Example 1: What is the smaller angle between the two hands of the clock at 5:00 O’clock?
Sol : At 5 O’clock, the hour hand will be at 5 and the minute hand will be at 12. Now, they have a gap of 5 hour spaces and each hour space is 30 degrees. That means the angle between the two hands will be 30 × 5 = 150 degrees.
Example 2: What is the smaller angle between the two hands of the clock at 2:20 pm?
Sol: At 2 O’clock the hour hand will be at 2 and the minute hand will be at 12. Now, they have a gap of 2 hour spaces and each hour space is 30 degrees. That means at 2:00 clock the angle will be = 30 × 2 = 60 degrees. After that in 20 min, the relative coverage of the minute hand will be 20 × 11/2 = 110 degrees. Now take the difference between the two angles, which will become our answer i.e. 110 - 60 = 50 degrees.
Besides that, there is another method. As discussed, in 12 hrs, hour hand travels 360 degrees. Thus, in 2 hrs 20 min i.e. 2 1/3 hrs it will travel (360/12) × (7/3) = 70 degrees.
Angle traced by min hand in 60 min = 360 degrees. Angle traced by it in 20 min = 20 × 360/60 = 120 degrees.
Required angle is the difference between the two = 120 – 70 = 50 degrees.
Example 3: At what time between 3 and 4 are the minute and hour hand together?
Sol: At 3 O’clock, the relative distance between the hour and the minute hand is 15 minutes. To catch up with the hour hand, the minute hand has to cover a relative distance of 15 minutes at the relative speed of 11/12 minutes per minute.
Thus, time required = 15/(11/12) = 15 × 12/11 = 180/11 = 16 4/11 min.
Example 4: The minute hand of a clock overtakes the hour hand at intervals of 63 minutes of correct time. How much does the clock lose or gain in a day?
Sol: In a correct clock, the minute hand and hour hand should meet after every 65 5/11 min. But we know that they are meeting after every 63 minutes. So gain in 63 minutes is 27/11 minutes.
Gain in 24 hours =(24 ×60/63) × (27/11) = 56 8/77 min.
Example 5: At what time between 4 and 5 are the minute and hour hand at right angles?
Sol: At 4 O’clock, the relative distance between the hour and the minute hand is 20 minutes. To make a 90–degree angle with the hour hand, the minute hand has to cover a relative distance of 5 minutes at the relative speed of (11/12) minutes per minute. Thus, time required = 5/(11/12) = 60/11 or 5 (5/11) minutes. As explained above in important points, there is still one more case. When a relative distance of 35 minutes has been covered, even then the angles would be at right angles. Time required = 35/(11/12) = 420/11 or 38(2/11) minutes
You can also think that the difference between the two right angles themselves be equal to 30 min. In one of the right angles, the minute hand will be 15 min before the hour hand and in the other; it will be 15 min ahead of the hour hand.
Example 6: A watch gains uniformly. It was observed that it was 6 min slow at 12 o’clock in the night on Monday. On Friday at 6 pm it was 4 min 48 second fast. When was it correct?
Sol:The time between 12 O’clock on Monday night and 6 pm on Friday = 90 hours. Now, the watch gains 6 + 4 4/5 min 6 + 24/5 = 54/5 minutes in 90 hours. So, the watch gains 6 minutes in (90×5×6)/54 = 50 hours
Add 50 hours in Monday 12’O clock night, thus the watch is correct at 2:00 am on Thursday.
Example 7: A clock is set right at 10 am. The clock gains 5 minutes in 12 hours. What will be the true time when the clock indicates 3 pm on the next day?
Sol: Time from 10 am to 3 pm on the following day is 29 hrs. Now, 12 hrs 5 min i.e. 145/12 hrs of this clock = 12 hours of correct clock. So, 29 hours of this clock is (29×12×12)/145 = 144/5 = 28 hours 48 minutes.
So, the time is 12 minutes before 3 pm.
Example 8: The minute hand of a clock overtakes the hour hand at intervals of 65 minutes of correct time. How much does the clock lose or gain in 12 hours?
Sol: In a correct clock, the minute hand and hour hand should meet after every 65 5/11 min. But we know that they are meeting after every 65 minutes. So, the gain in 65 minutes is 5/11 minutes.
Gain in 12 hours = (120×60/65) × (5/11) = 720/143 = 5 5/143 min.
Example 9: At what time between 4 and 5, the minute hand will be 2 minutes spaces ahead of hour hand?
Sol: At 4 O’ clock, the two hands are 20 min spaces apart. In this case, the min hand will have to gain (20 + 2) i.e. 22 – minute spaces. So, 22 – minute spaces will be gained in (60/55) × 22 = 24 min.
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Q:
| A) (9 + 10/11) min past 2 | B) (10 + 10/11) min past 2 |
| C) (11 + 10/11) min past 2 | D) (12 + 10/11) min past 2 |
Answer: B) (10 + 10/11) min past 2
Explanation:
At 2 o'clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min spaces apart.
To be together, the minute hand must gain 10 minutes over the hour hand.
Now, 55 minutes are gained by it in 60 min.
The hands will coincide at 101011 min. past 2.
University Grants Commission (Minimum Standards and Procedures for Award of Ph.D. Degree) Regulations, 2022 notified. As, per the new regulations, candidates with a 4 years Undergraduate degree with a minimum CGPA of 7.5 can enroll for PhD admissions. The UGC NET Final Result for merged cycles of December 2021 and June 2022 was released on 5th November 2022. Along with the results UGC has also released the UGC NET Cut-Off. With tis, the exam for the merged cycles of Dec 2021 and June 2022 have conclude. The notification for December 2022 is expected to be out soon. The UGC NET CBT exam consists of two papers - Paper I and Paper II. Paper I consists of 50 questions and Paper II consists of 100 questions. By qualifying this exam, candidates will be deemed eligible for JRF and Assistant Professor posts in Universities and Institutes across the country.