Assume that 1. the hour and minute hands of a clock move without jerking. 2. the clock shows a time between 8 o'clock and 9 o'clock. 3. the two hands of the clock are one above the other. After how many minutes (nearest integer) with the two hands will be again lying one above the other?

The minute hand moves at 360 degrees in 60 minutes, so its speed is 6 degrees per minute. The hour hand moves at 360 degrees in 12 hours (720 minutes), so its speed is 0.5 degrees per minute.

When the two hands are "one above the other," they are coincident (the angle between them is 0 degrees). For them to be one above the other again, the minute hand must complete one full lap (360 degrees) relative to the hour hand.

The relative speed of the minute hand with respect to the hour hand is 6 - 0.5 = 5.5 degrees per minute.

To gain 360 degrees at a relative speed of 5.5 degrees per minute, the time taken is: Time = Total degrees / Relative speed Time = 360 degrees / 5.5 degrees/minute Time = 360 / (11/2) minutes Time = (360 * 2) / 11 minutes Time = 720 / 11 minutes

Calculating 720 / 11: 720 / 11 = 65.4545... minutes

Rounding this to the nearest integer gives 65 minutes.

The conditions given (no jerking, time between 8 and 9 o'clock, hands already coincident) establish the scenario but do not change the fundamental time interval between two consecutive coincidences of the hands. This interval is always 720/11 minutes.

Analyzing the options: A) 60 minutes: This would mean the minute hand gained 360 degrees in 60 minutes, while the hour hand did not move (or moved at the same speed). Incorrect. B) 62 minutes: Too short. C) 65 minutes: This matches our calculation (65.45 rounded to 65). D) 67 minutes: Too long.

Therefore, after approximately 65 minutes, the two hands will again be lying one above the other.

The final answer is C) 65.