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DS problem

by prernamalhotra » Wed May 21, 2014 7:08 am
Please help with the below problem:

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
2) 3 is in K

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Prerna

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by [email protected] » Wed May 21, 2014 7:25 am
prernamalhotra wrote:
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
2) 3 is in K
Target question: Is 12 in K?

Statement 1: 2 is in K
By rule i, -2 is also in set K
By rule ii, if 2 and -2 are in set K, then -4 is in set K
By rule i, 4 is also in set K
By rule ii, if 2 and -4 are in set K, then -8 is in set K
By rule ii, if 2 and 4 are in set K, then 8 is in set K
etc..
So, all we can say is that the following numbers MUST be in K: ...-16, -8, -4, -2, 2, 4, 8, 16...
Since we don't know what other numbers might be in K, 12 might be K or 12 might NOT be in K.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 3 is in K
By rule i, -3 is also in set K
By rule ii, if 3 and -3 are in set K, then -9 is in set K
By rule i, 9 is also in set K
By rule ii, if 3 and -9 are in set K, then -27 is in set K
By rule ii, if 3 and 9 are in set K, then 27 is in set K
etc..
So, all we can say is that the following numbers MUST be in K: ...-27, -9, -3, 3, 9, 27,...
Since we don't know what other numbers might be in K, 12 might be K or 12 might NOT be in K.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that 4 must be in set K
Statement 2 tells us that 3 must be in set K
By rule ii, if 4 and 3 are in set K, then 12 is in set K
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
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by sukriti2hats » Mon Jun 02, 2014 6:27 pm
[email protected] wrote:
prernamalhotra wrote:
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
2) 3 is in K
Target question: Is 12 in K?

Statement 1: 2 is in K
By rule i, -2 is also in set K
By rule ii, if 2 and -2 are in set K, then -4 is in set K
By rule i, 4 is also in set K
By rule ii, if 2 and -4 are in set K, then -8 is in set K
By rule ii, if 2 and 4 are in set K, then 8 is in set K
etc..
So, all we can say is that the following numbers MUST be in K: ...-16, -8, -4, -2, 2, 4, 8, 16...
Since we don't know what other numbers might be in K, 12 might be K or 12 might NOT be in K.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 3 is in K
By rule i, -3 is also in set K
By rule ii, if 3 and -3 are in set K, then -9 is in set K
By rule i, 9 is also in set K
By rule ii, if 3 and -9 are in set K, then -27 is in set K
By rule ii, if 3 and 9 are in set K, then 27 is in set K
etc..
So, all we can say is that the following numbers MUST be in K: ...-27, -9, -3, 3, 9, 27,...
Since we don't know what other numbers might be in K, 12 might be K or 12 might NOT be in K.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that 4 must be in set K
Statement 2 tells us that 3 must be in set K
By rule ii, if 4 and 3 are in set K, then 12 is in set K
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
Hi Brent

I did not understand that how can -4 be in K and not other multiples of 2 like 6 etc? Why only exponential powers of 2 and 3 are in K?

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Sukriti

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by [email protected] » Mon Jun 02, 2014 7:06 pm
sukriti2hats wrote:
Hi Brent

I did not understand that how can -4 be in K and not other multiples of 2 like 6 etc? Why only exponential powers of 2 and 3 are in K?

Regards
Sukriti
Rules i and ii ensure that only POWERS of 2 are in K (from statement 1)
As you can see from the progression that I demonstrated, there's no way for other multiples of 2 to sneak into K.
Once we know that 2 and -2 are in set K, we've exhausted the limitations of rule i.
From this point, rule ii tells us that -4 is in set K.
And then rule i ensures that 4 is in set K.
At this point, we know that 2, -2, 4 and -4 are in set K.
Notice that neither rule i nor rule ii will allows to take these four values and somehow ensure that 6 is in set K

Cheers,
Brent
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