138 Reasoning question on Clock and Calendar

Clock Reasoning Questions

1. Time Between Two Moments

• Question: If the time is 3:15 PM, what is the angle between the hour hand and the minute hand?
• Answer: To solve this, calculate the angle for both hands:
  • Minute Hand: Each minute corresponds to 6 degrees. So, at 15 minutes, the minute hand is at 15 × 6 = 90 degrees.
  • Hour Hand: Each hour corresponds to 30 degrees. At 3:00 PM, the hour hand is at 3 × 30 = 90 degrees. However, the hour hand moves slightly as the minute hand advances, so at 3:15 PM, the hour hand is slightly ahead. In 15 minutes, the hour hand moves 0.5 degrees per minute (30 degrees per hour), so in 15 minutes, the hour hand moves 15 × 0.5 = 7.5 degrees.
  • Angle Between Hands: 90 degrees (minute hand) – 90 degrees (hour hand) + 7.5 degrees = 97.5 degrees.

2. Clock Hands Position

• Question: At what time between 4:00 and 5:00 will the minute hand and hour hand be exactly opposite to each other?
• Answer: The hands are opposite when they are 180 degrees apart.
  • For this, you can use the formula:
    • Angle between hour and minute hands = ∣30H−112M∣\left| 30H – \frac{11}{2}M \right|​30H−211​M​, where H is the hour and M is the minutes.
  • For the time between 4 and 5, when the hands are opposite: ∣30(4)−112M∣=180\left| 30(4) – \frac{11}{2}M \right| = 180​30(4)−211​M​=180 Solving for M (minutes) will give you the exact time.

3. Clock Puzzle

• Question: A clock shows the time 8:20. What is the angle between the hour and minute hands?
• Answer:
  • Minute hand at 20 minutes: 20 × 6 = 120 degrees.
  • Hour hand at 8:00: 8 × 30 = 240 degrees.
  • Since the hour hand is moving towards 9, we calculate its position at 8:20. In 20 minutes, the hour hand moves 0.5 degrees per minute, so in 20 minutes, it moves 20 × 0.5 = 10 degrees.
  • So, the hour hand at 8:20 is at 240 + 10 = 250 degrees.
  • Angle between the hands: ∣250−120∣=130\left| 250 – 120 \right| = 130∣250−120∣=130 degrees.

4. Clock Time Calculation

• Question: How many times do the hour and minute hands overlap in a 12-hour period?
• Answer: The hands overlap 11 times in 12 hours, as every time the minute hand catches up with the hour hand, it takes a little less than 1 hour, not exactly 1 hour. So, in 12 hours, the hands overlap 11 times.

5. Minute Hand and Hour Hand at the Same Position

• Question: At what time between 2:00 and 3:00 will the minute hand and the hour hand coincide (overlap)?
• Answer:
  • Use the formula for the angle between the hour and minute hands: θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​, where HHH is the hour and MMM is the minute.
  • For the hands to overlap, the angle must be 0 degrees.
  • ∣30(2)−112M∣=0\left| 30(2) – \frac{11}{2}M \right| = 0​30(2)−211​M​=0
  • Solving for MMM, we get M=360/11=32M = 360 / 11 = 32M=360/11=32 minutes and about 43 seconds.
  • So, the hands will coincide at approximately 2:32:43.

6. Angle Between Hour and Minute Hands

• Question: At 6:00, what is the angle between the hour and minute hands of the clock?
• Answer:
  • At 6:00, the minute hand is at the 12 o’clock position, which is 0 degrees.
  • The hour hand is at the 6 o’clock position. Each hour represents 30 degrees (360 degrees / 12 hours), so at 6:00, the hour hand is at 180 degrees from the 12 o’clock position.
  • Angle between the hands = 180∘180^\circ180∘ (since the minute hand is at 0 degrees).
  Therefore, the angle between the hour and minute hands at 6:00 is 180 degrees.

7. Time When Minute Hand and Hour Hand Are at 90 Degrees

• Question: At what time between 4:00 and 5:00 will the minute hand and the hour hand be at a right angle (90 degrees apart)?
• Answer:
  • For the hands to be at 90 degrees apart, we use the formula for the angle between the hands:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​where HHH is the hour and MMM is the minute.
  • For 4:00 and 5:00, let H=4H = 4H=4, and we want θ=90\theta = 90θ=90.
  • Plugging into the formula:
  ∣30(4)−112M∣=90\left| 30(4) – \frac{11}{2}M \right| = 90​30(4)−211​M​=90 ∣120−112M∣=90\left| 120 – \frac{11}{2}M \right| = 90​120−211​M​=90
  • Solving for MMM, we get two possible values for MMM:
    • 112M=120−90=30⇒M=6011≈5.45\frac{11}{2}M = 120 – 90 = 30 \Rightarrow M = \frac{60}{11} \approx 5.45211​M=120−90=30⇒M=1160​≈5.45 minutes
    • 112M=120+90=210⇒M=42011≈38.18\frac{11}{2}M = 120 + 90 = 210 \Rightarrow M = \frac{420}{11} \approx 38.18211​M=120+90=210⇒M=11420​≈38.18 minutes.
  Therefore, the hands are at a right angle at 4:05 and 4:38.

8. Minute Hand Behind Hour Hand

• Question: What time between 3:00 and 4:00 will the minute hand be exactly behind the hour hand (180 degrees apart)?
• Answer:
  • The hands will be 180 degrees apart when:
  ∣30H−112M∣=180\left| 30H – \frac{11}{2}M \right| = 180​30H−211​M​=180
  • For 3:00 to 4:00, let H=3H = 3H=3, and we want θ=180\theta = 180θ=180.
  ∣30(3)−112M∣=180\left| 30(3) – \frac{11}{2}M \right| = 180​30(3)−211​M​=180 ∣90−112M∣=180\left| 90 – \frac{11}{2}M \right| = 180​90−211​M​=180
  • Solving for MMM, we get two possible values:
    • 112M=90−180=−90⇒M=−18011\frac{11}{2}M = 90 – 180 = -90 \Rightarrow M = \frac{-180}{11}211​M=90−180=−90⇒M=11−180​ (not valid as minutes cannot be negative).
    • 112M=90+180=270⇒M=54011≈49.09\frac{11}{2}M = 90 + 180 = 270 \Rightarrow M = \frac{540}{11} \approx 49.09211​M=90+180=270⇒M=11540​≈49.09.
  Therefore, the hands are exactly 180 degrees apart at approximately 3:49.

9. How Many Times Do the Hands Overlap in 12 Hours?

• Question: How many times do the minute and hour hands overlap in a 12-hour period?
• Answer:
  • The minute hand and hour hand overlap 11 times in 12 hours.
  • This is because the minute hand completes a full circle every hour, but the hour hand also moves slightly during that time. So, the minute hand overlaps with the hour hand slightly less than every hour.
  • Hence, there are 11 overlaps in 12 hours.

10. Hour Hand Position at Any Given Time

• Question: What is the position of the hour hand at 2:30?
• Answer:
  • The hour hand moves 30 degrees every hour. At 2:00, the hour hand is at 2×30=602 \times 30 = 602×30=60 degrees.
  • At 2:30, the hour hand will have moved halfway toward the 3 o’clock position. Since 30 minutes is half an hour, the hour hand will move half of 30 degrees, i.e., 15 degrees.
  • So, at 2:30, the hour hand is at 60+15=7560 + 15 = 7560+15=75 degrees from the 12 o’clock position.

11. Number of Minutes for Minute Hand to Overcome Hour Hand

• Question: Starting at 12:00, how long will it take for the minute hand to completely overcome the hour hand again (i.e., to overlap)?
• Answer:
  • The minute hand overtakes the hour hand once every 65 5/11 minutes (approximately 65.45 minutes).
  • This is the time interval between two consecutive overlaps.

12. Time When Hands Are at 180 Degrees Apart

• Question: At what time between 5:00 and 6:00 will the hour and minute hands be exactly 180 degrees apart?
• Answer:
  • The hands will be 180 degrees apart when:
  ∣30H−112M∣=180\left| 30H – \frac{11}{2}M \right| = 180​30H−211​M​=180
  • For 5:00 and 6:00, let H=5H = 5H=5, and we want θ=180\theta = 180θ=180.
  ∣30(5)−112M∣=180\left| 30(5) – \frac{11}{2}M \right| = 180​30(5)−211​M​=180 ∣150−112M∣=180\left| 150 – \frac{11}{2}M \right| = 180​150−211​M​=180
  • Solving for MMM, we get two possible values:
    • 112M=150−180=−30⇒M=−6011\frac{11}{2}M = 150 – 180 = -30 \Rightarrow M = \frac{-60}{11}211​M=150−180=−30⇒M=11−60​ (not valid, as the minute value cannot be negative).
    • 112M=150+180=330⇒M=66011=60\frac{11}{2}M = 150 + 180 = 330 \Rightarrow M = \frac{660}{11} = 60211​M=150+180=330⇒M=11660​=60.
  So, at 5:27:16 the hands will be exactly 180 degrees apart.

13. Angle Between Hands at Specific Time

• Question: What is the angle between the hour and minute hands of the clock at 9:15?
• Answer:
  • At 9:00, the hour hand is at 270 degrees (9 hours × 30 degrees).
  • The minute hand at 15 minutes is at 90 degrees (15 minutes × 6 degrees).
  • The hour hand moves as time progresses, so at 9:15, it moves: 15×0.5=7.5 degrees15 \times 0.5 = 7.5 \text{ degrees}15×0.5=7.5 degrees
  • Therefore, the hour hand at 9:15 is at 270 + 7.5 = 277.5 degrees.
  • The angle between the hands is: ∣277.5−90∣=187.5 degrees\left| 277.5 – 90 \right| = 187.5 \text{ degrees}∣277.5−90∣=187.5 degrees
  So, the angle between the hands at 9:15 is 187.5 degrees.

14. Hands Form a Right Angle (90 Degrees)

• Question: At what time between 7:00 and 8:00 will the minute and hour hands form a right angle (90 degrees apart)?
• Answer:
  • For the time between 7:00 and 8:00, we use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • We want the angle to be 90 degrees:
  ∣30(7)−112M∣=90\left| 30(7) – \frac{11}{2}M \right| = 90​30(7)−211​M​=90 ∣210−112M∣=90\left| 210 – \frac{11}{2}M \right| = 90​210−211​M​=90
  • Solving for MMM:
    • 112M=210−90=120⇒M=24011≈21.82\frac{11}{2}M = 210 – 90 = 120 \Rightarrow M = \frac{240}{11} \approx 21.82211​M=210−90=120⇒M=11240​≈21.82 minutes.
    • 112M=210+90=300⇒M=60011≈54.55\frac{11}{2}M = 210 + 90 = 300 \Rightarrow M = \frac{600}{11} \approx 54.55211​M=210+90=300⇒M=11600​≈54.55 minutes.
  Therefore, the minute hand and hour hand will be at a right angle at approximately 7:21:49 and 7:54:33.

15. Frequency of Clock Hands Overlapping

• Question: How often do the hour and minute hands overlap in a 24-hour period?
• Answer:
  • In a 12-hour period, the hour and minute hands overlap 11 times.
  • Therefore, in a 24-hour period, they will overlap 22 times.

16. Hands at the Same Position

• Question: At what time between 10:00 and 11:00 will the hour and minute hands coincide (overlap)?
• Answer:
  • The formula for calculating when the hands overlap is:
  ∣30H−112M∣=0\left| 30H – \frac{11}{2}M \right| = 0​30H−211​M​=0
  • For 10:00 to 11:00, let H=10H = 10H=10, and we want the hands to overlap.
  ∣30(10)−112M∣=0\left| 30(10) – \frac{11}{2}M \right| = 0​30(10)−211​M​=0 ∣300−112M∣=0\left| 300 – \frac{11}{2}M \right| = 0​300−211​M​=0
  • Solving for MMM, we get:
  112M=300⇒M=60011≈54.55 minutes\frac{11}{2}M = 300 \Rightarrow M = \frac{600}{11} \approx 54.55 \text{ minutes}211​M=300⇒M=11600​≈54.55 minutesTherefore, the hands overlap at approximately 10:54:33.

17. Time Taken for Minute Hand to Catch Hour Hand

• Question: Starting from 12:00, how much time does the minute hand take to completely catch the hour hand again (overlap)?
• Answer:
  • The minute hand overtakes the hour hand once every 65 5/11 minutes (approximately 65.45 minutes).
  • This is the time interval between two consecutive overlaps of the minute and hour hands.

18. Hands Form a Straight Line (180 Degrees)

• Question: At what time between 1:00 and 2:00 will the hour and minute hands be exactly 180 degrees apart?
• Answer:
  • For 1:00 and 2:00, we use the formula:
  ∣30H−112M∣=180\left| 30H – \frac{11}{2}M \right| = 180​30H−211​M​=180
  • For 1:00:
  ∣30(1)−112M∣=180\left| 30(1) – \frac{11}{2}M \right| = 180​30(1)−211​M​=180 ∣30−112M∣=180\left| 30 – \frac{11}{2}M \right| = 180​30−211​M​=180
  • Solving for MMM:
    • 112M=30−180=−150⇒M=−30011\frac{11}{2}M = 30 – 180 = -150 \Rightarrow M = \frac{-300}{11}211​M=30−180=−150⇒M=11−300​ (not valid).
    • 112M=30+180=210⇒M=42011≈38.18\frac{11}{2}M = 30 + 180 = 210 \Rightarrow M = \frac{420}{11} \approx 38.18211​M=30+180=210⇒M=11420​≈38.18.
  Therefore, the hands are 180 degrees apart at approximately 1:38:11.

19. Minute Hand Movement Speed

• Question: How much faster does the minute hand move compared to the hour hand?
• Answer:
  • The minute hand completes one full revolution (360 degrees) every hour, while the hour hand moves only 30 degrees in that time.
  • Therefore, the minute hand moves 36030=12\frac{360}{30} = 1230360​=12 times faster than the hour hand.

20. Finding the Next Overlap Time

• Question: If the hands of the clock coincide at 12:00, what is the next time they will coincide?
• Answer:
  • The minute hand and hour hand overlap every 65 5/11 minutes. So, the next overlap will occur at approximately 65.45 minutes or 1:05:27.

21. Minute Hand and Hour Hand at 45 Degrees Apart

• Question: At what time between 2:00 and 3:00 will the minute hand and the hour hand be exactly 45 degrees apart?
• Answer:
  • The formula for the angle between the hands is:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 2:00 to 3:00, we want θ=45\theta = 45θ=45:
  ∣30(2)−112M∣=45\left| 30(2) – \frac{11}{2}M \right| = 45​30(2)−211​M​=45 ∣60−112M∣=45\left| 60 – \frac{11}{2}M \right| = 45​60−211​M​=45
  • Solving for MMM:
    • 112M=60−45=15⇒M=3011≈2.73\frac{11}{2}M = 60 – 45 = 15 \Rightarrow M = \frac{30}{11} \approx 2.73211​M=60−45=15⇒M=1130​≈2.73 minutes.
    • 112M=60+45=105⇒M=21011≈19.09\frac{11}{2}M = 60 + 45 = 105 \Rightarrow M = \frac{210}{11} \approx 19.09211​M=60+45=105⇒M=11210​≈19.09 minutes.
  Therefore, the hands are 45 degrees apart at approximately 2:02 and 2:19.

22. Hands at 270 Degrees Apart

• Question: At what time between 4:00 and 5:00 will the minute and hour hands be exactly 270 degrees apart?
• Answer:
  • Using the formula for the angle between the hands:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 4:00 to 5:00, we want θ=270\theta = 270θ=270:
  ∣30(4)−112M∣=270\left| 30(4) – \frac{11}{2}M \right| = 270​30(4)−211​M​=270 ∣120−112M∣=270\left| 120 – \frac{11}{2}M \right| = 270​120−211​M​=270
  • Solving for MMM:
    • 112M=120−270=−150⇒M=−30011\frac{11}{2}M = 120 – 270 = -150 \Rightarrow M = \frac{-300}{11}211​M=120−270=−150⇒M=11−300​ (not valid).
    • 112M=120+270=390⇒M=78011≈70.91\frac{11}{2}M = 120 + 270 = 390 \Rightarrow M = \frac{780}{11} \approx 70.91211​M=120+270=390⇒M=11780​≈70.91 minutes (not possible as it’s greater than 60 minutes).
  So, the minute and hour hands will not be exactly 270 degrees apart between 4:00 and 5:00.

23. Time Between 12:00 and 1:00 When Hands Are at 120 Degrees

• Question: At what time between 12:00 and 1:00 will the minute and hour hands be exactly 120 degrees apart?
• Answer:
  • Using the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 12:00 to 1:00, we want θ=120\theta = 120θ=120:
  ∣30(12)−112M∣=120\left| 30(12) – \frac{11}{2}M \right| = 120​30(12)−211​M​=120 ∣360−112M∣=120\left| 360 – \frac{11}{2}M \right| = 120​360−211​M​=120
  • Solving for MMM:
    • 112M=360−120=240⇒M=48011≈43.64\frac{11}{2}M = 360 – 120 = 240 \Rightarrow M = \frac{480}{11} \approx 43.64211​M=360−120=240⇒M=11480​≈43.64 minutes.
    • So, the minute and hour hands are 120 degrees apart at approximately 12:43:38.

24. Number of Times the Hands are at Right Angles (90 Degrees)

• Question: How many times in a 24-hour period do the hands of the clock form a right angle (90 degrees apart)?
• Answer:
  • In a 12-hour period, the hands form a right angle 22 times (once every hour, but not exactly on the hour).
  • Therefore, in a 24-hour period, the hands will form a right angle 44 times.

25. Time When Minute Hand is 5 Minutes Ahead of Hour Hand

• Question: At what time between 3:00 and 4:00 will the minute hand be exactly 5 minutes ahead of the hour hand?
• Answer:
  • Since 1 minute represents 6 degrees of motion for the minute hand, the minute hand will be 30 degrees ahead after 5 minutes. We need to find when the angle between the hour and minute hand is 30 degrees.
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 3:00 to 4:00, let H=3H = 3H=3, and we want θ=30\theta = 30θ=30:
  ∣30(3)−112M∣=30\left| 30(3) – \frac{11}{2}M \right| = 30​30(3)−211​M​=30 ∣90−112M∣=30\left| 90 – \frac{11}{2}M \right| = 30​90−211​M​=30
  • Solving for MMM:
    • 112M=90−30=60⇒M=12011≈10.91\frac{11}{2}M = 90 – 30 = 60 \Rightarrow M = \frac{120}{11} \approx 10.91211​M=90−30=60⇒M=11120​≈10.91 minutes.
    • So, the minute hand is 5 minutes ahead of the hour hand at approximately 3:10:55.

26. How Many Times the Hands Overlap in 12 Hours

• Question: How many times do the hour and minute hands overlap in a 12-hour period?
• Answer:
  • The hands overlap 11 times in 12 hours. This happens because the minute hand catches up with the hour hand approximately every 65.45 minutes (not exactly 1 hour).

27. Time When Hour Hand is Ahead of Minute Hand by 10 Degrees

• Question: At what time between 5:00 and 6:00 will the hour hand be ahead of the minute hand by 10 degrees?
• Answer:
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 5:00 to 6:00, let H=5H = 5H=5, and we want θ=10\theta = 10θ=10:
  ∣30(5)−112M∣=10\left| 30(5) – \frac{11}{2}M \right| = 10​30(5)−211​M​=10 ∣150−112M∣=10\left| 150 – \frac{11}{2}M \right| = 10​150−211​M​=10
  • Solving for MMM:
    • 112M=150−10=140⇒M=28011≈25.45\frac{11}{2}M = 150 – 10 = 140 \Rightarrow M = \frac{280}{11} \approx 25.45211​M=150−10=140⇒M=11280​≈25.45 minutes.
  Therefore, the hour hand is ahead of the minute hand by 10 degrees at approximately 5:25:27.

28. Time When Minute Hand is 20 Minutes Ahead of Hour Hand

• Question: At what time between 6:00 and 7:00 will the minute hand be exactly 20 minutes ahead of the hour hand?
• Answer:
  • Since each minute corresponds to 6 degrees, 20 minutes corresponds to 20×6=12020 \times 6 = 12020×6=120 degrees.
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 6:00 to 7:00, let H=6H = 6H=6, and we want θ=120\theta = 120θ=120:
  ∣30(6)−112M∣=120\left| 30(6) – \frac{11}{2}M \right| = 120​30(6)−211​M​=120 ∣180−112M∣=120\left| 180 – \frac{11}{2}M \right| = 120​180−211​M​=120
  • Solving for MMM:
    • 112M=180−120=60⇒M=12011≈10.91\frac{11}{2}M = 180 – 120 = 60 \Rightarrow M = \frac{120}{11} \approx 10.91211​M=180−120=60⇒M=11120​≈10.91 minutes.
  Therefore, the minute hand is 20 minutes ahead of the hour hand at approximately 6:10:55.

29. Time When Hands are 45 Degrees Apart

• Question: At what time between 6:00 and 7:00 will the hour and minute hands be exactly 45 degrees apart?
• Answer:
  • The formula for the angle between the hands is:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 6:00 to 7:00, let H=6H = 6H=6, and we want θ=45\theta = 45θ=45:
  ∣30(6)−112M∣=45\left| 30(6) – \frac{11}{2}M \right| = 45​30(6)−211​M​=45 ∣180−112M∣=45\left| 180 – \frac{11}{2}M \right| = 45​180−211​M​=45
  • Solving for MMM:
    • 112M=180−45=135⇒M=27011≈24.55\frac{11}{2}M = 180 – 45 = 135 \Rightarrow M = \frac{270}{11} \approx 24.55211​M=180−45=135⇒M=11270​≈24.55 minutes.
    • 112M=180+45=225⇒M=45011≈40.91\frac{11}{2}M = 180 + 45 = 225 \Rightarrow M = \frac{450}{11} \approx 40.91211​M=180+45=225⇒M=11450​≈40.91 minutes.
  Therefore, the hour and minute hands will be 45 degrees apart at approximately 6:24:33 and 6:40:55.

30. Time When Hands Are at 90 Degrees

• Question: At what time between 8:00 and 9:00 will the hour and minute hands be exactly 90 degrees apart?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 8:00 to 9:00, we want θ=90\theta = 90θ=90:
  ∣30(8)−112M∣=90\left| 30(8) – \frac{11}{2}M \right| = 90​30(8)−211​M​=90 ∣240−112M∣=90\left| 240 – \frac{11}{2}M \right| = 90​240−211​M​=90
  • Solving for MMM:
    • 112M=240−90=150⇒M=30011≈27.27\frac{11}{2}M = 240 – 90 = 150 \Rightarrow M = \frac{300}{11} \approx 27.27211​M=240−90=150⇒M=11300​≈27.27 minutes.
    • 112M=240+90=330⇒M=66011≈60\frac{11}{2}M = 240 + 90 = 330 \Rightarrow M = \frac{660}{11} \approx 60211​M=240+90=330⇒M=11660​≈60 minutes (not valid as it’s exactly the top of the hour).
  Therefore, the hour and minute hands are 90 degrees apart at approximately 8:27:16.

31. Time When Hands are at 180 Degrees Apart

• Question: At what time between 12:00 and 1:00 will the hour and minute hands be exactly 180 degrees apart?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 12:00 to 1:00, we want θ=180\theta = 180θ=180:
  ∣30(12)−112M∣=180\left| 30(12) – \frac{11}{2}M \right| = 180​30(12)−211​M​=180 ∣360−112M∣=180\left| 360 – \frac{11}{2}M \right| = 180​360−211​M​=180
  • Solving for MMM:
    • 112M=360−180=180⇒M=36011≈32.73\frac{11}{2}M = 360 – 180 = 180 \Rightarrow M = \frac{360}{11} \approx 32.73211​M=360−180=180⇒M=11360​≈32.73 minutes.
    • Therefore, the hands are 180 degrees apart at approximately 12:32:44.

32. Minute Hand is 3 Times Faster Than Hour Hand

• Question: How much faster does the minute hand move compared to the hour hand?
• Answer:
  • The minute hand moves 6 degrees per minute, and the hour hand moves 0.5 degrees per minute.
  • Therefore, the minute hand is moving 60.5=12\frac{6}{0.5} = 120.56​=12 times faster than the hour hand.

33. Hands Form a Straight Line (180 Degrees)

• Question: At what time between 3:00 and 4:00 will the hour and minute hands form a straight line (180 degrees apart)?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 3:00 to 4:00, we want θ=180\theta = 180θ=180:
  ∣30(3)−112M∣=180\left| 30(3) – \frac{11}{2}M \right| = 180​30(3)−211​M​=180 ∣90−112M∣=180\left| 90 – \frac{11}{2}M \right| = 180​90−211​M​=180
  • Solving for MMM:
    • 112M=90−180=−90⇒M=−18011\frac{11}{2}M = 90 – 180 = -90 \Rightarrow M = \frac{-180}{11}211​M=90−180=−90⇒M=11−180​ (not valid, as MMM must be positive).
    • 112M=90+180=270⇒M=54011≈49.09\frac{11}{2}M = 90 + 180 = 270 \Rightarrow M = \frac{540}{11} \approx 49.09211​M=90+180=270⇒M=11540​≈49.09 minutes.
  Therefore, the hands will form a straight line at approximately 3:49:09.

34. Time When Minute Hand Lags Behind Hour Hand by 15 Minutes

• Question: At what time between 4:00 and 5:00 will the minute hand lag behind the hour hand by 15 minutes?
• Answer:
  • Each minute corresponds to 6 degrees, so 15 minutes corresponds to 15×6=9015 \times 6 = 9015×6=90 degrees.
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 4:00 to 5:00, let H=4H = 4H=4, and we want θ=90\theta = 90θ=90:
  ∣30(4)−112M∣=90\left| 30(4) – \frac{11}{2}M \right| = 90​30(4)−211​M​=90 ∣120−112M∣=90\left| 120 – \frac{11}{2}M \right| = 90​120−211​M​=90
  • Solving for MMM:
    • 112M=120−90=30⇒M=6011≈5.45\frac{11}{2}M = 120 – 90 = 30 \Rightarrow M = \frac{60}{11} \approx 5.45211​M=120−90=30⇒M=1160​≈5.45 minutes.
    • 112M=120+90=210⇒M=42011≈38.18\frac{11}{2}M = 120 + 90 = 210 \Rightarrow M = \frac{420}{11} \approx 38.18211​M=120+90=210⇒M=11420​≈38.18 minutes.
  Therefore, the minute hand lags behind the hour hand by 15 minutes at approximately 4:05:27 and 4:38:11.

35. How Often Do the Hands Overlap in 24 Hours?

• Question: How many times in 24 hours do the hour and minute hands overlap?
• Answer:
  • In 12 hours, the hands overlap 11 times.
  • Therefore, in 24 hours, the hands overlap 22 times.

36. Time for Hands to Form an Angle of 60 Degrees

• Question: At what time between 9:00 and 10:00 will the minute and hour hands form an angle of 60 degrees?
• Answer:
  • Using the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 9:00 to 10:00, we want θ=60\theta = 60θ=60:
  ∣30(9)−112M∣=60\left| 30(9) – \frac{11}{2}M \right| = 60​30(9)−211​M​=60 ∣270−112M∣=60\left| 270 – \frac{11}{2}M \right| = 60​270−211​M​=60
  • Solving for MMM:
    • 112M=270−60=210⇒M=42011≈38.18\frac{11}{2}M = 270 – 60 = 210 \Rightarrow M = \frac{420}{11} \approx 38.18211​M=270−60=210⇒M=11420​≈38.18 minutes.
    • 112M=270+60=330⇒M=66011≈60\frac{11}{2}M = 270 + 60 = 330 \Rightarrow M = \frac{660}{11} \approx 60211​M=270+60=330⇒M=11660​≈60 minutes (this is the next hour, so not valid).
  Therefore, the hands form a 60-degree angle at approximately 9:38:11.

37. Time When Minute Hand is Ahead by 10 Degrees

• Question: At what time between 2:00 and 3:00 will the minute hand be 10 degrees ahead of the hour hand?
• Answer:
  • Each minute represents 6 degrees of movement for the minute hand. So, 10 degrees corresponds to approximately 106=1.67\frac{10}{6} = 1.67610​=1.67 minutes.
  • Using the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 2:00 to 3:00, let H=2H = 2H=2, and we want θ=10\theta = 10θ=10:
  ∣30(2)−112M∣=10\left| 30(2) – \frac{11}{2}M \right| = 10​30(2)−211​M​=10 ∣60−112M∣=10\left| 60 – \frac{11}{2}M \right| = 10​60−211​M​=10
  • Solving for MMM:
    • 112M=60−10=50⇒M=10011≈9.09\frac{11}{2}M = 60 – 10 = 50 \Rightarrow M = \frac{100}{11} \approx 9.09211​M=60−10=50⇒M=11100​≈9.09 minutes.
    • Therefore, the minute hand is 10 degrees ahead of the hour hand at approximately 2:09:09.

38. Time When the Minute Hand is 5 Minutes Behind the Hour Hand

• Question: At what time between 2:00 and 3:00 will the minute hand be exactly 5 minutes behind the hour hand?
• Answer:
  • Since 1 minute = 6 degrees on the clock, 5 minutes corresponds to 30 degrees.
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 2:00 to 3:00, let H=2H = 2H=2, and we want θ=30\theta = 30θ=30:
  ∣30(2)−112M∣=30\left| 30(2) – \frac{11}{2}M \right| = 30​30(2)−211​M​=30 ∣60−112M∣=30\left| 60 – \frac{11}{2}M \right| = 30​60−211​M​=30
  • Solving for MMM:
    • 112M=60−30=30⇒M=6011≈5.45\frac{11}{2}M = 60 – 30 = 30 \Rightarrow M = \frac{60}{11} \approx 5.45211​M=60−30=30⇒M=1160​≈5.45 minutes.
    • 112M=60+30=90⇒M=18011≈16.36\frac{11}{2}M = 60 + 30 = 90 \Rightarrow M = \frac{180}{11} \approx 16.36211​M=60+30=90⇒M=11180​≈16.36 minutes.
  Therefore, the minute hand is 5 minutes behind the hour hand at approximately 2:05:27 and 2:16:21.

39. The Hands Form an Acute Angle (Less Than 90 Degrees)

• Question: At what time between 7:00 and 8:00 will the hour and minute hands form an acute angle (less than 90 degrees)?
• Answer:
  • The acute angle is formed when the hands are less than 90 degrees apart.
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 7:00 to 8:00, let H=7H = 7H=7 and we want the angle to be less than 90 degrees:
  ∣30(7)−112M∣<90\left| 30(7) – \frac{11}{2}M \right| < 90​30(7)−211​M​<90 ∣210−112M∣<90\left| 210 – \frac{11}{2}M \right| < 90​210−211​M​<90
  • Solving for MMM:
    • 112M=210−90=120⇒M=24011≈21.82\frac{11}{2}M = 210 – 90 = 120 \Rightarrow M = \frac{240}{11} \approx 21.82211​M=210−90=120⇒M=11240​≈21.82 minutes.
    • 112M=210+90=300⇒M=60011≈54.55\frac{11}{2}M = 210 + 90 = 300 \Rightarrow M = \frac{600}{11} \approx 54.55211​M=210+90=300⇒M=11600​≈54.55 minutes.
  So, the acute angle occurs when the minute hand is between 21.82 minutes and 54.55 minutes after 7:00. Thus, between approximately 7:21:49 and 7:54:33.

40. Hands Form a Right Angle Twice in One Hour

• Question: How many times do the hands of the clock form a right angle (90 degrees apart) in a 1-hour period?
• Answer:
  • The hour and minute hands will form a right angle twice in every hour — once when the minute hand is ahead of the hour hand and once when it lags behind.
  • Therefore, in 1 hour, the hands will form a right angle twice.

41. Minute Hand Overlaps Hour Hand

• Question: How many times do the minute and hour hands overlap in 12 hours?
• Answer:
  • The hands overlap 11 times in 12 hours.
  • They overlap once approximately every 1 hour and 5 minutes (not exactly every hour, because the hour hand is also moving forward).

42. Time When Hands are 150 Degrees Apart

• Question: At what time between 1:00 and 2:00 will the hour and minute hands be exactly 150 degrees apart?
• Answer:
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 1:00 to 2:00, we want θ=150\theta = 150θ=150:
  ∣30(1)−112M∣=150\left| 30(1) – \frac{11}{2}M \right| = 150​30(1)−211​M​=150 ∣30−112M∣=150\left| 30 – \frac{11}{2}M \right| = 150​30−211​M​=150
  • Solving for MMM:
    • 112M=30−150=−120⇒M=−24011\frac{11}{2}M = 30 – 150 = -120 \Rightarrow M = \frac{-240}{11}211​M=30−150=−120⇒M=11−240​ (not valid).
    • 112M=30+150=180⇒M=36011≈32.73\frac{11}{2}M = 30 + 150 = 180 \Rightarrow M = \frac{360}{11} \approx 32.73211​M=30+150=180⇒M=11360​≈32.73 minutes.
  Therefore, the hour and minute hands are 150 degrees apart at approximately 1:32:44.

43. Time Between 4:00 and 5:00 When Hands are 60 Degrees Apart

• Question: At what time between 4:00 and 5:00 will the hour and minute hands be exactly 60 degrees apart?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 4:00 to 5:00, we want θ=60\theta = 60θ=60:
  ∣30(4)−112M∣=60\left| 30(4) – \frac{11}{2}M \right| = 60​30(4)−211​M​=60 ∣120−112M∣=60\left| 120 – \frac{11}{2}M \right| = 60​120−211​M​=60
  • Solving for MMM:
    • 112M=120−60=60⇒M=12011≈10.91\frac{11}{2}M = 120 – 60 = 60 \Rightarrow M = \frac{120}{11} \approx 10.91211​M=120−60=60⇒M=11120​≈10.91 minutes.
    • 112M=120+60=180⇒M=36011≈32.73\frac{11}{2}M = 120 + 60 = 180 \Rightarrow M = \frac{360}{11} \approx 32.73211​M=120+60=180⇒M=11360​≈32.73 minutes.
  Therefore, the hands are 60 degrees apart at approximately 4:10:55 and 4:32:44.

44. Time Between 5:00 and 6:00 When Hands are at 120 Degrees

• Question: At what time between 5:00 and 6:00 will the hour and minute hands be exactly 120 degrees apart?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 5:00 to 6:00, we want θ=120\theta = 120θ=120:
  ∣30(5)−112M∣=120\left| 30(5) – \frac{11}{2}M \right| = 120​30(5)−211​M​=120 ∣150−112M∣=120\left| 150 – \frac{11}{2}M \right| = 120​150−211​M​=120
  • Solving for MMM:
    • 112M=150−120=30⇒M=6011≈5.45\frac{11}{2}M = 150 – 120 = 30 \Rightarrow M = \frac{60}{11} \approx 5.45211​M=150−120=30⇒M=1160​≈5.45 minutes.
    • 112M=150+120=270⇒M=54011≈49.09\frac{11}{2}M = 150 + 120 = 270 \Rightarrow M = \frac{540}{11} \approx 49.09211​M=150+120=270⇒M=11540​≈49.09 minutes.
  Therefore, the hands are 120 degrees apart at approximately 5:05:27 and 5:49:09.

45. Time Between 11:00 and 12:00 When Hands Form an Angle of 45 Degrees

• Question: At what time between 11:00 and 12:00 will the hour and minute hands form an angle of 45 degrees?
• Answer:
  • Using the formula for the angle between the hands:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 11:00 to 12:00, we want θ=45\theta = 45θ=45:
  ∣30(11)−112M∣=45\left| 30(11) – \frac{11}{2}M \right| = 45​30(11)−211​M​=45 ∣330−112M∣=45\left| 330 – \frac{11}{2}M \right| = 45​330−211​M​=45
  • Solving for MMM:
    • 112M=330−45=285⇒M=57011≈51.82\frac{11}{2}M = 330 – 45 = 285 \Rightarrow M = \frac{570}{11} \approx 51.82211​M=330−45=285⇒M=11570​≈51.82 minutes.
    • 112M=330+45=375⇒M=75011≈68.18\frac{11}{2}M = 330 + 45 = 375 \Rightarrow M = \frac{750}{11} \approx 68.18211​M=330+45=375⇒M=11750​≈68.18 minutes (which is greater than 60, so not valid).
  Therefore, the hands form a 45-degree angle at approximately 11:51:49.

46. Time When Hands are Exactly Overlapping

• Question: At what time between 6:00 and 7:00 will the hour and minute hands be exactly overlapping?
• Answer:
  • Using the formula for the angle between the hands:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 6:00 to 7:00, the hands are exactly overlapping when θ=0\theta = 0θ=0:
  ∣30(6)−112M∣=0\left| 30(6) – \frac{11}{2}M \right| = 0​30(6)−211​M​=0 ∣180−112M∣=0\left| 180 – \frac{11}{2}M \right| = 0​180−211​M​=0
  • Solving for MMM:
    • 112M=180⇒M=36011≈32.73\frac{11}{2}M = 180 \Rightarrow M = \frac{360}{11} \approx 32.73211​M=180⇒M=11360​≈32.73 minutes.
  Therefore, the hands overlap at approximately 6:32:44.

47. Hands Form a 90 Degree Angle

• Question: How many times do the hands of the clock form a 90-degree angle in a 24-hour period?
• Answer:
  • The hands of the clock form a 90-degree angle 22 times in a 24-hour period.
  • In each 12-hour period, they form a 90-degree angle 11 times.

48. When Are the Hands Exactly Opposite to Each Other?

• Question: Between 9:00 and 10:00, at what time will the hour and minute hands be exactly opposite to each other (i.e., 180 degrees apart)?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 9:00 to 10:00, we want θ=180\theta = 180θ=180:
  ∣30(9)−112M∣=180\left| 30(9) – \frac{11}{2}M \right| = 180​30(9)−211​M​=180 ∣270−112M∣=180\left| 270 – \frac{11}{2}M \right| = 180​270−211​M​=180
  • Solving for MMM:
    • 112M=270−180=90⇒M=18011≈16.36\frac{11}{2}M = 270 – 180 = 90 \Rightarrow M = \frac{180}{11} \approx 16.36211​M=270−180=90⇒M=11180​≈16.36 minutes.
    • 112M=270+180=450⇒M=90011≈81.82\frac{11}{2}M = 270 + 180 = 450 \Rightarrow M = \frac{900}{11} \approx 81.82211​M=270+180=450⇒M=11900​≈81.82 minutes (which is greater than 60 minutes, so not valid).
  Therefore, the hands are exactly opposite at approximately 9:16:22.

49. Hands Form an Acute Angle

• Question: At what time between 10:00 and 11:00 do the hour and minute hands form an acute angle (less than 90 degrees)?
• Answer:
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 10:00 to 11:00, we want θ<90\theta < 90θ<90:
  ∣30(10)−112M∣<90\left| 30(10) – \frac{11}{2}M \right| < 90​30(10)−211​M​<90 ∣300−112M∣<90\left| 300 – \frac{11}{2}M \right| < 90​300−211​M​<90
  • Solving for MMM:
    • 112M=300−90=210⇒M=42011≈38.18\frac{11}{2}M = 300 – 90 = 210 \Rightarrow M = \frac{420}{11} \approx 38.18211​M=300−90=210⇒M=11420​≈38.18 minutes.
    • 112M=300+90=390⇒M=78011≈70.91\frac{11}{2}M = 300 + 90 = 390 \Rightarrow M = \frac{780}{11} \approx 70.91211​M=300+90=390⇒M=11780​≈70.91 minutes.
  Therefore, the acute angle occurs when the minute hand is between 38.18 minutes and 70.91 minutes after 10:00. Thus, the acute angle occurs between 10:38:11 and 10:70:55.

50. Hands Form a 45 Degree Angle

• Question: At what time between 2:00 and 3:00 do the hour and minute hands form a 45-degree angle?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 2:00 to 3:00, we want θ=45\theta = 45θ=45:
  ∣30(2)−112M∣=45\left| 30(2) – \frac{11}{2}M \right| = 45​30(2)−211​M​=45 ∣60−112M∣=45\left| 60 – \frac{11}{2}M \right| = 45​60−211​M​=45
  • Solving for MMM:
    • 112M=60−45=15⇒M=3011≈2.73\frac{11}{2}M = 60 – 45 = 15 \Rightarrow M = \frac{30}{11} \approx 2.73211​M=60−45=15⇒M=1130​≈2.73 minutes.
    • 112M=60+45=105⇒M=21011≈19.09\frac{11}{2}M = 60 + 45 = 105 \Rightarrow M = \frac{210}{11} \approx 19.09211​M=60+45=105⇒M=11210​≈19.09 minutes.
  Therefore, the hands are 45 degrees apart at approximately 2:02:44 and 2:19:05.

51. The Minute Hand Catches Up with the Hour Hand

• Question: How often does the minute hand catch up with the hour hand in a 12-hour period?
• Answer:
  • The minute hand catches up with the hour hand 11 times in a 12-hour period.
  • The minute hand overlaps the hour hand once approximately every 65 minutes.

52. Hands Form a 30 Degree Angle

• Question: At what time between 12:00 and 1:00 do the hour and minute hands form a 30-degree angle?
• Answer:
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 12:00 to 1:00, we want θ=30\theta = 30θ=30:
  ∣30(12)−112M∣=30\left| 30(12) – \frac{11}{2}M \right| = 30​30(12)−211​M​=30 ∣360−112M∣=30\left| 360 – \frac{11}{2}M \right| = 30​360−211​M​=30
  • Solving for MMM:
    • 112M=360−30=330⇒M=66011≈60\frac{11}{2}M = 360 – 30 = 330 \Rightarrow M = \frac{660}{11} \approx 60211​M=360−30=330⇒M=11660​≈60 minutes.
    • 112M=360+30=390⇒M=78011≈70.91\frac{11}{2}M = 360 + 30 = 390 \Rightarrow M = \frac{780}{11} \approx 70.91211​M=360+30=390⇒M=11780​≈70.91 minutes (which is greater than 60 minutes, so not valid).
  Therefore, the hands are 30 degrees apart at exactly 12:00 (this happens at the start of each hour).

53. Hands Form a 180 Degree Angle

• Question: Between 6:00 and 7:00, at what time will the hour and minute hands be 180 degrees apart?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 6:00 to 7:00, we want θ=180\theta = 180θ=180:
  ∣30(6)−112M∣=180\left| 30(6) – \frac{11}{2}M \right| = 180​30(6)−211​M​=180 ∣180−112M∣=180\left| 180 – \frac{11}{2}M \right| = 180​180−211​M​=180
  • Solving for MMM:
    • 112M=180−180=0⇒M=0\frac{11}{2}M = 180 – 180 = 0 \Rightarrow M = 0211​M=180−180=0⇒M=0 (which means at exactly 6:00).
    • 112M=180+180=360⇒M=72011≈65.45\frac{11}{2}M = 180 + 180 = 360 \Rightarrow M = \frac{720}{11} \approx 65.45211​M=180+180=360⇒M=11720​≈65.45 minutes.
  Therefore, the hands are exactly 180 degrees apart at 6:00 and approximately 6:65:27 (or 7:05:27).

54. Time Between 3:00 and 4:00 When Hands Are 15 Degrees Apart

• Question: At what time between 3:00 and 4:00 do the hour and minute hands form an angle of 15 degrees?
• Answer:
  • Using the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 3:00 to 4:00, we want θ=15\theta = 15θ=15:
  ∣30(3)−112M∣=15\left| 30(3) – \frac{11}{2}M \right| = 15​30(3)−211​M​=15 ∣90−112M∣=15\left| 90 – \frac{11}{2}M \right| = 15​90−211​M​=15
  • Solving for MMM:
    • 112M=90−15=75⇒M=15011≈13.64\frac{11}{2}M = 90 – 15 = 75 \Rightarrow M = \frac{150}{11} \approx 13.64211​M=90−15=75⇒M=11150​≈13.64 minutes.
    • 112M=90+15=105⇒M=21011≈19.09\frac{11}{2}M = 90 + 15 = 105 \Rightarrow M = \frac{210}{11} \approx 19.09211​M=90+15=105⇒M=11210​≈19.09 minutes.
  Therefore, the hands are 15 degrees apart at approximately 3:13:38 and 3:19:05.

55. Hands Form a 180 Degree Angle

• Question: Between 2:00 and 3:00, at what time will the hour and minute hands form an angle of 180 degrees?
• Answer:
  • Using the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 2:00 to 3:00, we want θ=180\theta = 180θ=180:
  ∣30(2)−112M∣=180\left| 30(2) – \frac{11}{2}M \right| = 180​30(2)−211​M​=180 ∣60−112M∣=180\left| 60 – \frac{11}{2}M \right| = 180​60−211​M​=180
  • Solving for MMM:
    • 112M=60−180=−120⇒M=−24011\frac{11}{2}M = 60 – 180 = -120 \Rightarrow M = \frac{-240}{11}211​M=60−180=−120⇒M=11−240​ (not valid).
    • 112M=60+180=240⇒M=48011≈43.64\frac{11}{2}M = 60 + 180 = 240 \Rightarrow M = \frac{480}{11} \approx 43.64211​M=60+180=240⇒M=11480​≈43.64 minutes.
  Therefore, the hands are 180 degrees apart at approximately 2:43:38.

56. Hands Form a 15 Degree Angle

• Question: At what time between 9:00 and 10:00 will the hour and minute hands form an angle of 15 degrees?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 9:00 to 10:00, we want θ=15\theta = 15θ=15:
  ∣30(9)−112M∣=15\left| 30(9) – \frac{11}{2}M \right| = 15​30(9)−211​M​=15 ∣270−112M∣=15\left| 270 – \frac{11}{2}M \right| = 15​270−211​M​=15
  • Solving for MMM:
    • 112M=270−15=255⇒M=51011≈46.36\frac{11}{2}M = 270 – 15 = 255 \Rightarrow M = \frac{510}{11} \approx 46.36211​M=270−15=255⇒M=11510​≈46.36 minutes.
    • 112M=270+15=285⇒M=57011≈51.82\frac{11}{2}M = 270 + 15 = 285 \Rightarrow M = \frac{570}{11} \approx 51.82211​M=270+15=285⇒M=11570​≈51.82 minutes.
  Therefore, the hands are 15 degrees apart at approximately 9:46:21 and 9:51:49.

57. Time Between 10:00 and 11:00 When Hands Are 120 Degrees Apart

• Question: At what time between 10:00 and 11:00 will the hour and minute hands form a 120-degree angle?
• Answer:
  • Using the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 10:00 to 11:00, we want θ=120\theta = 120θ=120:
  ∣30(10)−112M∣=120\left| 30(10) – \frac{11}{2}M \right| = 120​30(10)−211​M​=120 ∣300−112M∣=120\left| 300 – \frac{11}{2}M \right| = 120​300−211​M​=120
  • Solving for MMM:
    • 112M=300−120=180⇒M=36011≈32.73\frac{11}{2}M = 300 – 120 = 180 \Rightarrow M = \frac{360}{11} \approx 32.73211​M=300−120=180⇒M=11360​≈32.73 minutes.
    • 112M=300+120=420⇒M=84011≈76.36\frac{11}{2}M = 300 + 120 = 420 \Rightarrow M = \frac{840}{11} \approx 76.36211​M=300+120=420⇒M=11840​≈76.36 minutes (not valid).
  Therefore, the hands are 120 degrees apart at approximately 10:32:44.

58. The Hands Form an Acute Angle (Less Than 90 Degrees)

• Question: At what time between 11:00 and 12:00 will the hour and minute hands form an acute angle (less than 90 degrees)?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 11:00 to 12:00, we want θ<90\theta < 90θ<90.
  • Solving for the angle when θ\thetaθ is less than 90 degrees:
    • After solving the equation for the valid range, we get that the acute angle occurs between 11:00 and 11:20.
  Therefore, the hands are acute (less than 90 degrees apart) between 11:00 and 11:20.

59. Hands Are Exactly Overlapping

• Question: At what time between 1:00 and 2:00 will the hour and minute hands be exactly overlapping?
• Answer:
  • The hands are overlapping when the angle between the hour and minute hands is 0 degrees.
  • Use the angle formula:
  θ=∣30H−112M∣=0\theta = \left| 30H – \frac{11}{2}M \right| = 0θ=​30H−211​M​=0
  • For 1:00 to 2:00, we set the formula equal to 0:
  ∣30(1)−112M∣=0\left| 30(1) – \frac{11}{2}M \right| = 0​30(1)−211​M​=0 ∣30−112M∣=0\left| 30 – \frac{11}{2}M \right| = 0​30−211​M​=0
  • Solving for MMM:
    • 112M=30⇒M=6011≈5.45\frac{11}{2}M = 30 \Rightarrow M = \frac{60}{11} \approx 5.45211​M=30⇒M=1160​≈5.45 minutes.
  Therefore, the hands are exactly overlapping at approximately 1:05:27.

60. Hands Form a 45 Degree Angle

• Question: At what time between 4:00 and 5:00 will the hour and minute hands form a 45-degree angle?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 4:00 to 5:00, we want θ=45\theta = 45θ=45:
  ∣30(4)−112M∣=45\left| 30(4) – \frac{11}{2}M \right| = 45​30(4)−211​M​=45 ∣120−112M∣=45\left| 120 – \frac{11}{2}M \right| = 45​120−211​M​=45
  • Solving for MMM:
    • 112M=120−45=75⇒M=15011≈13.64\frac{11}{2}M = 120 – 45 = 75 \Rightarrow M = \frac{150}{11} \approx 13.64211​M=120−45=75⇒M=11150​≈13.64 minutes.
    • 112M=120+45=165⇒M=33011≈30\frac{11}{2}M = 120 + 45 = 165 \Rightarrow M = \frac{330}{11} \approx 30211​M=120+45=165⇒M=11330​≈30 minutes.
  Therefore, the hands are 45 degrees apart at approximately 4:13:38 and 4:30:00.

61. Time Between 5:00 and 6:00 When Hands Are 30 Degrees Apart

• Question: At what time between 5:00 and 6:00 will the hour and minute hands be 30 degrees apart?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 5:00 to 6:00, we want θ=30\theta = 30θ=30:
  ∣30(5)−112M∣=30\left| 30(5) – \frac{11}{2}M \right| = 30​30(5)−211​M​=30 ∣150−112M∣=30\left| 150 – \frac{11}{2}M \right| = 30​150−211​M​=30
  • Solving for MMM:
    • 112M=150−30=120⇒M=24011≈21.82\frac{11}{2}M = 150 – 30 = 120 \Rightarrow M = \frac{240}{11} \approx 21.82211​M=150−30=120⇒M=11240​≈21.82 minutes.
    • 112M=150+30=180⇒M=36011≈32.73\frac{11}{2}M = 150 + 30 = 180 \Rightarrow M = \frac{360}{11} \approx 32.73211​M=150+30=180⇒M=11360​≈32.73 minutes.
  Therefore, the hands are 30 degrees apart at approximately 5:21:49 and 5:32:44.

62. When are the Minute and Hour Hands 180 Degrees Apart?

• Question: Between 6:00 and 7:00, at what time will the minute and hour hands be exactly 180 degrees apart?
• Answer:
  • Use the angle formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 6:00 to 7:00, we want θ=180\theta = 180θ=180:
  ∣30(6)−112M∣=180\left| 30(6) – \frac{11}{2}M \right| = 180​30(6)−211​M​=180 ∣180−112M∣=180\left| 180 – \frac{11}{2}M \right| = 180​180−211​M​=180
  • Solving for MMM:
    • 112M=180−180=0⇒M=0\frac{11}{2}M = 180 – 180 = 0 \Rightarrow M = 0211​M=180−180=0⇒M=0 (not valid as it’s 6:00 exactly).
    • 112M=180+180=360⇒M=72011≈65.45\frac{11}{2}M = 180 + 180 = 360 \Rightarrow M = \frac{720}{11} \approx 65.45211​M=180+180=360⇒M=11720​≈65.45 minutes.
  Therefore, the hands are 180 degrees apart at approximately 6:05:27.

63. Hands Are 30 Minutes Ahead of the Hour Hand

• Question: Between 4:00 and 5:00, at what time will the minute hand be exactly 30 minutes ahead of the hour hand?
• Answer:
  • Since 30 minutes = 180 degrees, use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 4:00 to 5:00, we want θ=180\theta = 180θ=180:
  ∣30(4)−112M∣=180\left| 30(4) – \frac{11}{2}M \right| = 180​30(4)−211​M​=180 ∣120−112M∣=180\left| 120 – \frac{11}{2}M \right| = 180​120−211​M​=180
  • Solving for MMM:
    • 112M=120−180=−60⇒M=−12011\frac{11}{2}M = 120 – 180 = -60 \Rightarrow M = \frac{-120}{11}211​M=120−180=−60⇒M=11−120​ (not valid).
    • 112M=120+180=300⇒M=60011≈54.55\frac{11}{2}M = 120 + 180 = 300 \Rightarrow M = \frac{600}{11} \approx 54.55211​M=120+180=300⇒M=11600​≈54.55 minutes.
  Therefore, the minute hand is 30 minutes ahead of the hour hand at approximately 4:54:33.

64. Time Between 8:00 and 9:00 When Hands are 45 Degrees Apart

• Question: Between 8:00 and 9:00, at what time will the hour and minute hands be 45 degrees apart?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 8:00 to 9:00, we want θ=45\theta = 45θ=45:
  ∣30(8)−112M∣=45\left| 30(8) – \frac{11}{2}M \right| = 45​30(8)−211​M​=45 ∣240−112M∣=45\left| 240 – \frac{11}{2}M \right| = 45​240−211​M​=45
  • Solving for MMM:
    • 112M=240−45=195⇒M=39011≈35.45\frac{11}{2}M = 240 – 45 = 195 \Rightarrow M = \frac{390}{11} \approx 35.45211​M=240−45=195⇒M=11390​≈35.45 minutes.
    • 112M=240+45=285⇒M=57011≈51.82\frac{11}{2}M = 240 + 45 = 285 \Rightarrow M = \frac{570}{11} \approx 51.82211​M=240+45=285⇒M=11570​≈51.82 minutes.
  Therefore, the hands are 45 degrees apart at approximately 8:35:27 and 8:51:49.

65. Minute and Hour Hands Form an Acute Angle

• Question: At what time between 3:00 and 4:00 will the minute and hour hands form an acute angle (less than 90 degrees)?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 3:00 to 4:00, we want θ<90\theta < 90θ<90. Let’s check when the angle becomes acute:
    • The angle will be acute from 3:00 to approximately 3:15, and the minute hand will form a progressively smaller acute angle with the hour hand during that time.
  Therefore, the hands are in an acute angle between 3:00 and 3:15.

66. Hands Form a 60-Degree Angle

• Question: At what time between 9:00 and 10:00 will the minute and hour hands form an exact 60-degree angle?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 9:00 to 10:00, we want θ=60\theta = 60θ=60:
  ∣30(9)−112M∣=60\left| 30(9) – \frac{11}{2}M \right| = 60​30(9)−211​M​=60 ∣270−112M∣=60\left| 270 – \frac{11}{2}M \right| = 60​270−211​M​=60
  • Solving for MMM:
    • 112M=270−60=210⇒M=42011≈38.18\frac{11}{2}M = 270 – 60 = 210 \Rightarrow M = \frac{420}{11} \approx 38.18211​M=270−60=210⇒M=11420​≈38.18 minutes.
    • 112M=270+60=330⇒M=66011≈60\frac{11}{2}M = 270 + 60 = 330 \Rightarrow M = \frac{660}{11} \approx 60211​M=270+60=330⇒M=11660​≈60 minutes (not valid).
  Therefore, the hands are 60 degrees apart at approximately 9:38:11.

67. The Hands Form a Right Angle (90 Degrees)

• Question: At what time between 1:00 and 2:00 will the minute and hour hands form a 90-degree angle?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 1:00 to 2:00, we want θ=90\theta = 90θ=90:
  ∣30(1)−112M∣=90\left| 30(1) – \frac{11}{2}M \right| = 90​30(1)−211​M​=90 ∣30−112M∣=90\left| 30 – \frac{11}{2}M \right| = 90​30−211​M​=90
  • Solving for MMM:
    • 112M=30−90=−60⇒M=−12011\frac{11}{2}M = 30 – 90 = -60 \Rightarrow M = \frac{-120}{11}211​M=30−90=−60⇒M=11−120​ (not valid).
    • 112M=30+90=120⇒M=24011≈21.82\frac{11}{2}M = 30 + 90 = 120 \Rightarrow M = \frac{240}{11} \approx 21.82211​M=30+90=120⇒M=11240​≈21.82 minutes.
  Therefore, the hands are 90 degrees apart at approximately 1:21:49.

68. Time Between 6:00 and 7:00 When Hands are 150 Degrees Apart

• Question: At what time between 6:00 and 7:00 will the hour and minute hands be exactly 150 degrees apart?
• Answer:
  • Use the formula:
  θ=∣30H−112M∣\theta = \left| 30H – \frac{11}{2}M \right|θ=​30H−211​M​
  • For 6:00 to 7:00, we want θ=150\theta = 150θ=150:
  ∣30(6)−112M∣=150\left| 30(6) – \frac{11}{2}M \right| = 150​30(6)−211​M​=150 ∣180−112M∣=150\left| 180 – \frac{11}{2}M \right| = 150​180−211​M​=150
  • Solving for MMM:
    • 112M=180−150=30⇒M=6011≈5.45\frac{11}{2}M = 180 – 150 = 30 \Rightarrow M = \frac{60}{11} \approx 5.45211​M=180−150=30⇒M=1160​≈5.45 minutes.
    • 112M=180+150=330⇒M=66011≈60\frac{11}{2}M = 180 + 150 = 330 \Rightarrow M = \frac{660}{11} \approx 60211​M=180+150=330⇒M=11660​≈60 minutes.
  Therefore, the hands are 150 degrees apart at approximately 6:05:27.

Also read : Verbal reasoning question blood relations

Calendar Reasoning Questions

1. Day of the Week Calculation

• Question: What day of the week will it be on January 1, 2025?
• Answer: To calculate the day of the week for any given date, you can use a calendar formula or a day calculation method like Zeller’s congruence. Alternatively, you can use a known reference point and work your way forward.
• Answer Calculation: January 1, 2025, will fall on a Wednesday.

2. Leap Year Calculation

• Question: Is the year 2024 a leap year? How many days are in February in that year?
• Answer: A year is a leap year if:
  • It is divisible by 4, and
  • It is not divisible by 100, unless also divisible by 400.
  • 2024 is divisible by 4, and not by 100, so it is a leap year.
  • February will have 29 days in 2024.

3. Number of Days in a Given Period

• Question: How many days are there between March 1, 2023, and March 1, 2024?
• Answer: From March 1, 2023, to March 1, 2024, there is a full year, and since 2024 is a leap year, it will have 366 days.

4. Finding the Day of the Week for a Given Date

• Question: What day of the week was August 15, 1947?
• Answer: Using a formula like Zeller’s congruence or applying a reference date, you can determine that August 15, 1947, was a Friday.

5. Months with 31 Days

• Question: How many months in a year have 31 days?
• Answer: Seven months in a year have 31 days: January, March, May, July, August, October, and December.

6. How Many Odd Days Are There in 100 Years?

• Question: How many odd days are there in 100 years?
• Answer:
  • In 100 years, there are 36525 days (since every 4th year is a leap year, except for years divisible by 100 but not 400).
  • To find the number of odd days, divide the total number of days by 7 (since a week has 7 days) and take the remainder:
  365257=5217 weeks and 6 odd days\frac{36525}{7} = 5217 \text{ weeks and } 6 \text{ odd days}736525​=5217 weeks and 6 odd daysTherefore, in 100 years, there are 6 odd days.

7. Find the Day of the Week on 1st January 2000

• Question: What day of the week was 1st January 2000?
• Answer:
  • The reference date is 1st January 1900, which was a Monday.
  • To calculate the day of the week for 1st January 2000, count the total number of odd days in 100 years (from 1900 to 2000).
  • 100 years contain 6 odd days (from the previous problem).
  • Starting from Monday (1900), after 6 odd days, the day on 1st January 2000 will be Saturday.

8. Number of Leap Years Between 1900 and 2020

• Question: How many leap years are there between 1900 and 2020 (inclusive)?
• Answer:
  • A leap year occurs if the year is divisible by 4, except for years divisible by 100 but not by 400.
  • The leap years from 1900 to 2020 are: 1904, 1908, 1912, 1916, 1920, 1924, 1928, 1932, 1936, 1940, 1944, 1948, 1952, 1956, 1960, 1964, 1968, 1972, 1976, 1980, 1984, 1988, 1992, 1996, 2000, 2004, 2008, 2012, 2016, and 2020.
  • Therefore, there are 30 leap years between 1900 and 2020.

9. What is the Day of the Week on 15th August 1947?

• Question: What was the day of the week on 15th August 1947?
• Answer:
  • Start by finding the number of odd days from 1st January 1900 to 15th August 1947.
  • For the years from 1900 to 1947:
    • First, calculate the odd days for the complete years (1900 to 1946).
    • 1900 to 1946: There are 12 leap years between 1900 and 1946.
    • Then, for 1947, count the odd days for each month:
      • January 1st to August 15th.
  • After applying the calculation, we determine that 15th August 1947 was a Friday.

10. How Many Days Are There in 2000?

• Question: How many days are there in the year 2000?
• Answer:
  • The year 2000 was a leap year (divisible by 4 and divisible by 400).
  • Since a leap year has 366 days, there are 366 days in the year 2000.

11. What Day of the Week Was 31st December 1999?

• Question: What day of the week was 31st December 1999?
• Answer:
  • Start by finding the number of odd days from 1st January 1900 to 31st December 1999.
  • From 1900 to 1999, we have 24 leap years and 75 regular years.
  • Counting the odd days gives us a remainder of 5 odd days.
  • Since 1st January 1900 was a Monday, and after 5 odd days, the day on 31st December 1999 was a Friday.

12. Find the Day of the Week on 1st January 1900

• Question: What day of the week was 1st January 1900?
• Answer:
  • To determine the day of the week on 1st January 1900, start by noting that 1st January 1900 is a Monday.
  • Using known facts or a reference calendar, we can verify that 1st January 1900 was indeed a Monday.

13. How Many Sundays Are There in a Leap Year?

• Question: How many Sundays are there in a leap year?
• Answer:
  • A leap year has 366 days, which is exactly 52 weeks and 2 extra days.
  • The 2 extra days could be any two consecutive days.
  • If the year starts on a Sunday, there will be 53 Sundays (because the extra days will include a Sunday).
  • If the year starts on a Monday, the 2 extra days will be Monday and Sunday, and there will be 53 Sundays.
  • In all other cases, there will be 52 Sundays.
  • Hence, a leap year generally has 52 Sundays, but it could have 53 Sundays if it starts on Sunday or Monday.

14. How Many Days Are There in the Month of February in a Leap Year?

• Question: How many days are there in the month of February in a leap year?
• Answer:
  • In a leap year, February has 29 days instead of the usual 28 days in a common year.

15. When Will the 1st of January Fall on a Sunday Again?

• Question: After the year 2000, in which year will 1st January fall on a Sunday again?
• Answer:
  • The pattern of days shifts by one day for each regular year and two days for each leap year.
  • To find the next occurrence of 1st January on a Sunday, check the years following 2000:
    • 2000 was a leap year, so 1st January 2001 was a Monday.
    • 2001 was a regular year, so 1st January 2002 was a Tuesday.
    • This pattern continues until 1st January 2006, which falls on a Sunday.
  Therefore, 1st January will fall on a Sunday again in 2006.

16. How Many Days Are There Between 1st January 2000 and 1st January 2020?

• Question: How many days are there between 1st January 2000 and 1st January 2020?
• Answer:
  • From 1st January 2000 to 1st January 2020, there are 20 years.
  • The leap years between 2000 and 2020 are: 2000, 2004, 2008, 2012, 2016, and 2020 (total of 6 leap years).
  • The number of regular years is 20 – 6 = 14.
  • The total number of days is: (6 leap years×366 days)+(14 regular years×365 days)=2196+5110=7306 days.(6 \text{ leap years} \times 366 \text{ days}) + (14 \text{ regular years} \times 365 \text{ days}) = 2196 + 5110 = 7306 \text{ days}.(6 leap years×366 days)+(14 regular years×365 days)=2196+5110=7306 days.
  Therefore, there are 7306 days between 1st January 2000 and 1st January 2020.

17. What Day of the Week Was 1st January 1900?

• Question: What day of the week was 1st January 1900?
• Answer:
  • From known historical facts, 1st January 1900 was a Monday.
  This can also be verified using a formula or by referring to an established calendar.

18. How Many Days Are There in 400 Years?

• Question: How many days are there in 400 years?
• Answer:
  • A leap year occurs every 4 years, except for years divisible by 100 but not 400.
  • Over a period of 400 years, there are 100 leap years (years divisible by 4).
  • However, the years divisible by 100 but not 400 (like 1700, 1800, and 1900) are not leap years.
  • This reduces the leap years by 3, leaving 97 leap years.
  • Therefore, in 400 years: (97 leap years×366 days)+(303 regular years×365 days)=35442+110595=146037 days.(97 \text{ leap years} \times 366 \text{ days}) + (303 \text{ regular years} \times 365 \text{ days}) = 35442 + 110595 = 146037 \text{ days}.(97 leap years×366 days)+(303 regular years×365 days)=35442+110595=146037 days.
  So, there are 146,037 days in 400 years.

19. How Many Days Are There in a Century (100 Years)?

• Question: How many days are there in 100 years?
• Answer:
  • In 100 years, there are typically 24 leap years (since every 4th year is a leap year, but subtract the century years that are not divisible by 400).
  • Hence, the number of regular years is 100 – 24 = 76.
  • The total number of days is: (24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.(24 \text{ leap years} \times 366 \text{ days}) + (76 \text{ regular years} \times 365 \text{ days}) = 8784 + 27740 = 36524 \text{ days}.(24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.
  Therefore, there are 36,524 days in 100 years.

20. How Many Days Are There Between 15th August 1947 and 15th August 2022?

• Question: How many days are there between 15th August 1947 and 15th August 2022?
• Answer:
  • From 15th August 1947 to 15th August 2022, there are 75 years.
  • The leap years between 1947 and 2022 are: 1952, 1956, 1960, 1964, 1968, 1972, 1976, 1980, 1984, 1988, 1992, 1996, 2000, 2004, 2008, 2012, 2016, and 2020 (total of 18 leap years).
  • The number of regular years is 75 – 18 = 57.
  • The total number of days is: (18 leap years×366 days)+(57 regular years×365 days)=6588+20705=27293 days.(18 \text{ leap years} \times 366 \text{ days}) + (57 \text{ regular years} \times 365 \text{ days}) = 6588 + 20705 = 27293 \text{ days}.(18 leap years×366 days)+(57 regular years×365 days)=6588+20705=27293 days.
  Therefore, there are 27,293 days between 15th August 1947 and 15th August 2022.

21. What Day of the Week Was 15th August 1947?

• Question: What day of the week was 15th August 1947?
• Answer:
  • From known historical references or through day calculations, 15th August 1947 was a Friday.

22. How Many Days Are There in a Non-Leap Year?

• Question: How many days are there in a non-leap year?
• Answer:
  • A non-leap year has 365 days. This is because there are 12 months, and the total number of days in these months equals 365.

23. When Was the Last Leap Year?

• Question: When was the last leap year?
• Answer:
  • The most recent leap year, as of 2025, was 2020. Leap years occur every 4 years, and 2024 will be the next leap year.

24. How Many Mondays Are There in a Leap Year?

• Question: How many Mondays are there in a leap year?
• Answer:
  • A leap year has 366 days.
  • There are 52 full weeks in a leap year (52 weeks x 7 days = 364 days), plus 2 extra days.
  • Depending on which day of the week the leap year starts, the extra days can be any two consecutive days.
  • If a leap year starts on Monday, then the 2 extra days are Monday and Tuesday, resulting in 53 Mondays.
  • If it starts on any other day, there will be 52 Mondays.

25. How Many Days Are There in February in a Century (100 Years)?

• Question: How many days are there in February in a century (100 years)?
• Answer:
  • In a century, there are usually 24 leap years (since every 4th year is a leap year, except for century years that are not divisible by 400).
  • Hence, there are 24 leap years in a century.
  • In each leap year, February has 29 days, while in regular years, February has 28 days.
  • The total number of days in February over 100 years is: (24 leap years×29 days)+(76 regular years×28 days)=696+2128=2824 days.(24 \text{ leap years} \times 29 \text{ days}) + (76 \text{ regular years} \times 28 \text{ days}) = 696 + 2128 = 2824 \text{ days}.(24 leap years×29 days)+(76 regular years×28 days)=696+2128=2824 days.
  Therefore, there are 2824 days in February in 100 years.

26. How Many Days Are There Between 1st January 1900 and 31st December 1999?

• Question: How many days are there between 1st January 1900 and 31st December 1999?
• Answer:
  • From 1st January 1900 to 31st December 1999, there are 100 years.
  • There are 24 leap years in these 100 years.
  • The number of regular years is 100 – 24 = 76.
  • The total number of days is: (24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.(24 \text{ leap years} \times 366 \text{ days}) + (76 \text{ regular years} \times 365 \text{ days}) = 8784 + 27740 = 36524 \text{ days}.(24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.
  Therefore, there are 36,524 days between 1st January 1900 and 31st December 1999.

27. How Many Leap Years Are There Between 2000 and 2100?

• Question: How many leap years are there between 2000 and 2100 (inclusive)?
• Answer:
  • Leap years are divisible by 4, except for century years that are not divisible by 400.
  • Between 2000 and 2100, the leap years are 2000 and 2004, 2008, 2012, 2016, 2020, and so on, up to 2096.
  • 2100 is not a leap year because it is divisible by 100 but not by 400.
  • Therefore, there are 25 leap years between 2000 and 2100.

28. What Day of the Week Was 1st January 1990?

• Question: What day of the week was 1st January 1990?
• Answer:
  • Start from the known fact that 1st January 1900 was a Monday.
  • Count the total odd days from 1st January 1900 to 1st January 1990.
  • There are 22 leap years between 1900 and 1990.
  • The total number of odd days is 4 odd days after 90 years.
  • Starting from Monday and moving forward 4 odd days, 1st January 1990 was a Monday.

29. How Many Mondays Are There in a Regular Year?

• Question: How many Mondays are there in a regular year?
• Answer:
  • A regular year has 365 days, which is 52 full weeks and 1 extra day.
  • The extra day can be any day of the week.
  • Therefore, a regular year will have 52 Mondays plus the extra day.
  • If the extra day is a Monday, there will be 53 Mondays; otherwise, there will be 52 Mondays.
  • Generally, there are 52 Mondays in a regular year, and 53 Mondays if the year starts on a Monday.

30. What is the Day of the Week on 31st December 2024?

• Question: What day of the week will it be on 31st December 2024?
• Answer:
  • Start from the known reference that 1st January 1900 was a Monday.
  • Count the number of odd days from 1st January 1900 to 31st December 2024.
  • There are 31 leap years between 1900 and 2024.
  • The number of odd days is 5 odd days.
  • Starting from 1st January 1900 (Monday) and counting forward 5 odd days, 31st December 2024 will fall on a Tuesday.

31. How Many Days Are There in a Year If 1st January is a Sunday?

• Question: How many days are there in a year if 1st January is a Sunday?
• Answer:
  • If 1st January is a Sunday, the year is either a regular year or a leap year, depending on the century and the number of leap years.
  • A regular year starting on Sunday will have 365 days, as the 1st January being a Sunday means there are 52 Sundays, and the extra day falls on Sunday.

32. How Many Odd Days Are There in 250 Years?

• Question: How many odd days are there in 250 years?
• Answer:
  • In 250 years, there are 62 leap years (since every 4th year is a leap year, except century years that are not divisible by 400).
  • The total number of days in 250 years is: (62 leap years×366 days)+(188 regular years×365 days)=22692+68620=91312 days.(62 \text{ leap years} \times 366 \text{ days}) + (188 \text{ regular years} \times 365 \text{ days}) = 22692 + 68620 = 91312 \text{ days}.(62 leap years×366 days)+(188 regular years×365 days)=22692+68620=91312 days.
  • To find the number of odd days, divide the total days by 7 and take the remainder: 913127=13044 weeks and 4 odd days.\frac{91312}{7} = 13044 \text{ weeks and } 4 \text{ odd days}.791312​=13044 weeks and 4 odd days.
  Therefore, there are 4 odd days in 250 years.

33. What Day of the Week Was 15th August 1947?

• Question: What day of the week was 15th August 1947?
• Answer:
  • By using the known reference that 15th August 1947 was the date of India’s independence, the day of the week can be verified.
  • 15th August 1947 was a Friday.

34. How Many Days Are There in the Month of March in a Leap Year?

• Question: How many days are there in the month of March in a leap year?
• Answer:
  • The month of March always has 31 days, whether it is a leap year or not.

35. How Many Days Are There in a Year If 1st January is a Friday?

• Question: How many days are there in a year if 1st January is a Friday?
• Answer:
  • If 1st January is a Friday, it can be a regular year or a leap year depending on the year.
  • A regular year starting on a Friday will have 365 days.
  • A leap year starting on a Friday will have 366 days.

36. In Which Year Will the Next Leap Year Be?

• Question: After 2024, which year will be the next leap year?
• Answer:
  • A leap year occurs every 4 years, except century years that are not divisible by 400.
  • Since 2024 is a leap year, the next leap year will be 2028.

37. How Many Leap Years Are There Between 1000 and 2000?

• Question: How many leap years are there between 1000 and 2000 (inclusive)?
• Answer:
  • A leap year occurs every 4 years, except century years that are not divisible by 400.
  • Between 1000 and 2000, the leap years are: 1004, 1008, 1012, …, 1996.
  • Century years like 1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, and 1900 are not leap years, but 1600 is a leap year because it is divisible by 400.
  • Therefore, there are 242 leap years between 1000 and 2000.

38. What Day of the Week Was 1st January 2000?

• Question: What day of the week was 1st January 2000?
• Answer:
  • From known historical references or day calculation, 1st January 2000 was a Saturday.

39. How Many Days Are There Between 1st January 2010 and 31st December 2019?

• Question: How many days are there between 1st January 2010 and 31st December 2019?
• Answer:
  • From 1st January 2010 to 31st December 2019, there are 10 years.
  • The leap years in this period are: 2012, 2016 (total of 2 leap years).
  • The number of regular years is 10 – 2 = 8.
  • The total number of days is: (2 leap years×366 days)+(8 regular years×365 days)=732+2920=3652 days.(2 \text{ leap years} \times 366 \text{ days}) + (8 \text{ regular years} \times 365 \text{ days}) = 732 + 2920 = 3652 \text{ days}.(2 leap years×366 days)+(8 regular years×365 days)=732+2920=3652 days.
  Therefore, there are 3652 days between 1st January 2010 and 31st December 2019.

40. How Many Leap Years Are There Between 2000 and 2100?

• Question: How many leap years are there between 2000 and 2100?
• Answer:
  • Leap years are divisible by 4, except century years that are not divisible by 400.
  • Between 2000 and 2100, the leap years are 2000, 2004, 2008, 2012, 2016, 2020, and so on.
  • The year 2100 is not a leap year since it is divisible by 100 but not by 400.
  • Therefore, there are 25 leap years between 2000 and 2100.

41. How Many Days Are There in a Leap Year?

• Question: How many days are there in a leap year?
• Answer:
  • A leap year has 366 days. This is because February has 29 days in a leap year instead of the usual 28 days.

42. How Many Saturdays Are There in a Leap Year?

• Question: How many Saturdays are there in a leap year?
• Answer:
  • A leap year has 366 days (52 weeks + 2 extra days).
  • In a leap year, there are 52 Saturdays. If the leap year starts on a Saturday, there will be 53 Saturdays because the extra days will be Saturday and Sunday.
  • Generally, there will be 52 Saturdays in a leap year.

43. How Many Days Are There in the Month of February in a Leap Year?

• Question: How many days are there in the month of February in a leap year?
• Answer:
  • In a leap year, February has 29 days, which is one extra day compared to regular years, where February has 28 days.

44. What Was the Day of the Week on 15th August 1947?

• Question: What day of the week was 15th August 1947?
• Answer:
  • 15th August 1947, the day India gained independence, was a Friday.

45. How Many Odd Days Are There in 100 Years?

• Question: How many odd days are there in 100 years?
• Answer:
  • In 100 years, there are 24 leap years and 76 regular years.
  • The total number of days in 100 years is: (24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.(24 \text{ leap years} \times 366 \text{ days}) + (76 \text{ regular years} \times 365 \text{ days}) = 8784 + 27740 = 36524 \text{ days}.(24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.
  • Divide by 7 to find the remainder (odd days): 365247=5217 weeks and 5 odd days.\frac{36524}{7} = 5217 \text{ weeks} \text{ and } 5 \text{ odd days}.736524​=5217 weeks and 5 odd days.
  Therefore, there are 5 odd days in 100 years.

46. How Many Days Are There in 10 Years, Including Leap Years?

• Question: How many days are there in 10 years, including leap years?
• Answer:
  • There are typically 2 leap years in every 10 years.
  • Therefore, the number of days in 10 years is: (2 leap years×366 days)+(8 regular years×365 days)=732+2920=3652 days.(2 \text{ leap years} \times 366 \text{ days}) + (8 \text{ regular years} \times 365 \text{ days}) = 732 + 2920 = 3652 \text{ days}.(2 leap years×366 days)+(8 regular years×365 days)=732+2920=3652 days.
  Hence, there are 3652 days in 10 years including leap years.

47. What Day of the Week Will 1st January 2050 Be?

• Question: What day of the week will 1st January 2050 be?
• Answer:
  • From known references or by calculating odd days, 1st January 2050 will fall on a Saturday.

48. How Many Days Are There in the Month of February in a Regular Year?

• Question: How many days are there in the month of February in a regular year?
• Answer:
  • In a regular year, February has 28 days. In a leap year, February has 29 days.

49. How Many Days Are There in a Century (100 Years)?

• Question: How many days are there in a century (100 years)?
• Answer:
  • A century typically has 24 leap years and 76 regular years.
  • The total number of days in a century is: (24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.(24 \text{ leap years} \times 366 \text{ days}) + (76 \text{ regular years} \times 365 \text{ days}) = 8784 + 27740 = 36524 \text{ days}.(24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.
  Therefore, there are 36,524 days in 100 years.

50. How Many Days Are There in the Month of April in a Leap Year?

• Question: How many days are there in the month of April in a leap year?
• Answer:
  • The month of April always has 30 days, regardless of whether it is a leap year or not.

51. What Day of the Week Was 1st January 1800?

• Question: What day of the week was 1st January 1800?
• Answer:
  • Using date calculation or references, 1st January 1800 was a Friday.

52. How Many Days Are There Between 1st January 2020 and 31st December 2020?

• Question: How many days are there between 1st January 2020 and 31st December 2020?
• Answer:
  • Since 2020 is a leap year, it has 366 days.

53. How Many Days Are There Between 1st January 2000 and 31st December 2024?

• Question: How many days are there between 1st January 2000 and 31st December 2024?
• Answer:
  • From 1st January 2000 to 31st December 2024, there are 25 years.
  • Leap years between 2000 and 2024 are: 2000, 2004, 2008, 2012, 2016, 2020, 2024 (7 leap years).
  • The total number of days is: (7 leap years×366 days)+(18 regular years×365 days)=2562+6570=9132 days.(7 \text{ leap years} \times 366 \text{ days}) + (18 \text{ regular years} \times 365 \text{ days}) = 2562 + 6570 = 9132 \text{ days}.(7 leap years×366 days)+(18 regular years×365 days)=2562+6570=9132 days.
  Therefore, there are 9132 days between 1st January 2000 and 31st December 2024.

54. How Many Days Are There in the Month of October?

• Question: How many days are there in the month of October?
• Answer:
  • The month of October always has 31 days, regardless of whether it’s a leap year or not.

55. How Many Leap Years Are There Between 1st January 1000 and 31st December 2000?

• Question: How many leap years are there between 1st January 1000 and 31st December 2000?
• Answer:
  • Leap years are those divisible by 4, except for century years that are not divisible by 400.
  • Between 1000 and 2000, the leap years are: 1004, 1008, 1012, …, 1996 (with some exclusions like 1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, and 1900).
  • The total number of leap years is 242.

56. How Many Odd Days Are There in 200 Years?

• Question: How many odd days are there in 200 years?
• Answer:
  • In 200 years, there are 48 leap years and 152 regular years.
  • The total number of days in 200 years is: (48 leap years×366 days)+(152 regular years×365 days)=17568+55480=73048 days.(48 \text{ leap years} \times 366 \text{ days}) + (152 \text{ regular years} \times 365 \text{ days}) = 17568 + 55480 = 73048 \text{ days}.(48 leap years×366 days)+(152 regular years×365 days)=17568+55480=73048 days.
  • Divide by 7 to find the remainder (odd days): 730487=10435 weeks and 3 odd days.\frac{73048}{7} = 10435 \text{ weeks} \text{ and } 3 \text{ odd days}.773048​=10435 weeks and 3 odd days.
  Therefore, there are 3 odd days in 200 years.

57. How Many Days Are There Between 1st January 1500 and 31st December 2000?

• Question: How many days are there between 1st January 1500 and 31st December 2000?
• Answer:
  • From 1st January 1500 to 31st December 2000, there are 501 years.
  • Leap years between 1500 and 2000: calculate by checking divisibility by 4 and excluding century years that are not divisible by 400.
  • The total number of leap years is 120.
  • The total number of regular years is 501 – 120 = 381.
  • The total number of days is: (120 leap years×366 days)+(381 regular years×365 days)=43920+139965=183885 days.(120 \text{ leap years} \times 366 \text{ days}) + (381 \text{ regular years} \times 365 \text{ days}) = 43920 + 139965 = 183885 \text{ days}.(120 leap years×366 days)+(381 regular years×365 days)=43920+139965=183885 days.
  Therefore, there are 183,885 days between 1st January 1500 and 31st December 2000.

58. What Day of the Week Was 31st December 1999?

• Question: What day of the week was 31st December 1999?
• Answer:
  • 31st December 1999 was a Friday.

59. How Many Leap Years Are There Between 1900 and 2000?

• Question: How many leap years are there between 1900 and 2000?
• Answer:
  • Leap years are divisible by 4, except century years that are not divisible by 400.
  • Between 1900 and 2000, 1900 is not a leap year, but 1996 is a leap year.
  • Therefore, there is only 1 leap year between 1900 and 2000 (which is 1996).

60. How Many Sundays Are There in the Year 2024?

• Question: How many Sundays are there in the year 2024?
• Answer:
  • Since 2024 is a leap year, it has 366 days.
  • A leap year typically has 52 Sundays, plus one extra Sunday if the year starts on a Sunday.
  • 2024 begins on a Monday, so there are 52 Sundays in 2024.

61. How Many Days Are There in the Month of September?

• Question: How many days are there in the month of September?
• Answer:
  • The month of September always has 30 days.

62. How Many Leap Years Are There Between 2001 and 2100?

• Question: How many leap years are there between 2001 and 2100?
• Answer:
  • Leap years are divisible by 4, except century years that are not divisible by 400.
  • Between 2001 and 2100, the leap years are: 2004, 2008, 2012, 2016, 2020, 2024, 2028, 2032, 2036, …, 2096.
  • 2100 is not a leap year since it is divisible by 100 but not by 400.
  • Therefore, there are 25 leap years between 2001 and 2100.

63. What Was the Day of the Week on 1st January 1900?

• Question: What was the day of the week on 1st January 1900?
• Answer:
  • 1st January 1900 was a Monday.

64. How Many Days Are There Between 1st January 1800 and 31st December 1900?

• Question: How many days are there between 1st January 1800 and 31st December 1900?
• Answer:
  • From 1st January 1800 to 31st December 1900, there are 100 years.
  • Leap years between 1800 and 1900: 1804, 1808, 1812, …, 1896 (there are 24 leap years).
  • Regular years are 100 – 24 = 76.
  • The total number of days is: (24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.(24 \text{ leap years} \times 366 \text{ days}) + (76 \text{ regular years} \times 365 \text{ days}) = 8784 + 27740 = 36524 \text{ days}.(24 leap years×366 days)+(76 regular years×365 days)=8784+27740=36524 days.
  Therefore, there are 36,524 days between 1st January 1800 and 31st December 1900.

65. How Many Odd Days Are There in 400 Years?

• Question: How many odd days are there in 400 years?
• Answer:
  • In 400 years, there are 97 leap years and 303 regular years.
  • The total number of days in 400 years is: (97 leap years×366 days)+(303 regular years×365 days)=35442+110595=146037 days.(97 \text{ leap years} \times 366 \text{ days}) + (303 \text{ regular years} \times 365 \text{ days}) = 35442 + 110595 = 146037 \text{ days}.(97 leap years×366 days)+(303 regular years×365 days)=35442+110595=146037 days.
  • Dividing by 7 to find the remainder (odd days): 1460377=20862 weeks and 3 odd days.\frac{146037}{7} = 20862 \text{ weeks} \text{ and } 3 \text{ odd days}.7146037​=20862 weeks and 3 odd days.
  Therefore, there are 3 odd days in 400 years.

66. What Day of the Week Was 15th August 1947?

• Question: What day of the week was 15th August 1947?
• Answer:
  • 15th August 1947 was a Friday, the day India gained independence.

67. How Many Days Are There Between 1st January 1995 and 31st December 2025?

• Question: How many days are there between 1st January 1995 and 31st December 2025?
• Answer:
  • From 1st January 1995 to 31st December 2025, there are 31 years.
  • Leap years between 1995 and 2025: 1996, 2000, 2004, 2008, 2012, 2016, 2020, 2024 (8 leap years).
  • The total number of days is: (8 leap years×366 days)+(23 regular years×365 days)=2928+8395=11323 days.(8 \text{ leap years} \times 366 \text{ days}) + (23 \text{ regular years} \times 365 \text{ days}) = 2928 + 8395 = 11323 \text{ days}.(8 leap years×366 days)+(23 regular years×365 days)=2928+8395=11323 days.
  Therefore, there are 11,323 days between 1st January 1995 and 31st December 2025.

68. How Many Days Are There in the Month of February in a Century Year (e.g., 1900)?

• Question: How many days are there in the month of February in a century year (e.g., 1900)?
• Answer:
  • February in a century year (like 1900) is 28 days, since century years divisible by 100 are not leap years unless divisible by 400.
  • Therefore, February 1900 had 28 days.

69. How Many Leap Years Are There Between 2001 and 2100?

• Question: How many leap years are there between 2001 and 2100?
• Answer:
  • Leap years are divisible by 4, except for century years not divisible by 400.
  • Leap years between 2001 and 2100 are: 2004, 2008, 2012, 2016, 2020, 2024, 2028, 2032, 2036, …, 2096.
  • 2100 is not a leap year because it is divisible by 100 but not by 400.
  • Therefore, there are 25 leap years between 2001 and 2100.

70. What Day of the Week Was 1st January 2020?

• Question: What day of the week was 1st January 2020?
• Answer:
  • 1st January 2020 was a Wednesday.

71. How Many Days Are There in the Month of February in a Leap Year?

• Question: How many days are there in the month of February in a leap year?
• Answer:
  • In a leap year, February has 29 days.

72. What Was the Day of the Week on 14th July 1789 (The French Revolution)?

• Question: What was the day of the week on 14th July 1789 (the day of the French Revolution)?
• Answer:
  • 14th July 1789 was a Tuesday.

73. How Many Days Are There in the Month of May?

• Question: How many days are there in the month of May?
• Answer:
  • The month of May always has 31 days, regardless of whether it is a leap year or not.

74. How Many Sundays Are There in the Year 2024?

• Question: How many Sundays are there in the year 2024?
• Answer:
  • Since 2024 is a leap year, it has 366 days.
  • A leap year typically has 52 Sundays, plus one extra Sunday if the year starts on a Sunday.
  • 2024 starts on a Monday, so there are 52 Sundays in 2024.

75. What Day of the Week Was 1st January 1970?

• Question: What day of the week was 1st January 1970?
• Answer:
  • 1st January 1970 was a Thursday.

76. How Many Days Are There in the Month of December?

• Question: How many days are there in the month of December?
• Answer:
  • The month of December always has 31 days.

77. How Many Leap Years Are There Between 2000 and 2100?

• Question: How many leap years are there between 2000 and 2100?
• Answer:
  • Leap years are divisible by 4, except century years not divisible by 400.
  • The leap years between 2000 and 2100 are: 2000, 2004, 2008, 2012, 2016, 2020, …, 2096.
  • 2100 is not a leap year.
  • Therefore, there are 25 leap years between 2000 and 2100.

78. How Many Days Are There Between 1st January 2000 and 31st December 2099?

• Question: How many days are there between 1st January 2000 and 31st December 2099?
• Answer:
  • From 1st January 2000 to 31st December 2099, there are 100 years.
  • The leap years between 2000 and 2099 are: 2000, 2004, 2008, 2012, …, 2096 (25 leap years).
  • The total number of days is: (25 leap years×366 days)+(75 regular years×365 days)=9150+27375=36525 days.(25 \text{ leap years} \times 366 \text{ days}) + (75 \text{ regular years} \times 365 \text{ days}) = 9150 + 27375 = 36525 \text{ days}.(25 leap years×366 days)+(75 regular years×365 days)=9150+27375=36525 days.
  Therefore, there are 36,525 days between 1st January 2000 and 31st December 2099.

79. How Many Leap Years Are There Between 1901 and 2000?

• Question: How many leap years are there between 1901 and 2000?
• Answer:
  • Leap years are divisible by 4, except century years that are not divisible by 400.
  • The leap years between 1901 and 2000 are: 1904, 1908, 1912, …, 1996 (24 leap years).
  • Therefore, there are 24 leap years between 1901 and 2000.

80. How Many Days Are There Between 1st January 2021 and 31st December 2021?

• Question: How many days are there between 1st January 2021 and 31st December 2021?
• Answer:
  • Since 2021 is a common year, it has 365 days.

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