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I dont think C is correct , let set={5,6,10,15,19,23,25,10000000001} and let x=10 then (x+5) 15 is in Set and (x-5) 5 is in set but the set doesnt contain all the multiples of 5.

What is the source of this question? It's very unclear...

If S is a set of integers and 5 is in S, is every multiple of 5 in S?

1) if x is in S, then x+5 is in S
2) if x is in S, then x-5 is in S

This question could be interpreted in a number of ways....

First, I picked E. 1) Cause let's say that x is 5, then 5 and 10 will be in the set. Not sufficient. 2) Let's say that x is 5, then 5, and 0 will be in the set. Not sufficient. 3) Again, let's use 5.. then 5, 10, 0 will be in the set. Not sufficient. Also, you can use other numbers that are not mutliples of 5.

To accurately get the answer C, then the question needs to be reworded to something like this... For every member of set S (called x), then x+5 is in S. Or something to that "ETS verbage" stuff. The question as it is, implies that x is only one member of the set, and you use the +5 or -5 rule to only one member, not all the members.

What is the source of this question? It's very unclear...

If S is a set of integers and 5 is in S, is every multiple of 5 in S?

1) if x is in S, then x+5 is in S 2) if x is in S, then x-5 is in S

This question could be interpreted in a number of ways....

First, I picked E. 1) Cause let's say that x is 5, then 5 and 10 will be in the set. Not sufficient. 2) Let's say that x is 5, then 5, and 0 will be in the set. Not sufficient. 3) Again, let's use 5.. then 5, 10, 0 will be in the set. Not sufficient. Also, you can use other numbers that are not mutliples of 5.

To accurately get the answer C, then the question needs to be reworded to something like this... For every member of set S (called x), then x+5 is in S. Or something to that "ETS verbage" stuff. The question as it is, implies that x is only one member of the set, and you use the +5 or -5 rule to only one member, not all the members.

I am sorry but I do not agree with you. The given wording of the problem is accurate.

When the statement says that if x is in S, then x+5 is in S, then you are invited to think recursively.

We have already been given that 5 is in S, so based on the first statement 5, 10, 15, ..... infinite all are members of S. Similarly the second statment also apllies. The answer should be C.

It's news to me that 0 is a multiple of 5
This was not the case where I went to school

Though this expains the answer to one of the Kaplin
maths probems in the 2004 GMAT book
Question 13 in the maths section practice test states that 0 is a multiple
of 5