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15 Jun 2020, 15:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
63% (00:31) correct
38%
(01:11)
wrong
based on 8
sessions
History
Date
Time
Result
Not Attempted Yet
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
I. 3
II. 12
III. 24
A. None
B. I only
C. II only
D. I and II only
E. II and III only
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15 Jun 2020, 15:56
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hashim88
The only numbers that are specifically stated to be members of the set are 2 and 6. Since the product of these two numbers is 2 × 6 = 12, the number 12 must also be in the set. Because 12 is in the set, eliminate any choice that does not include Roman Number II. Therefore, eliminate choices A and B. The two remaining possibilities are 3 and 24. Determine whether products of the three numbers now known to be in the set can be made to form 3 or 24. Since 2 × 12 = 24, 24 is in the set. Eliminate any choice that does not include Roman Number III, so eliminate choices C and D. Only one choice remains, so there is no need to go further. However, to understand why 3 does not need to be in the set, remember that if two numbers are in the set, their product is also in the set. The reverse is not true. If a number is in the set, there is no reason to assume that its factors are in the set. The only way to prove that 3 must be in the set would be to have 2 numbers whose product is 3 already in the set. Since this is not the case, 3 need not be in the set and the correct answer is choice E.
