a) x is factor of 7200.
if x=2, 1 prime factor. if x=6, 2 prime factors
Insufficient
b)180 is factor of x.
x=180*11 or x=180*11*13.. insufficient

a&b) 7200/180 = 40. the factors of 40 (2,5) are already in 180.
thus sufficient
IMO C

(1) x is a factor of 7200
7200=2^5*3^2*5^2

Basically x contains any of the prime numbers 2,3 or 5 in some combination. Could contain all of them, could contain two of them or all three of them. example x could be 8 i.e. 2^3 or x could be 15 i.e. 3*5 etc. No fixed value of x can be deduced. INSUFFICIENT

(2)180 is a factor of x meaning
x=2^2*3^2*5*unknown
if unknown=1 then x has 3 prime factors
if unknown=7 then x has 4 prime factors etc. We get different answers so INSUFFICIENT

Combining both,
From(1) we have that, x contains only the prime factors 2,5,3 in some combination and NO OTHER PRIME example if x contained a 7 then 7200 wouldn't be divisible by x. Also, from statement (2) x=2^2*3^2*5*unknown and unknown can't contain any other prime as this would make 7200 indivisible by x. So unknown has to contain some other power of 2,3,5 which makes 7200 perfectly divisible by x. So this narrows down the prime factorisation of x to 2 AND 3 AND 5 with different/same powers for each. So x has three prime factors. Hence C