QUESTION

CSAT

Hard

Maths

Prelims 2014

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?

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Explanation

Given conditions:

  • The time is between 8:00 and 9:00.
  • The hour and minute hands coincide (one above the other).
  • We need to find when they will coincide again.

Step 1: Finding the first coincidence At 8:00, the hour hand is at 8×30=2408 \times 30 = 240^\circ. The minute hand starts at 0° and moves 360° per hour (6° per minute). The hour hand moves 30° per hour (0.5° per minute).

Let tt be the time in minutes after 8:00 when they coincide: 240+0.5t=6t240 + 0.5t = 6t 240=5.5t240 = 5.5t

t=240/5.5=43.64t = 240 / 5.5 = 43.64

Step 2: Finding the next coincidence The hands coincide every 360/(5.5 per minute)65.45 minutes360^\circ / (5.5^\circ \text{ per minute}) \approx 65.45 \text{ minutes}.

Rounding to the nearest integer: 65 minutes65 \text{ minutes}.

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