II. |x+3|>2

x+3 > 2 -> x > -1 (or)
-x-3 > 2 -> x < -5

Since one of the range(x<-5) satisfies the inequality, this statement can be considered true.

So both II and III must be true. DIMO

from the inequality you can deduce x<-5;
you are clear about II;
now if x<-5; for any value of x; |x+3|>2 will be true; substitute any value of x below -5 and see
say for x=-6 ; |-6+3|= 3 which is >2

thus II and III

so I can ignore the x>-1 in III?

The trick in this problem is you are given the condition 4<(7-x)/3 is valid and not asked if its true

If you had to see whether the condition satisfies for the 3 statements then statement II need not hold good always as it would fail for x>-1.

But again remember you are given 4<(7-x)/3 is true, so all values satisfying this condition will definitely lie only in the range x < -5. Thats why we can ignore the other range. If the questions was asked the other way around, only III would hold true. GMAT always comes up with these smart tricks, be especially careful when you find answers too easy.

kavn wrote:so I can ignore the x>-1 in III?