At any point of time, let x be the smaller of the two angles made by the hour hand with the minute hand on an analogue clock (in degrees). During the time interval from 2:30 p.m. to 3:00 p.m., what is the minimum possible value of x?
- 45
- 105
- 90
- 0
- 75
The Correct Option is C
Solution and Explanation
The formula for the angle between the hour and minute hand is:
\[ \theta = \left| 30H - 5.5M \right| \]
where \(H\) = hour and \(M\) = minutes.
Step 1: Express position at 2:30
At 2:30, \(H = 2, M = 30\).
Hour hand = \(30H + 0.5M = 30 \times 2 + 0.5 \times 30 = 60 + 15 = 75^\circ\).
Minute hand = \(6M = 6 \times 30 = 180^\circ\).
Angle = \(|180 - 75| = 105^\circ\).
Step 2: Express position at 3:00
At 3:00, \(H = 3, M = 0\).
Hour hand = \(30 \times 3 + 0.5 \times 0 = 90^\circ\).
Minute hand = 0.
Angle = \(|90 - 0| = 90^\circ\).
Step 3: General expression between 2:30 and 3:00
Let \(M\) = minutes after 2:00. So between 2:30 and 3:00, \(M\) varies from 30 to 60.
The formula is: \[ \theta = |30 \times 2 - 5.5M| = |60 - 5.5M| \]
Step 4: Check when hands overlap
For overlap, \(\theta = 0 \Rightarrow 60 - 5.5M = 0\).
\(\Rightarrow M = \frac{60}{5.5} \approx 10.91 \, \text{minutes after 2:00}\).
That is at approximately 2:10:55, which is before 2:30. So check again carefully.
Step 5: Careful adjustment
Actually, formula for general angle is: \[ \theta = |30H + 0.5M - 6M| \] For \(H = 2\): \[ \theta = |60 + 0.5M - 6M| = |60 - 5.5M| \] At 2:30, we found \(\theta = 105^\circ\).
At 3:00, \(\theta = 90^\circ\).
But the hour hand also progresses towards 3 during these 30 minutes, so overlap happens again after 3. Between 2:30 and 3:00, the angle decreases continuously from 105° to 90°. Hence, the minimum possible angle is 90°.
Final Answer:
\[ \boxed{90^\circ} \]
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