From S1:
HCF(a,64) = 4
So a contains a common factor 4.
a can be 4, 12, 20, 28, 44 and so on.
No info about b.
INSUFFICIENT.
From S2:
HCF(b,64) = 1
There are no common factors between b and 64.
b can be 3, 5, 7, 11, 15, 21 and so on.
No info about a.
INSUFFICIENT.
Combining both:
If (a,b) = (4,3), then HCF = 1
If (a,b) = (60,15), then HCF = 15
Yes or no.
Hence INSUFFICIENT.
E is the answer.
HCF(a,64) = 4
So a contains a common factor 4.
a can be 4, 12, 20, 28, 44 and so on.
No info about b.
INSUFFICIENT.
From S2:
HCF(b,64) = 1
There are no common factors between b and 64.
b can be 3, 5, 7, 11, 15, 21 and so on.
No info about a.
INSUFFICIENT.
Combining both:
If (a,b) = (4,3), then HCF = 1
If (a,b) = (60,15), then HCF = 15
Yes or no.
Hence INSUFFICIENT.
E is the answer.









