n=X^2?

From stmt1,

If n is divisible by p, then it is also divisible by p^2

n=16,16 is divisible by 2 and 4.n is square of an integer

n=27,27 is divisible by 3 and 9.n is even.n is not square of an integer.

Insufficient.

From stmt2,

If sqrt(n) is an integer,then must certainly be certainly square of an integer.

sufficient.

Hence B

a little bit more about st 1 ( hope my reasoning is valid)

(1) we are given that
n=p*k, and n=p^2*a, so
pk=p^2*a. cancel p, and left with k=pa. let us insert this in, n=pk
n=p*p*a=p^2*a,
the value of n depends on a p^2 is perfect square, but a may be yes or not be perfect square
so 1 st insufficient

From stmnt 1 :

n can be = p x p (since both p & p^2 are divisors)
but it can also be n = p x p x Y (where Y is any other integer, both p & p^2 are still divisors)

so from stmnt 1 we cannot conclude if n = integer ^2

From stmnt 2:

more straight forward ... if sqrt n = integer, then n = integer ^ 2

So the answer is (B)


comments?

Q

stmt1

if p=3,n=9 cond satisfies
p=5,n=75 cond not satisfied

insuff


stmt2
suff

B