$40 = 2*2*2*5$ and $75 = 3*5*5$

For $x^2$ to be divisible by 40 and 75, its prime-factorization must include at least three 2's (since there are three 2's within 40), at least one 3 (since
there is one 3 within 75), and at least two 5's (since there are two 5's within 75):
$2^3 * 3^1 * 5^2$

However, since $x^2$ is a perfect square, its prime-factorization must have an EVEN NUMBER of every prime factor.
Since the prime-factorization of x must include $2^3$, $3^1$ and $5^2$ -- but $x$ must have an even number of each of these prime factors -- the least possible option for $x^2$ is as follows:
$2^4 * 3^2 * 5^2$
Since the least possible option for $x^2 = 2^4 * 3^2 * 5^2$, the least possible option for $x = 2^2 * 3 * 5 = 60$.

Implication:
$x$ must be a MULTIPLE OF 60.
In addtion, since $x$ must have exactly three distinct prime factors, it cannot be divisible by any prime number other than 2, 3 and 5.

Since 30 and 200 are not divisible by 60, eliminate A and C.
Since 420 is divisible by 7 -- a prime number other than 2, 3 and 5 -- eliminate E.