Question:
K is a set of numbers such that
(i) if x is in K, then -x is in K, and
(ii) if each of x and y is in K, then xy is in K
Is 12 in K?
(1) 2 is in K
(3) 3 is in K
Couldn't the set therefore be:
-6 -3 -2 0 2 3 6 ?? Which shows 12 isn't in set and it's easy to find a set with 12 in it.
Why is the correct answer C and not E. The books explanation was at most vague.
Thanks!
Set K
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- knight247
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(1) Statement 1 says 2 is in k. Obviously implying that -2 is in k. So if 2 and -2 are in k then -4 also has to be in k(rememeber if x and y are in k then xy also is in k)
So, we have 2,-2 and -4. But this info is insufficient to prove 12 is in k.
(2)3 is in k. So -3 has to be in k. Also -9 has to be in k. So k has 3,-3 and -9. However, this is also insufficient to prove 12 is in k.
Combining both we know that k has 2,-2 -4 and 3,-3,-9. Consider the case of only -4 and -3. If x and y are in k then xy is in k. So if -4 and -3 are in k then (-4)*(-3)=12 will also be in k.
Hence C
So, we have 2,-2 and -4. But this info is insufficient to prove 12 is in k.
(2)3 is in k. So -3 has to be in k. Also -9 has to be in k. So k has 3,-3 and -9. However, this is also insufficient to prove 12 is in k.
Combining both we know that k has 2,-2 -4 and 3,-3,-9. Consider the case of only -4 and -3. If x and y are in k then xy is in k. So if -4 and -3 are in k then (-4)*(-3)=12 will also be in k.
Hence C
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- Brent@GMATPrepNow
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This set is incomplete.DevitaN wrote:
Couldn't the set therefore be:
-6 -3 -2 0 2 3 6 ?? Which shows 12 isn't in set and it's easy to find a set with 12 in it.
Rule ii says that, if each of x and y is in K, then xy is in K. So, since you have 2 and 6 in the set, it must also be true that (2)(6) is in the set. So, 12 is definitely in the set.
knight247's solution is great.
Cheers,
Brent
Thanksknight247 wrote:(1) Statement 1 says 2 is in k. Obviously implying that -2 is in k. So if 2 and -2 are in k then -4 also has to be in k(rememeber if x and y are in k then xy also is in k)
So, we have 2,-2 and -4. But this info is insufficient to prove 12 is in k.
(2)3 is in k. So -3 has to be in k. Also -9 has to be in k. So k has 3,-3 and -9. However, this is also insufficient to prove 12 is in k.
Combining both we know that k has 2,-2 -4 and 3,-3,-9. Consider the case of only -4 and -3. If x and y are in k then xy is in k. So if -4 and -3 are in k then (-4)*(-3)=12 will also be in k.
Hence C
Forgive my stupidity, but couldn't x or y be -2 or 2? So x*y = -4 if y=2 and x=-2 or x=-2 and y=2. So by saying 12 must be in the equation, isn't that like saying xy*(Y (3 or -3) or X (3 or -3)) so it would be x*y^2?
So i suppose basically my question is why must either x or y be -4. I feel like in the end 12 can be in the equation and doesn't have to be in the equation, thereby proving insufficient.
Sorry again for my thick headedness. I appreciate you taking the time to help me with this!
Okay, so basically it's saying if 2 numbers are in a set, than their products are in the set...it's not having you specify an x and a y and ensuring their products are in the set? It's simply referring to any 2 numbers in the set?Brent@GMATPrepNow wrote:This set is incomplete.DevitaN wrote:
Couldn't the set therefore be:
-6 -3 -2 0 2 3 6 ?? Which shows 12 isn't in set and it's easy to find a set with 12 in it.
Rule ii says that, if each of x and y is in K, then xy is in K. So, since you have 2 and 6 in the set, it must also be true that (2)(6) is in the set. So, 12 is definitely in the set.
knight247's solution is great.
Cheers,
Brent
If my statement is correct then I understand and thank you both!!
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- Brent@GMATPrepNow
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Edit: Sorry, it looks like I was still writing this when you made your last post (which is absolutely correct)DevitaN wrote:Forgive my stupidity, but couldn't x or y be -2 or 2? So x*y = -4 if y=2 and x=-2 or x=-2 and y=2. So by saying 12 must be in the equation, isn't that like saying xy*(Y (3 or -3) or X (3 or -3)) so it would be x*y^2?
So i suppose basically my question is why must either x or y be -4. I feel like in the end 12 can be in the equation and doesn't have to be in the equation, thereby proving insufficient.
Sorry again for my thick headedness. I appreciate you taking the time to help me with this!
There's no "thickheadedness" on your part. It's easy to confuse the roles that variables play in different settings.
The part where it says "if each of x and y is in K, then xy is in K" isn't assigning values to the variables x and y. It is using x and y to help phrase the rule.
This part could have been written without variables as well. If could have been written as "if there are two different numbers in set K, then it must also be true that the product of those two numbers is also in set K" This sentence is equivalent to the original sentence. It's just that the original version is more concise.
The point here is that we aren't saying that x and y are assigned certain values, we're just using the variables to better explain a rule.
Cheers,
Brent