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07 Jan 2005, 16:41
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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:
65%
(hard)
Question Stats:
52% (01:37) correct
48%
(01:33)
wrong
based on 488
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Set T is an infinite sequence of positive integers. A "superset" is a sequence in which there is a finite number of multiples of three. Is T a superset?
(1) The first six integers in T are multiples of three.
(2) An infinite number of integers in T are multiples of four.
(1) The first six integers in T are multiples of three.
(2) An infinite number of integers in T are multiples of four.
Isn't the right answer A? If the first six integers in T are multiples of three, doesn't that make T a "superset?"
According to the PrincetonReview, where I swiped this question, the answer is E. Their reasoning:
thanks,
- kevin
According to the PrincetonReview, where I swiped this question, the answer is E. Their reasoning:
Quote:
thanks,
- kevin
08 Jan 2005, 12:31
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Quote:
If we have established that the first six integers of set T are multiples of three, then does it matter whether the other positive integers are? Are not the first six integers a finite sequence of multiples of three?
For example, if we have a set of numbers: 6, 7, 8 & 9. This set of positive integers contains multiples of three. Does that not make this set a "superset?" Or do all the numbers have to be multiples of three?
Taken from: https://www.gmatclub.com/phpbb/viewtopic ... t=superset
Quote:
Why is this insufficient? Because the other numbers in the infinite set may not be multiples of three?
Confused and bitter about standardized tests,
- kevin
08 Jan 2005, 16:32
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One thing to remember in DS questions in GMAT is that, you pick a choice if it gives you one and only one solution if there is two or more possibilities then you dont pick the choice.
A simple illustration:
What is the value of x?
1) x=2
2) x^2=4
Here the answer is A because in statement 2, x can either be +2 or -2.
That said, in this problem it says that T is a set with infinite infinite +ve integers. T is a superset if it has a finite number of multiples of 3.
S[1]: The first six integers in T are multiples of three
Let's look at two examples for T.
a) T = [6,9,12,15,24,27,31,40,49,...]
This set above has no pattern but has infinite positive integers. Let's assume however that after the 1st 6 numbers which are multiples of 3 there are no other multiples of 3 in set T. i.e, T has a finite number(=6) of multiples of 3. So, T is a superset.
b) T = [6,9,12,15,18,21,24,27,30,33,...infinity]
In this you can see that set T is a infinite set of multiples of 3. This set satisfies the S[1] in that the first 6 integers in T are multiples of 3. But this set has an infinite number of multiples of 3, so T cannot be a Superset.
Based on the two examples of T that satisfy S[1] you cannot state conclusively that T is a Superset or T is not a Superset. Hence it is Insufficient.
HTH.
A simple illustration:
What is the value of x?
1) x=2
2) x^2=4
Here the answer is A because in statement 2, x can either be +2 or -2.
That said, in this problem it says that T is a set with infinite infinite +ve integers. T is a superset if it has a finite number of multiples of 3.
S[1]: The first six integers in T are multiples of three
Let's look at two examples for T.
a) T = [6,9,12,15,24,27,31,40,49,...]
This set above has no pattern but has infinite positive integers. Let's assume however that after the 1st 6 numbers which are multiples of 3 there are no other multiples of 3 in set T. i.e, T has a finite number(=6) of multiples of 3. So, T is a superset.
b) T = [6,9,12,15,18,21,24,27,30,33,...infinity]
In this you can see that set T is a infinite set of multiples of 3. This set satisfies the S[1] in that the first 6 integers in T are multiples of 3. But this set has an infinite number of multiples of 3, so T cannot be a Superset.
Based on the two examples of T that satisfy S[1] you cannot state conclusively that T is a Superset or T is not a Superset. Hence it is Insufficient.
HTH.
