A clock is observed. The hour hand is exactly at the minute mark, and the minute hand is six minutes ahead of it. Later, the clock is observed again. This time, the hour hand is exactly on a different minute mark, and the minute hand is seven minutes ahead of it. What could be the least time difference between first and second observations?Post the solution also....

1) 12 Minutes
2) 1 Hour 12 Minutes
3) 2 Hours 12 Minutes
4) Data Insufficient :P

• Solution for #14
  
  The hour hand is exactly on a minute mark five times per hour -- on the hour, twelve minutes past the hour, twenty four minutes past, thirty six minutes past, and forty eight minutes past.
  
  Let X be the number of hours, and Y be the number of minutes past the hour. When the hour hand is on a minute mark, the position of the hour hand is 5X + Y/12, and the position of the minute hand is Y. On the first occasion, Y = 5X + Y/12 + 6. This is equivalent to 60X = 11Y - 72. Since Y can only take one of the values in the set { 0, 12, 24, 36, 48 }, it can be determined that the only legal values for the equation are X = 1 and Y = 12. So the time is 1:12.
  
  Similarly, the second occasion's equation is 60X = 11Y - 84. The only legal values here are X = 3 and Y = 24. So the time is 3:24.
  
  Between 1:12 and 3:24, two hours and twelve minutes have elapsed
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