if x is in the set, then x-3 is also is in the set

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For a certain set of numbers, if x is in the set, then x-3 is also is in the set. If the number of 1 is in the set, which of the following must also be in the set?

a) 4

b) -1

c) -5

A) a only

B) b only

C) c only

D) a & b only

E) b & c only

OAC

Please explain

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by regor60 » Wed Sep 14, 2016 9:02 am
Needgmat wrote:For a certain set of numbers, if x is in the set, then x-3 is also is in the set. If the number of 1 is in the set, which of the following must also be in the set?

a) 4

b) -1

c) -5

A) a only

B) b only

C) c only

D) a & b only

E) b & c only

OAC

Please explain

This has recently been discussed. I'll assume you're confused as to why 4 is not correct.

It is not correct because while the statement specifically states successively lower numbers are in the set, this statement doesn't permit the inference that the presence of a number in the set necessarily means the next higher number HAS to be.

In other words, for all we know, the set began at 1.

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by [email protected] » Wed Sep 14, 2016 9:45 am
Hi Needgmat,

The logic that this question is based on is the same type of logic that appears in "causality-based" CR questions. (re: X --> Y).

We're told that the number "1" is in the set, so (1-3) = -2 must also be in the set. Since -2 is in the set, (-2 - 3) = -5 must also be in the set... and so on. The logic does NOT 'work backwards' though, it only works 'forwards', thus we don't know whether the number 4 appears in the set or not.

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by Matt@VeritasPrep » Thu Sep 15, 2016 6:53 pm
The easiest way to think of this is to assume that you've just started this set, and the only number you've been given is 1.

From there, you're told that the set has to contain (1 - 3), so you add -2 to the set. Then, you must have (-2 - 3), so -5 appears in the set, etc.

So we've got 1, -2, -5, ... all the way to -∞.

It's certainly possible that 4 is in the set, but it doesn't HAVE to be: we only know that 1 is there, and that all the lesser terms (1 - 3, 1 - 3 - 3, etc.) must follow along with it.

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by Jeff@TargetTestPrep » Tue Dec 19, 2017 7:16 am
For a certain set of numbers, if x is in the set, then x-3 is also is in the set. If the number of 1 is in the set, which of the following must also be in the set?

a) 4

b) -1

c) -5

A) a only

B) b only

C) c only

D) a & b only

E) b & c only
If we let x = 1, then (1 - 3) = -2 is in the set. If -2 is in the set, then (-2 - 3) = -5 is in the set. We can see that if we keep subtracting 3, we will get terms such as -8, -11, -14, etc. We see that -5 is definitely in the set, but -1 isn't, since if 1 and -2 are in the set, then any numbers between them can't be in the set.

Note: some people might argue that 4 is also in the set, since if we let x - 3 = 1, then x = 4. Of course, the answer choices don't have 'a and c only' as an option. The problem says: "if x is in the set, then x - 3 is also in the set." The problem doesn't say: "if x - 3 is in the set, then x is also in the set." So, when the problem says "if the number 1 is in the set," we have to assume that 1 is the value of x, and we have to subtract 3 and keep subtracting 3 to get subsequent terms. We can't assume that 1 is the value of x - 3.

Answer: C

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