---ASIDE---------------------
Take a moment to learn the following property; it will provide a useful way to think of divisibility questions.

For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------

If x/3, x/5, and x/4 are integers, then x is divisible by 3, 5 and 4.
This means there must be a 3, a 5 and a 4 hiding in the prime factorization of x.
Since 4 = (2)(2), the prime factorization of x looks something like this: x = (2)(2)(3)(5)(?)(?)(?) [The additional (?)'s represents other possible prime factors of x]

If x = (2)(2)(3)(5)(?)(?)(?), we can see a 10, 15, 20 and 30 hiding in the prime factorization of x, which means A, B, C and E must be integers.
Answer: D

ALTERNATE APPROACH
If x/3, x/5, and x/4 are integers, then it could be the case that x = 60
Now plug x = 60 into each answer choice...
(A) 60/10 = 6, which is an integer.
(B) 60/15 = 4, which is an integer.
(C) 60/20 = 3, which is an integer.
(D) 60/24 = 2.5, which is NOT an integer.
(E) 60/30 = 2, which is an integer.
Answer: D