. Concept / Approach: Known counts per 12-hour cycle: 11 overlaps and 11 oppositions. Multiply by 2 for 24 hours. Step-by-Step Solution: Overlaps in 12 h: 11 ⇒ in 24 h: 22.Oppositions in 12 h: 11 ⇒ in 24 h: 22.Total straight-line events per day = 22 + 22 = 44. Verification / Alternative check: The average spacing between overlaps is ~65 5/11 minutes; doubling counts for 24 h yields the standard results used in clock problems. Why Other Options Are Wrong: 24/22/42 are partial counts; 44 uniquely includes both types for the entire day. Common Pitfalls: Counting only overlaps or only oppositions, or assuming one per hour. Final Answer: 44

Question 20

Uniformly gaining watch corrected time: A watch is 2 minutes slow at noon on Sunday and 4 minutes 48 seconds fast at 2 pm the following Sunday (uniform gain). When was it correct?

Accepted Answer

Introduction / Context: A uniform-gain (or uniform-loss) watch changes its error at a constant rate. If we know the error at two instants and the elapsed real time between them, we can compute the rate of change of error and then find when the error would be exactly zero (i.e., watch shows correct time). Given Data / Assumptions: Error at start (Sun 12:00): −2 min. Error at end (next Sun 2:00 pm): +4 min 48 s = +4.8 min. Elapsed real time: 7 days + 2 hours = 170 hours. Concept / Approach: Change in error = (+4.8) − (−2) = +6.8 min over 170 h ⇒ rate = 6.8/170 = 0.04 min/h. To go from −2 min to 0, the watch must gain 2 minutes at this rate. Step-by-Step Solution: Rate = 0.04 min per hour.Time to correct = 2 / 0.04 = 50 hours after Sun noon.Sun 12:00 + 48 h = Tue 12:00; +2 h = Tue 2:00 pm. Verification / Alternative check: From Tue 2 pm to next Sun 2 pm is 120 h; error accrued = 120 × 0.04 = 4.8 min (fits final reading). Why Other Options Are Wrong: Other listed times do not lie exactly 50 h from the initial reference. Common Pitfalls: Converting 48 seconds as 0.48 min instead of 0.8 min; forgetting the sign of the initial error (slow). Final Answer: 2 pm on Tuesday

Calendar Questions