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hardnstrong
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Splendid...+1! Dint look at it that way. Thanks for the explanation.
OA Please! (Not that I doubt it is D)
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Bunuel
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The set S of numbers has the following properties:
I) If x is in S, then 1/x is in S.
II) If both x and y are in S, then so is x + y.

Is 3 in S?

(1) 1/3 is in S.
(2) 1 is in S.

OA will be posted later.

The wording is a little bit strange but anyway:

(1) 1/3 is in S --> according to (i) \(\frac{1}{\frac{1}{3}}=3\) must also be in S. Sufficient.

(1) 1 is in S --> according to (i) \(\frac{1}{1}=1\) (another 1) must also be in S, then according to (ii) \(1+1=2\) must also be in S --> \(1+2=3\) must also be in S. Sufficient.

Answer: D.


OA is D

But my point is (in stmt 2), if 1 is in set then according to (i) we have 1/1 in set, that mean another 1 . so x=1 and y =1
Now acc. to (ii) we have 1+1 =2 (x + y = z) in set
Now, can we again take the value of x or y as 2
How can x+y = x or y when neither x nor y is 0

Can we assume having more than one value of x or y?
or does it mean x and y are any two values in set? But question dosent say any two values, it just say x and y (two specific values)
Please explain?
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hardnstrong
Bunuel
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The set S of numbers has the following properties:
I) If x is in S, then 1/x is in S.
II) If both x and y are in S, then so is x + y.

Is 3 in S?

(1) 1/3 is in S.
(2) 1 is in S.

OA will be posted later.

The wording is a little bit strange but anyway:

(1) 1/3 is in S --> according to (i) \(\frac{1}{\frac{1}{3}}=3\) must also be in S. Sufficient.

(1) 1 is in S --> according to (i) \(\frac{1}{1}=1\) (another 1) must also be in S, then according to (ii) \(1+1=2\) must also be in S --> \(1+2=3\) must also be in S. Sufficient.

Answer: D.


OA is D

But my point is (in stmt 2), if 1 is in set then according to (i) we have 1/1 in set, that mean another 1 . so x=1 and y =1
Now acc. to (ii) we have 1+1 =2 (x + y = z) in set
Now, can we again take the value of x or y as 2
How can x+y = x or y when neither x nor y is 0

Can we assume having more than one value of x or y?
or does it mean x and y are any two values in set? But question dosent say any two values, it just say x and y (two specific values)
Please explain?

As I said the wording is quite strange, GMAT won't use such language. But as OA given as D, then the stem is meant to be understood as I did in my solution. One note though: x and y are variables not exact values.
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Thanks for your explanation
It was helpful as always :-D
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Technically in a set you can't have repeated values, which is why GMAT stats questions which permit repeated values always talk about "lists" of numbers, or "data sets" (which can have repeated values). So here, when you use Statement 2, you don't need to use the property that "If x is in S, then 1/x is in S" to prove that you have another '1' in the set -- you already know '1' is in the set, and there can only be one '1' in a set. The only question is, reading this, "If both x and y are in S, then so is x + y", whether x and y can represent the exact same value from the set. If so, then when 1 is in the set, so is 1+1 and so is thus 2+1, but if not, the set could just be {1}. Because of the word "both", I'd be inclined to think that x and y need to be different values, but there's no unambiguous answer to that question.

So really we're just left to guess what the question writer meant. Bunuel has certainly correctly interpreted the intentions of the question writer, but the wording of the question doesn't convey those intentions properly. There are official questions (rare ones) that test a similar concept that have none of the issues that this question has, and those might be worth looking at, but this question is not.
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Bunuel
hardnstrong
The set S of numbers has the following properties:
I) If x is in S, then 1/x is in S.
II) If both x and y are in S, then so is x + y.

Is 3 in S?

(1) 1/3 is in S.
(2) 1 is in S.

OA will be posted later.

The wording is a little bit strange but anyway:

(1) 1/3 is in S --> according to (i) \(\frac{1}{\frac{1}{3}}=3\) must also be in S. Sufficient.




(1) 1 is in S --> according to (i) \(\frac{1}{1}=1\) (another 1) must also be in S, then according to (ii) \(1+1=2\) must also be in S --> \(1+2=3\) must also be in S. Sufficient.

Answer: D.


Hi Bunuel,

If the answer is D then both the condition A & B should answer the question independently but you have use condition A & B together to answer the question.

Is it right ?



"(1) 1 is in S --> according to (i) \(\frac{1}{1}=1\) (another 1) must also be in S, then according to (ii) \(1+1=2\) must also be in S --> \(1+2=3\) must also be in S. Sufficient.
"
as per the condition B only one value is given So we can consider it either as X or as Y.
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Bunuel
hardnstrong
The set S of numbers has the following properties:
I) If x is in S, then 1/x is in S.
II) If both x and y are in S, then so is x + y.


Is 3 in S?

(1) 1/3 is in S.
(2) 1 is in S.

OA will be posted later.

The wording is a little bit strange but anyway:

(1) 1/3 is in S --> according to (i) \(\frac{1}{\frac{1}{3}}=3\) must also be in S. Sufficient.




(1) 1 is in S --> according to (i) \(\frac{1}{1}=1\) (another 1) must also be in S, then according to (ii) \(1+1=2\) must also be in S --> \(1+2=3\) must also be in S. Sufficient.

Answer: D.


Hi Bunuel,

If the answer is D then both the condition A & B should answer the question independently but you have use condition A & B together to answer the question.

Is it right ?



"(1) 1 is in S --> according to (i) \(\frac{1}{1}=1\) (another 1) must also be in S, then according to (ii) \(1+1=2\) must also be in S --> \(1+2=3\) must also be in S. Sufficient.
"
as per the condition B only one value is given So we can consider it either as X or as Y.

Highlighted text above is a part of the stem, NOT the statements. The statements are below, (1) and (2).
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This helped I had a similar doubt, thanks

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