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Updated on: 26 Oct 2025, 23:41
Originally posted by kalravaibhav on 03 Oct 2013, 12:21.
Last edited by Bunuel on 26 Oct 2025, 23:41, edited 3 times in total.
Last edited by Bunuel on 26 Oct 2025, 23:41, edited 3 times in total.
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Difficulty:
35%
(medium)
Question Stats:
63% (00:51) correct
37%
(01:07)
wrong
based on 425
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A set of numbers is defined such that if x and y are in the set, xy is also in the set. If 2 and 6 are in the set, which of the following must also be in the set?
I. 3
II. 12
III. 24
A. None
B. I only
C. II only
D. I and II only
E. II and III only
Hey Guys,
I am confused here. Please help. According to me the following is the logic:
2, 6 are in the set, then 2*6 = 12 is also in the set
if 6 and 2 are in the set then 2*3=6 ==> 3 must be also in the set
if 12 and 2 are in the set then 2*12 = 24 should also be in the set
So according to me all 3 options are possible in the set. However such an option does not exist. Please clarify.
Thanks.
I. 3
II. 12
III. 24
A. None
B. I only
C. II only
D. I and II only
E. II and III only
Hey Guys,
I am confused here. Please help. According to me the following is the logic:
2, 6 are in the set, then 2*6 = 12 is also in the set
if 6 and 2 are in the set then 2*3=6 ==> 3 must be also in the set
if 12 and 2 are in the set then 2*12 = 24 should also be in the set
So according to me all 3 options are possible in the set. However such an option does not exist. Please clarify.
Thanks.
04 Oct 2013, 13:10
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AccipiterQ
Dear AccipiterQ,
First of all, I am just going to recommend a grammar correction in what you have written:
"You don't know whether 2 or 6 are x or y."
That's a very typical GMAT SC trap. See:
https://magoosh.com/gmat/2012/gmat-sente ... s-whether/
I would say --- the wording of this question is very clear. I would call this a high quality, well-written question.
The issue you raise ---- we don't know whether the numbers given equal x or y --- this betrays a subtle misunderstanding. In some algebra problems, the "solve for x" variety of problems, x or y or any variable will have one definite value, or perhaps two possible values, and the point is to find those individual values. In many other algebraic situations, the variables are general variables that don't have specific values. For example, when we have the equation of the graph of a line in the x-y plane, such as y = 2x - 3, the x & y don't have individual specific values --- rather, the x & y coordinates of every point on the line, the whole continuous infinity of them, satisfy the equation. Thus, the x & y could be any one of a whole infinity of values, and yet it's very precise and clear which x' & y's could or couldn't fit the pattern.
In this set, it is unambiguous from context --- x and y are general variables that could equal any member of the set. If we know a number is in the set, that number could equal x or y. At the outset, the only known members of the set are 2 & 6, so either could be x, and either could be y. That's how we know 12 must be in the set. There's nothing saying that x and y have to be different, so if we pick x = y = 2, then 4 has to be in the set, and if we pick x = y = 6, then 36 has to be in the set. With just this analysis, we see the members of the set must include {2, 4, 6, 12, 36}. Now, any of those five numbers could be x, and any of the five could be y. Thus, if we pick x = 2 and y = 12, we see that 24 must be in the set. In fact, for what it's worth, here is the full set of all numbers less than 100 that must be in the set
{2, 4, 6, 8, 12, 16, 24, 32, 36, 48, 64, 72, 96}
Many other numbers could be in the set, but those are the ones that absolutely have to be in the set, no question.
It's true --- many other element could be in the set, but that's not the question. The question is unambiguously clear: "which of the following must also be in the set?" From the analysis above, we see that 12 and 24 absolutely must be in the set, no doubt. It's true, 3 could be in the set, but that's strictly irrelevant to this particular question.
My friend, it is crucially important for you to understand: any ambiguity you experience in this question does not lie in the wording of the question itself, but in subtle gaps in your own understanding. I want you to understand this, so you resolve these issues and don't have this same ambiguous experience on test day.
Let me know if you have any further questions.
Mike
03 Oct 2013, 13:44
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kalravaibhav
Dear kalravaibhav,
Be careful about switching parts of the premise and conclusion in an IF-THEN statement. The statement If P, then Q does not imply If Q, then P. Similarly, the statement If P and Q, then T does not implies If P and T, then Q. That's the transformation you did.
TRUE: if x and y are in the set, the xy is also in the set
FALSE: if x and xy are in the set, then y is also in the set
That's precisely the logic you followed in the line of yours that I marked in red. That's not true ---- in fact, that was precisely the trap of the question, inducing folks to make this incorrect logical deduction.
Does all this make sense?
Mike
