Solution:

Given that HCF(x,y) * LCM(x,y) = 91

Let HCF(x,y) = a and LCM(x,y) = b

Then a*b = 91

As 91 is a product of two distinct prime numbers 7 and 13, there are only two pairs of factors of 91: (1,91) and (7,13)

Case 1: a = 1 and b = 91

In this case, x and y have no common factors other than 1, so they must be coprime.

Number of coprime pairs of positive integers less than or equal to 91 is given by Euler's totient function φ(91) = (7-1)(13-1) = 72

Therefore, there are 72 pairs of positive integers x and y such that HCF(x,y) = 1 and LCM(x,y) = 91.

Case 2: a = 7 and b = 13

In this case, x and y have a common factor of 7, and their LCM is 91 = 7*13.

Let x = 7p and y = 7q, where p and q are coprime.

Then LCM(x,y) = 7pq = 7*13, so pq = 13.

The only pairs of coprime positive integers whose product is 13 are (1,13) and (13,1).

Therefore, there are two pairs of positive integers x and y such that HCF(x,y) = 7 and LCM(x,y) = 91.

Total number of pairs of positive integers x and y such that HCF(x,y) * LCM(x,y) = 91 is 72 + 2 = 74.

But we need to divide this by 2, since each pair of positive integers is counted twice (once as (x,y) and once as (y,x)).

Therefore, the number of pairs of positive integers x and y such that HCF(x,y) * LCM(x,y) = 91 is 74/2 = 37.

Answer: B) 8