A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8, thumb 9, then back to the index finger for 10, middle finger for 11, and so on. She counted up to 1994. She ended on her.
Explanation:
It is clear that after a particular amount of time P and Q are equidistant from A and B respectively and speed of Q is twice the speed of P, therefore, in the remaining time distance moved by Q will be twice than P. Hence, they would meet closer to A.
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