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Bunuel
If S is a set of odd integers and 3 and –1 are in S, is –15 in S ?

(1) 5 is in S.
(2) Whenever two numbers are in S, their product is in S.


Given: S is a set of odd integers and 3 and –1 are in S

Target question: Is –15 in S ?

Statement 1: 5 is in S
So far, set S looks like this: {-1, 3, 5, . . . .}
So, -15 may or may not be in set S
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Whenever two numbers are in S, their product is in S.
So far, set S looks like this: {-1, 3, -3, -9, 9, 27, -27, -81, 81, . . . . , }
So, -15 may or may not be in set S
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
If we have the set {-1, 3, 5, . . . .} AND we know that, whenever two numbers are in S, their product is in S, then we can see that 15 (the product of 3 and 5) is also in set S.
If 15 is in set S, then we can see that -15 (the product of -1 and 15) is also in set S.
The answer to the target question is YES, -15 IS in set S
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Hi All,

We're told that Set S is a set of ODD INTEGERS and 3 and -1 are in S. We're asked if -15 in S. This question is more about logic and basic Arithmetic than anything else, so taking the proper notes - and thinking about the possibilities - is all that's really needed to beat it.

(1) 5 is in S.

Based on the information in Fact 1, we now know that at least 3 numbers are in Set S: -1, +3 and +5. This is clearly not enough information to determine whether -15 is also in the Set or not.
Fact 1 is INSUFFICIENT

(2) Whenever two numbers are in S, their product is in S.

With the information in Fact 2, we can determine some of the additional numbers in Set S. With -1 and +3, we know that (-1)(+3) = -3 is also in the Set. With -3 included, we also know that (-3)(+3) = -9 is in the set. With that integer we also have (-1)(-9) = +9 as well as -27 and +27. You might recognize that we'll end up with a bunch of numbers that are 'powers of 3' and their negative equivalents. This still doesn't tell us whether -15 is in the Set or not.
Fact 2 is INSUFFICIENT

Combined, we know...
(1) 5 is in S.
(2) Whenever two numbers are in S, their product is in S.

From Fact 2, we know that -3 is in the Set, so since +5 is also in the set, we know for sure that (-3)(+5) = 15 will be in the Set (and the answer to the question is ALWAYS YES).
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Bunuel
If S is a set of odd integers and 3 and –1 are in S, is –15 in S ?

(1) 5 is in S.
(2) Whenever two numbers are in S, their product is in S.



DS10602.01
OG2020 NEW QUESTION

We are given that S is a set of odd integers, and 3 and –1 are in S. We need to determine whether –15 is in S.

Statement One Alone:

5 is in S.

Although, we know 3, -1, and 5 are in S, we do not know whether -15 is in S.

Statement one alone is not sufficient to answer the question.

Statement Two Alone:

Whenever two numbers are in S, their product is in S.

We see that 3 x -1 = -3 is in S, and therefore, -3 x 3 = -9 is in S, and so on. However, we still do not have enough information to determine whether -15 is in S. Statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Since 3, -1, and 5 are in S, and since whenever two numbers are in S, their product is in S, we see that:

3 x -1 = -3 and -3 x 5 = -15, so we know that -15 IS indeed in set S.

Answer: C
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Solution



Steps 1 & 2: Understand Question and Draw Inferences

In this question, we are given
    • S is a set of odd integers
    • The numbers 3 and -1 are in S

We need to determine
    • Whether the number -15 is in S or not.

As we do not have any other information about the elements present in S, let us now analyse the individual statements.

Step 3: Analyse Statement 1

As per the information given in statement 1, the number 5 is in the set S.
    • However, from this statement we cannot say whether the number -15 is present in S or not.
Hence, statement 1 is not sufficient to answer the question.

Step 4: Analyse Statement 2


As per the information given in statement 2, whenever two numbers are present in S, their product is also present in S.
We already know that 3 and -1 are present in S.

    • Therefore, we can say that (3 × -1) or -3 is also present in S.
    • However, we can’t say whether -15 is present in S or not.
      o As to get -15, we must have a 5 present in S.
      o But we don’t have sufficient information about the presence of 5.

Hence, statement 2 is not sufficient to answer the question.

Step 5: Combine Both Statements Together (If Needed)


From statements 1 and 2, we can say
    • 3, -1, -3 and 5 are present in set S.
    • Also, for any two elements present in S, their product is also present in S.
    • Therefore, we can say (-3 × 5) or -15 is also present in S.

As we can determine that -15 is present in S by combining both statements, the correct answer is option C.

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Bunuel
If S is a set of odd integers and 3 and –1 are in S, is –15 in S ?

(1) 5 is in S.
(2) Whenever two numbers are in S, their product is in S.


Given: S is a set of odd integers and 3 and –1 are in S

Target question: Is –15 in S ?

Statement 1: 5 is in S
So far, set S looks like this: {-1, 3, 5, . . . .}
So, -15 may or may not be in set S
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Whenever two numbers are in S, their product is in S.
So far, set S looks like this: {-1, 3, -3, -9, 9, 27, -27, -81, 81, . . . . , }
So, -15 may or may not be in set S
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
If we have the set {-1, 3, 5, . . . .} AND we know that, whenever two numbers are in S, their product is in S, then we can see that 15 (the product of 3 and 5) is also in set S.
If 15 is in set S, then we can see that -15 (the product of -1 and 15) is also in set S.
The answer to the target question is YES, -15 IS in set S
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

Dear Brent/Rich/Scott

I do have a query regarding statement 2.

As -15 can not be derived with condition of statement 2 (as mentioned in above set numbers), why can't we consider answer as B.

Thanks,
Raxit.
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Hi Raxit,

The initial prompt tells us two of the numbers are in Set S (re: +3 and -1), but that does NOT mean that those are the ONLY numbers in Set S. There might be other numbers - but we do not know for sure (and if there actually are other numbers in Set S, then we do not know what those numbers are). We have to consider those possibilities when dealing with Fact 2....

IF.... -5 is also in Set S, then the information in Fact 2 means that -15 would also be in Set S --> and the answer to the question would be YES.
IF.... -5 is NOT in Set S and -15 isn't already in Set S, then the answer to the question would be NO.

Thus, the answer to the question is inconsistent and Fact 2 is INSUFFICIENT.

GMAT assassins aren't born, they're made,
Rich
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Raxit85

Dear Brent/Rich/Scott

I do have a query regarding statement 2.

As -15 can not be derived with condition of statement 2 (as mentioned in above set numbers), why can't we consider answer as B.

Thanks,
Raxit.

The reason B is not the correct answer is because even when we assume statement II, -15 may or may not be in S. I am assuming we have established that -15 may not be in S; but it is also possible that -15 is actually in the set S. The questions stem says -1 and 3 are in set S and statement II says whenever two numbers are in S, so is their product; but we have no information from which we can deduce -15 is not in the set. A number of people wrote the numbers such as -3, -9, 27 etc. which are definitely in set S; but it doesn't mean that set S does not contain any other elements besides the above. That's why we can't know for sure -15 is in (or is not in) set S by assuming only statement II.
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chetan2u Bunuel VeritasKarishma GMATBusters nick1816

Hi experts -- i was about to select B because i thought I could answer as a definitive "No" [its a Yes/No question]

But then i realized, perhaps i don't have the entire set of values in set S, just from S2

Question : if S1 was instead, lets say 9 (instead of the given 5) , keeping S2 the same

Between C and E -- what would be the answer choice in this scenario ?

In that case, would E be the answer choice because we don't know per S1 or S2 if we know all the values of set S ?

Thank you !
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chetan2u Bunuel VeritasKarishma GMATBusters nick1816

Hi experts -- i was about to select B because i thought I could answer as a definitive "No" [its a Yes/No question]

But then i realized, perhaps i don't have the entire set of values in set S, just from S2

Question : if S1 was instead, lets say 9 (instead of the given 5) , keeping S2 the same

Between C and E -- what would be the answer choice in this scenario ?

In that case, would E be the answer choice because we don't know per S1 or S2 if we know all the values of set S ?

Thank you !

The question only says that "3 and - 1 are in S". It does not say what other numbers there may be in S.

Again statement I says "5 is in S". This does not mean other numbers are not in S. Even if you replace 5 by 9, you still don't know which other numbers are in S.

If we know that 5, 3 and -1 are in S, then we know that 15, -3, -5, -15, 45 ... etc are in S. So -15 is in S.
If we know that 9, 3 and -1 are in S, we don't know whether -15 is in S or not. You can get a definitive "Yes" as above but not a definitive "No".

So if you replace 5 by 9, you get (E) as your answer.
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EMPOWERgmatRichC
Hi All,

We're told that Set S is a set of ODD INTEGERS and 3 and -1 are in S. We're asked if -15 in S. This question is more about logic and basic Arithmetic than anything else, so taking the proper notes - and thinking about the possibilities - is all that's really needed to beat it.

(1) 5 is in S.

Based on the information in Fact 1, we now know that at least 3 numbers are in Set S: -1, +3 and +5. This is clearly not enough information to determine whether -15 is also in the Set or not.
Fact 1 is INSUFFICIENT

(2) Whenever two numbers are in S, their product is in S.

With the information in Fact 2, we can determine some of the additional numbers in Set S. With -1 and +3, we know that (-1)(+3) = -3 is also in the Set. With -3 included, we also know that (-3)(+3) = -9 is in the set. With that integer we also have (-1)(-9) = +9 as well as -27 and +27. You might recognize that we'll end up with a bunch of numbers that are 'powers of 3' and their negative equivalents. This still doesn't tell us whether -15 is in the Set or not.
Fact 2 is INSUFFICIENT

Combined, we know...
(1) 5 is in S.
(2) Whenever two numbers are in S, their product is in S.

From Fact 2, we know that -3 is in the Set, so since +5 is also in the set, we know for sure that (-3)(+5) = 15 will be in the Set (and the answer to the question is ALWAYS YES).
Combined, SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

There is one question in one of the quant official guides which has "t + 3 is in the set" as one of the statements (I can't find this question now). The question asked if one of the negative numbers is also part of the set. This negative number satisfied the statement, however the explanation said that it isn't correct because it mentions "t+3" and not "t-3". In this case only the product is mentioned, and not the division (??). So, how do we decide if these conditions are for infinite sets or not? (the OG question did not restrict the number of elements in the set to a finite number)
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Hi NKJJ1990,

Based on your general description, is this the prompt that you are referring to?

https://gmatclub.com/forum/a-set-of-num ... 68728.html

In both that prompt and this DS prompt, the same 'rule' applies. We can only go "forwards" with the information that we are given. In that PS question, we're told that when a number "T" is in the set, then the number "T+2" is ALSO in the set. We can only go 'forwards' with this because the description ONLY applies to adding 2 to an existing number that's in the set.

Thus, if we have -1 in the set, then we can add an infinite number of +2s to find other values in the set - specifically 1, 3, 5, 7, 9, 11.... because those numbers fit what we're told. We CANNOT 'go backwards' though because the question does not tell us to do so.

This is a rare concept on Test Day (you likely will NOT see it on your Official GMAT), but if you find the concept a little confusing, then you might think in terms of what MUST be in the Set vs. what MIGHT be in the Set. The ONLY numbers that we know for sure are in the Set are the ones that fit the limited description that we're given: "T+2"... meaning that we can ONLY ADD to find other numbers in the Set. There might be other values (that don't fit that given rule), but we cannot say for sure whether those numbers are in the Set or not.

GMAT assassins aren't born, they're made,
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Bunuel
If S is a set of odd integers and 3 and –1 are in S, is –15 in S ?

(1) 5 is in S.
(2) Whenever two numbers are in S, their product is in S.


Answer: Option C

Video solution by GMATinsight

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avigutman. Sir isn't B sufficient?
Coz no matter how we multiply 3 and -1, -15 is not gonna be in the set,
coz there needs to be a 5 or a -5 in the set for us to get -15.
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Arsalan24
avigutman. Sir isn't B sufficient?
Coz no matter how we multiply 3 and -1, -15 is not gonna be in the set,
coz there needs to be a 5 or a -5 in the set for us to get -15.
If you interpreted statement (2) to mean that ONLY 3 and -1 are in the set, Arsalan24, that would contradict statement (1). Since the statements are always truthful, that contradiction is a way for you to realize that you misinterpreted.
With statement (2) on its own, we don’t know whether 5 is in the set. It may or may not be a member of the set, so we are unable to answer the question.

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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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