Given b=pq where p&q are prime numbers ≥3 and 1≤pq≤60
If p=3,q=3,5,7,11,13,17,19
And if p=5q=5,7,11 and so on,
So, b will be 3×3,3×5,3×7,3×11,3×13,3×17,3×19,5×5,5×7,5×11,7×7
i.e. total 11 possibilities
Now a can be 1,2,⋯,60
i.e. total 60 possibilities
Hence, the number of relations =60×11=660