Maximum Persons Standing Ahead in Queue

Let the number of persons standing ahead of Pinky be $3k$ and behind her be $5k$. Total number of persons in the queue $= 3k + 5k + 1 = 8k + 1$ (including Pinky).

We are given: $8k + 1 < 300$ $8k < 299$ $k < 37.375$

Since $k$ is an integer, the maximum value of $k$ is $37$.

The maximum possible number of persons standing ahead of Pinky is: $3k = 3 \times 37 = 111$