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Set integer/multiple of Q

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Set integer/multiple of Q

by milanproda » Mon Dec 27, 2010 12:34 pm
Q is a set of integers and 11 is in Q. Let q (small q) denote any element in the Q set. Is every positive multiple of 11 in Q?

1 q+11 is in Q

2 q-11 is in Q

Answer: 1- Statement 1 is sufficient, but statment 2 is not sufficient

Thanks! If anyone has time, please explain the answer thoroughly becuase I am having trouble understanding it.
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by Anurag@Gurome » Mon Dec 27, 2010 12:46 pm
milanproda wrote:Q is a set of integers and 11 is in Q. Let q (small q) denote any element in the Q set. Is every positive multiple of 11 in Q?

1 q+11 is in Q

2 q-11 is in Q
Given: Q is a set of integers and 11 is in Q.
Now q denote any element in the Q set. Note that this means any element of set Q can be taken as q.

Statement 1: (q + 11) is in Q
We know that 11 is in set Q. Therefore we can say q = 11. This implies (q + 11) = 22 is in Q. Now we can say q = 22, thus (q + 11) = 33 is in Q. Thus 44, 55, 66 etc all are in Q. Which means all positive multiples of 11 are in Q.

Sufficient

Statement 2: (q - 11) is in Q
We know that 11 is in set Q. Therefore we can say q = 11. This implies (q - 11) = 0 is in Q. Now we can say q = 0, thus (q - 11) = -11 is in Q. Thus -22, -33, -44 etc all are in Q. Which means all non-positive multiples of 11 are in Q. But nothing can be said about positive multiples of 11.

Not sufficient

The correct answer is A.
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by milanproda » Mon Dec 27, 2010 12:52 pm
Thanks!

I had trouble understanding what the question wanted, and I did not realize that I should substitute 11 as q, which was my biggest fault.

For arguements sake, because the question said that q can denote any element in Q set, could I assume that any positive integer is in Q set? Or does the 1st stem indicate that it must be a multiple of 11?
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by clock60 » Mon Dec 27, 2010 1:06 pm
hi Anurag
honestly did not get nether problem nor your explanation,
q can be any any element in Q, and the terms in set Q are not specified
q=1, q+11=1+11=12 is not multiple of 11
q=11, 11+11=22 multiple of 11
deeply confused....?

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by milanproda » Mon Dec 27, 2010 1:07 pm
That is what I thought also...
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by Anurag@Gurome » Mon Dec 27, 2010 2:04 pm
clock60 wrote:hi Anurag
honestly did not get nether problem nor your explanation,
q can be any any element in Q, and the terms in set Q are not specified
q=1, q+11=1+11=12 is not multiple of 11 (we don't have to bother about it)
q=11, 11+11=22 multiple of 11 (we know 11 is there. Thus all multiples of 11 is there)
deeply confused....?
Yes, there may or may not be other integers in set Q.
But 11 is certainly in Q and thus we can take 11 as q. Therefore for statement 1, all the positive multiples of 11 are there in Q. We are sure about it. Now there may be other integers, but we don't have to bother about it as the question is not asking whether Q is the set of all the positive multiples of 11. The question asks whether Q contains them or not.
milanproda wrote:For arguements sake, because the question said that q can denote any element in Q set, could I assume that any positive integer is in Q set? Or does the 1st stem indicate that it must be a multiple of 11?
Yes there may be other integers in Q, but you cannot assume which one is there. But we have the information that 11 is there and we have to proceed accordingly.
Last edited by Anurag@Gurome on Mon Dec 27, 2010 2:08 pm, edited 3 times in total.
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by milanproda » Mon Dec 27, 2010 2:05 pm
Thanks again
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by clock60 » Mon Dec 27, 2010 2:20 pm
great thanks Anurad
i got it, pretty convoluated question. don`t like it

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by goyalsau » Tue Jan 04, 2011 2:51 am
milanproda wrote:Q is a set of integers and 11 is in Q. Let q (small q) denote any element in the Q set. Is every positive multiple of 11 in Q?

1 q+11 is in Q

2 q-11 is in Q
It looks like a real Gmat question. What's the source?/ : : : : : : :
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by ashforgmat » Tue Jan 04, 2011 3:35 am
@Anurag: hi! Sorry but your explanation is even more confusing...
If we can have other elements apart from multiples of 11 in Q ...then we can have zero and other -ve nos. as well in Q???
Please can you help here.

Thanks

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by nipunkathuria » Tue Jan 04, 2011 5:09 pm
Anurag@Gurome wrote:
clock60 wrote:hi Anurag
honestly did not get nether problem nor your explanation,
q can be any any element in Q, and the terms in set Q are not specified
q=1, q+11=1+11=12 is not multiple of 11 (we don't have to bother about it)
q=11, 11+11=22 multiple of 11 (we know 11 is there. Thus all multiples of 11 is there)
deeply confused....?
Yes, there may or may not be other integers in set Q.
But 11 is certainly in Q and thus we can take 11 as q. Therefore for statement 1, all the positive multiples of 11 are there in Q. We are sure about it. Now there may be other integers, but we don't have to bother about it as the question is not asking whether Q is the set of all the positive multiples of 11. The question asks whether Q contains them or not.
milanproda wrote:For arguements sake, because the question said that q can denote any element in Q set, could I assume that any positive integer is in Q set? Or does the 1st stem indicate that it must be a multiple of 11?
Yes there may be other integers in Q, but you cannot assume which one is there. But we have the information that 11 is there and we have to proceed accordingly.

Hi Anurag,

I have one scenario:

If we have q=-22
q+11=-11,etc
so the elements can be
-22.-11,0,11,etc
here -22 is (-2)times 11---->negative multiple

If this is true , we cannot proceed with A as sufficient
But u have taken only positive number...here too we find that 11 is a member of the group
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by Geva@EconomistGMAT » Wed Jan 05, 2011 12:45 am
ashforgmat wrote:@Anurag: hi! Sorry but your explanation is even more confusing...
If we can have other elements apart from multiples of 11 in Q ...then we can have zero and other -ve nos. as well in Q???
Please can you help here.

Thanks
Just to reiterate the point made by Anurag: read the question carefully. The question asks "is every positive multiple of 11 in Q", not "does Q hold ONLY the positive multiples of 11"? As long as Q holds all positive multiples of 11, the answer to the question is "yes", even if Q holds negative multiples of 11 as well - which is why stat. (1) is sufficient.

Stat. (2) is insufficient because out of the positive multiples, we can only really be sure that 11 is in the set: Q will hold all negative multiples of 11, but we're not sure about the positive ones - Q MAY or MAY NOT hold the positive multiples of 11.

BTW, I would be very surprised if this were a real GMAT question. Too many abstract concepts - for one, an infinite set is highly un-GMAT, IMHO.
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by thebigkats » Sun Jan 09, 2011 11:05 am
Q is a set of integers and 11 is in Q. This tells us that (a) Q is not necessarily a finite set of integers - we don't know and (b) Q contains 11

STAT #1 - q+11 is in Q

we know that 11 is there in set, so 11+11 is also in set (amont other integers that we don;t care much about because Q only asks about positive multiples of 11)
However if 11+11 is there in te set then (11+11) + 11 is also there in the set (for any q, q+11 is also in set)
and same goes for further values also.
There may be other integers in set but forsure 11, 22, 33, 44, 55.... are all there
So it is SUFFICIENT

STAT #2 - q-11 is in Q
LEts start from what we know from stem - 11 is in set.
so forsure 11-11 (= 0) is in the set
and hence 0-11 (-11) is in set and so forth.... so the set at least has 11, 0, -11, -22, -33....
it doesn;t prove anything about POSITIVE MULTIPLES of 11

Hence A

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by Taran » Sat Feb 05, 2011 7:35 am
Hi Guys,

What is the source of this question? Anyhow, i found the question very vaguely drafted.

1. It asks whether the Set Q contains 'every' positive multiple of 11. What if the set is {11, 22, 33, 45, 21, so on......}. In this case, we can assume q to be any integer. Considering this, we will always have positive multiples of 11. My doubt, here is that what if q is 2 and all q+11's are in the set, but only 100 multiples of 11 are there. I hope my question is clear to all.

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by cyrwr1 » Sat Feb 05, 2011 6:15 pm
Yes I agree with Anurag:

when you set 11 as a q, then q+11 is in the set ===>( q+11)+11 is in as well and as a transitive property this is infinite.

11, 11+q, 11+2q,...... 11+nq when n is any positive integer, hence all of these elements are positive and are multiples of 11.

With 2),

11-11, leads to 0 and when you continue this. Numbers you get are negative and this only shows 0, 11, and negative multiples of 11.

Be careful and not select D!!

The answer is A only!

I hope I helped.