Question

At what time, in minutes, between 3 o'clock and 4 o'clock, both the needles will coincide each other?
 

• A. 5 and 5/11 minutes past 3
• B. 12 and 4/11 minutes past 3
• C. 16 and 4/11 minutes past 3
• D. 13 and 4/11 minutes past 3

Hint:

A quick and reliable formula to find the time \( M \) minutes past H o'clock when the hands of a clock coincide is \( M = \frac{60 \times H}{11} \). For H=3, this gives \( M = \frac{60 \times 3}{11} = \frac{180}{11} = 16 \frac{4}{11} \) minutes.

Answer 1

The Correct Option is C

To find the exact time when the hour and minute hands of a clock coincide, we can analyze their relative speed.

Step 1: Determine the speeds of the hands. - The minute hand completes a full 360\(^{\circ}\) circle in 60 minutes. Its speed is \( \frac{360^{\circ}}{60 \text{ min}} = 6^{\circ} \) per minute. - The hour hand completes a full 360\(^{\circ}\) circle in 12 hours (720 minutes). Its speed is \( \frac{360^{\circ}}{720 \text{ min}} = 0.5^{\circ} \) per minute.

Step 2: Calculate the relative speed. The minute hand is faster. The relative speed at which the minute hand "gains" on the hour hand is \( 6^{\circ} - 0.5^{\circ} = 5.5^{\circ} \) per minute.

Step 3: Find the initial separation angle at 3 o'clock. At 3:00, the minute hand is at the 12 (0\(^{\circ}\)) and the hour hand is at the 3. The angle between them is \( 3 \times 30^{\circ} = 90^{\circ} \).

Step 4: Calculate the time to coincide. For the hands to coincide, the minute hand must cover this 90\(^{\circ}\) gap at its relative speed. Time required = \( \frac{\text{Angle to be gained}}{\text{Relative speed}} \) \[ \text{Time} = \frac{90^{\circ}}{5.5^{\circ}/\text{min}} = \frac{90}{11/2} = \frac{180}{11} \text{ minutes} \]

Step 5: Convert the improper fraction to a mixed fraction. To convert \( \frac{180}{11} \) to a mixed fraction, we divide 180 by 11. \[ 180 \div 11 = 16 \text{ with a remainder of } 4 \] So, the time is \( 16 \frac{4}{11} \) minutes past 3 o'clock. This corresponds to option (C).