Question 1

What is the angle between the minute and hour hands of a clock at 4:20?

Accepted Answer

Problem restatement Find the smaller angle between the hour and minute hands at 4:20. Given data Hour hand speed = 0.5°/minute (30° per hour). Minute hand speed = 6°/minute (360° per hour). Concept/Approach At H:M, the angle is |30H − 5.5M| degrees (since 5.5 = 11/2). Step-by-step calculation Angle = |30×4 − 5.5×20| = |120 − 110| = 10° Verification/Alternative Direct positions: Hour hand at 4:20 is at 4×30° + 20×0.5° = 120° + 10° = 130°. Minute hand at 20 minutes is 20×6° = 120°. Difference = 10°. Common pitfalls Using 30H − 6M (ignoring hour-hand movement during the minutes). Final Answer 10°

Question 2

At what angle are the hands of a clock inclined at 15 minutes past 5?

Accepted Answer

Problem restatement Compute the smaller angle between the hands at 5:15. Given data Hour-hand position at time H:M = 30H + 0.5M (degrees from 12). Minute-hand position at time H:M = 6M (degrees from 12). Concept/Approach Use angle = |(30H + 0.5M) − 6M| = |30H − 5.5M|. Step-by-step calculation Angle = |30×5 − 5.5×15| = |150 − 82.5| = 67.5° Verification/Alternative Hour hand at 5:15 is 150° + 7.5° = 157.5°. Minute hand is 15×6° = 90°. Difference = 67.5°. Final Answer 67.5°

Question 3

A watch gains 5 seconds every 3 minutes. It was set right at 7:00 A.M. the same day. In the afternoon, when the watch shows 4:15 P.M., what is the true time?

Accepted Answer

Problem restatement The watch runs fast: it gains 5 s every 3 minutes. After it shows 4:15 P.M., find the true clock time. Given data Gain = 5 s per 3 min. Set right at 7:00 A.M.; indicated time later is 4:15 P.M. Concept/Approach In 3 true minutes (180 s), the watch indicates 185 s. Hence the indicated-to-true time ratio is 185:180 = 37:36. So true elapsed time = indicated elapsed × (36/37). Step-by-step calculation Indicated elapsed from 7:00 A.M. to 4:15 P.M. = 9 h 15 m = 555 minutes True elapsed = 555 × (36/37) minutes Since 555/37 = 15, true elapsed = 15 × 36 = 540 minutes = 9 hours True time = 7:00 A.M. + 9:00 = 4:00 P.M. Verification/Alternative Forward check: Over 9 true hours, the watch gains (9 h × 60/3)×5 s = 180×5 s = 900 s = 15 min. Thus it would indicate 9 h 15 m → 4:15 P.M. (consistent). Common pitfalls Subtracting 15 minutes directly from the indicated time without aligning start time and total elapsed. Final Answer 4:00 P.M.

Question 4

How many times in a day are the hands of a clock straight (in the same line, either together or opposite)?

Accepted Answer

Problem restatement Count the number of times in 24 hours that the hour and minute hands are collinear: either overlapping (0°) or opposite (180°). Concept/Approach In 12 hours, the hands overlap 11 times and are opposite 11 times. These are distinct events (no double-counting). Over 24 hours, the counts double. Step-by-step reasoning Overlaps in 12 h = 11; oppositions in 12 h = 11 Total in 12 h = 11 + 11 = 22 Total in 24 h = 2 × 22 = 44 Verification/Alternative Between any two consecutive overlaps (~65 5/11 minutes apart), there is exactly one opposition (~32 8/11 minutes after an overlap). This yields paired counts of 11 each per 12 hours. Common pitfalls Confusing “straight” with “overlap only”; “straight” includes both 0° and 180°. Assuming 24 or 48 based on hours rather than relative motion counts. Final Answer 44 times

Question 5

Between 4 and 5 o'clock, at what exact time will the hands of a clock point in opposite directions (180°)?

Accepted Answer

Problem restatement Find the moment between 4:00 and 5:00 when the minute and hour hands are exactly opposite (angle = 180°). Given data At t minutes after 4, hour-hand angle = 120° + 0.5t°. Minute-hand angle = 6t°. Opposite direction → absolute difference = 180°. Concept/Approach Use relative speed of hands (5.5° per minute) or set up the angle-difference equation and solve for t. Step-by-Step calculation |6t − (120 + 0.5t)| = 1806t − 120 − 0.5t = 180 → 5.5t = 300t = 300 ÷ 5.5 = 54 6⁄11 minutesTime = 4 : 54 6⁄11 Verification/Alternative Relative angular speed = 6 − 0.5 = 5.5°/min. From 4:00, the hands are 120° apart; to be opposite they need a 300° separation change: 300 ÷ 5.5 = 54 6⁄11 min. Common pitfalls Using 60/11 formula (overlap) instead of the 180° condition; forgetting the hour-hand motion (0.5°/min). Final Answer 4:54 6⁄11

Question 6

Between 7 and 8 o'clock, at what time will the clock hands be in a straight line but not together (i.e., 180° apart)?

Accepted Answer

Problem restatement Find the time between 7 and 8 when the hands are 180° apart. Given data At t minutes after 7, hour-hand angle = 210° + 0.5t°. Minute-hand angle = 6t°. Required: |difference| = 180°. Concept/Approach Solve |6t − (210 + 0.5t)| = 180 for t; discard any t outside 0–60. Step-by-Step calculation 6t − 210 − 0.5t = ±180Case 1: 5.5t = 390 → t = 70 10⁄11 (invalid > 60)Case 2: 5.5t = 30 → t = 5 5⁄11Time = 7 : 05 5⁄11 Verification/Alternative Relative speed = 5.5°/min; using a diagram or quick check confirms the second solution is within the hour. Common pitfalls Accepting the extraneous solution t > 60 minutes; mixing up 0° (together) with 180° (opposite). Final Answer 7:05 5⁄11

Question 7

How many times are the hands of a clock at right angles (90°) in a 24-hour day?

Accepted Answer

Problem restatement Count the number of 90° configurations between the hour and minute hands in 24 hours. Given data In 12 hours, hands are at right angles 22 times. Concept/Approach Right angles occur twice in most hours, except slightly irregular spacing due to the hour-hand motion. The well-known count is 22 in 12 hours. Step-by-Step reasoning Right angles in 12 hours = 22Hence in 24 hours = 2 × 22 = 44 Common pitfalls Assuming exactly two per hour (would give 48) and ignoring the skipped/shifted occurrences.

Question 8

Between 3 and 4 o'clock, after how many minutes will the two hands coincide?

Accepted Answer

Problem restatement Find the exact minutes past 3 when the hour and minute hands overlap. Given data At t minutes after 3, hour-hand angle = 90° + 0.5t°. Minute-hand angle = 6t°. Coincide → angles equal. Concept/Approach Set 6t = 90 + 0.5t and solve for t. Step-by-Step calculation 6t = 90 + 0.5t5.5t = 90 → t = 90 ÷ 5.5 = 16 4⁄11 minutes Verification/Alternative General formula: minutes past h o’clock for overlap = (60h) ÷ 11; for h = 3, t = 180÷11 = 16 4⁄11. Common pitfalls Using 60/11 blindly (that gives overlap after 12 o’clock), not adjusting for h. Final Answer 16 4⁄11 minutes

Question 9

What is the angle between the hour and minute hands at 8:30?

Accepted Answer

Problem restatement Compute the smaller angle between hands at 8:30. Given data Hour-hand angle = 30° per hour + 0.5° per minute. Minute-hand angle = 6° per minute. Concept/Approach Find absolute difference of the two angular positions. Step-by-Step calculation Hour hand at 8:30: 8×30 + 0.5×30 = 240 + 15 = 255°Minute hand at 30: 30×6 = 180°Angle = |255 − 180| = 75° Common pitfalls Ignoring the hour-hand's extra 0.5°/min during the half hour.

Question 10

How many times do the hands of a clock coincide (overlap) in a 24-hour day?

Accepted Answer

Problem restatement Count the number of overlaps of the hour and minute hands in 24 hours. Concept/Approach In 12 hours, the hands coincide 11 times (not 12), due to the hour hand's continuous motion. Therefore, in 24 hours they coincide twice that. Step-by-Step reasoning Coincidences in 12 hours = 11Hence in 24 hours = 2 × 11 = 22 Common pitfalls Assuming one overlap per hour (would give 24) instead of 11 per 12 hours.

Question 11

A watch gains uniformly. It is 2 minutes slow at noon on Monday and 4 minutes 48 seconds fast at 2 p.m. the following Monday. When was it correct?

Accepted Answer

Problem restatement The watch's error changes uniformly from −2 min at Monday 12:00 to +4 min 48 s at the next Monday 14:00. Find the instant when the error was zero (i.e., watch showed correct time). Given data Start error (E0) = −2 min. End error (E1) = +4 min 48 s = +4.8 min. Elapsed real time between readings: 7 days + 2 hours = 170 hours. Concept/Approach With uniform gain, error increases linearly. The fraction of the interval to reach zero error is |E0| ÷ (E1 − E0). Step-by-Step calculation Total error change = 4.8 − (−2) = 6.8 minutesFraction to get to zero from start = 2 ÷ 6.8 = 5 ÷ 17Time from Monday 12:00 to correction = 170 h × (5 ÷ 17) = 50 hours50 hours after Monday 12:00 = Wednesday 14:00 Verification/Alternative Remaining time to the second reading: 170 − 50 = 120 h. Error accrued during first 50 h: 2 min (to reach zero). During next 120 h: gain = (6.8 − 2) = 4.8 min, matching the final fastness. Common pitfalls Converting 48 s incorrectly (must be 0.8 minute) or forgetting the 2 extra hours in the 7-day span. Final Answer Wednesday 2 p.m.

Question 12

An accurate clock shows 8:00 a.m. How many degrees does the hour hand rotate by the time the clock shows 2:00 p.m. the same day?

Accepted Answer

Problem restatement Compute the hour hand's rotation from 8 a.m. to 2 p.m. Given data Hour hand rotates 360° in 12 hours → 30°/hour. Elapsed time = 6 hours (8 a.m. to 2 p.m.). Step-by-Step calculation Rotation = 30°/h × 6 h = 180° Common pitfalls Using the minute hand's 6°/min rate instead of the hour hand's 30°/h.

Question 13

Between 5:30 and 6:00, at what time will the hands of the clock be at right angles (90°)?

Accepted Answer

Problem restatement Find the exact time between 5:30 and 6:00 when the angle between the hands is 90°. Given data At t minutes after 5, hour-hand angle = 150° + 0.5t°. Minute-hand angle = 6t°. Right angle → |difference| = 90°. Concept/Approach Solve |6t − (150 + 0.5t)| = 90 and select the solution between t = 30 and t = 60. Step-by-Step calculation |5.5t − 150| = 90Case 1: 5.5t − 150 = 90 → t = 240 ÷ 5.5 = 43 7⁄11Case 2: 5.5t − 150 = −90 → t = 60 ÷ 5.5 = 10 10⁄11 (not in 5:30–6:00)Required time = 5 : 43 7⁄11 Verification/Alternative Relative speed method confirms one valid right angle in this half-hour window. Common pitfalls Forgetting to restrict to the stated interval and accepting the earlier solution. Final Answer 5:43 7⁄11

Question 14

What is the reflex angle between the hands of a clock at 10:25?

Accepted Answer

Problem restatement Find the reflex angle (> 180°) between the hands at 10:25. Given data Hour-hand angle at 10:25: 10×30 + 0.5×25 = 300 + 12.5 = 312.5°. Minute-hand angle: 25×6 = 150°. Concept/Approach The smaller angle is the absolute difference; the reflex angle is 360° minus that. Step-by-Step calculation Smaller angle = |312.5 − 150| = 162.5°Reflex angle = 360° − 162.5° = 197.5° Common pitfalls Reporting the smaller 162.5° instead of the requested reflex angle.

Question 15

A clock is started at noon. By 5:10, through how many degrees has the hour hand turned?

Accepted Answer

Problem restatement Find the total rotation of the hour hand from 12:00 noon to 5:10. Given data Hour hand speed = 0.5° per minute (30° per hour). Elapsed time from 12:00 to 5:10 = 5 hours 10 minutes = 310 minutes. Concept/Approach The hour hand advances continuously. Rotation = speed × time. Step-by-Step calculation Rotation = 0.5°/min × 310 min = 155° Verification/Alternative Break into hours + minutes: in 5 hours the hand turns 5 × 30° = 150°. In 10 minutes, it turns 10 × 0.5° = 5°. Total = 150° + 5° = 155°. Common pitfalls Ignoring the extra 10 minutes of hour-hand motion. Final Answer 155°

Question 16

At 3:40, what is the smaller angle between the hour hand and the minute hand?

Accepted Answer

Problem restatement Compute the smaller angle between the clock hands at 3:40. Given data Hour-hand position at H:M = 30H + 0.5M (degrees from 12). Minute-hand position at H:M = 6M (degrees from 12). Concept/Approach Angle formula: |30H − 5.5M| degrees gives the smaller angle for most times. Step-by-Step calculation Angle = |30 × 3 − 5.5 × 40| = |90 − 220| = 130° Verification/Alternative Explicit positions: hour hand = 30×3 + 0.5×40 = 120 + 20 = 140°; minute hand = 6×40 = 240°; difference = |240 − 140| = 100°? Careful—hour hand at 3:40 is actually 30×3 + 0.5×40 = 90 + 20 = 110°. Minute hand = 240°. Difference = 130°. (Corrected arithmetic.) Common pitfalls Forgetting the hour hand also moves during the minutes. Final Answer 130°

Question 17

Between 9 and 10 o'clock, at what time will the hands of a watch be together?

Accepted Answer

Problem restatement Find the exact time between 9 and 10 when the hour and minute hands coincide. Given data Hour-hand position at time t minutes after 9: 30×9 + 0.5t = 270 + 0.5t (degrees). Minute-hand position: 6t (degrees). Concept/Approach At coincidence: positions equal. Solve 270 + 0.5t = 6t. Step-by-Step calculation 270 + 0.5t = 6t 270 = 5.5t = (11/2)t t = 270 × (2/11) = 540/11 = 49 1/11 minutes Verification/Alternative Time = 9:49 1/11, i.e., 9:49 and 5 5/11 seconds (since 1/11 min = 60/11 s ≈ 5.4545 s). Common pitfalls Using 65 5/11 minutes after 9:00 (that gives the next overlap overall, not within each hour). Final Answer 9:49 1/11

Question 18

A watch has its hands coincide every 64 minutes. How much does the watch gain or lose per day?

Accepted Answer

Problem restatement The normal interval between successive coincidences (overlaps) of hands on a correct clock is 720/11 minutes ≈ 65 5/11 minutes. Here, the watch has overlaps every 64 minutes (of true time). Determine the daily gain or loss. Concept/Approach If the watch runs at a uniform rate k (indicated minutes per true minute), both hands speed up or slow down by the same factor. The true interval between overlaps becomes (720/11) × (1/k). We are told this true interval equals 64 minutes. Step-by-Step calculation (720/11) × (1/k) = 64 1/k = 64 × (11/720) = 704/720 = 88/90 k = 90/88 = 45/44 > 1 (watch runs fast) Daily gain = (k − 1) × 24 × 60 minutes = (1/44) × 1440 = 1440/44 = 32 8/11 minutes Verification/Alternative Because the observed interval (64) is less than the standard 65 5/11, the watch must be gaining time. A shorter overlap interval implies a faster indicated time scale. Common pitfalls Assuming a loss because 64 is smaller; in fact, a smaller interval indicates a gain. Confusing indicated versus true minutes; always anchor equations in true time. Final Answer Gains 32 8/11 minutes/day

Question 19

How many times in a day are the hands of a clock in a straight line but opposite in direction?

Accepted Answer

Problem restatement Count how often, in 24 hours, the hands are collinear with 180° between them (opposite direction). Concept/Approach In any 12-hour span, the hands are opposite exactly 11 times (roughly every 65 5/11 minutes offset by half a cycle). Over 24 hours, this doubles. Step-by-Step reasoning Oppositions in 12 hours = 11 Oppositions in 24 hours = 2 × 11 = 22 Common pitfalls Including overlaps (0°) as well; the question asks for opposite direction (180°) only. Final Answer 22 times

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