There are three points P, Q and R on a straight line such that PQ : QR = 3 : 5, If n is the number of possible values of PQ : PR, then what is n equal to?

The correct option is 2.

Let us analyze the possible arrangements of the three points P, Q, and R on a straight line.

We are given that the ratio of the distances between the points is:
$PQ:QR=3:5$

This implies that for some positive constant k:
$PQ=3k$
$QR=5k$

Since P, Q, and R are collinear, the point Q can either lie between P and R, or P can lie between Q and R. (Note that R cannot lie between P and Q because QR = 5k is greater than PQ = 3k).

Case 1: Point Q lies between P and R
In this arrangement, the total distance PR is the sum of PQ and QR:
$PR=PQ+QR=3k+5k=8k$
Therefore, the ratio of PQ to PR is:
$\frac{PQ}{PR}=\frac{3k}{8k}=\frac{3}{8}$

Case 2: Point P lies between Q and R
In this arrangement, QR is the distance between Q and R, so:
$QR=QP+PR$
$5k=3k+PR$
$PR=5k-3k=2k$
Therefore, the ratio of PQ to PR is:
$\frac{PQ}{PR}=\frac{3k}{2k}=\frac{3}{2}$

Thus, there are 2 possible values for the ratio PQ : PR, which are $3:8$ and $3:2$.
Hence, n = 2.