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What is the average of the terms in set J?

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What is the average of the terms in set J? [#permalink]

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What is the average of the terms in set J?

(1) The sum of any three terms in set J is 21.
(2) Set J consists of 12 total terms.

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[Reveal] Spoiler: OA

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Re: What is the average of the terms in set J? [#permalink]

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New post 08 Oct 2015, 07:35
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1. Sa+Sb+Sc= 21
Average = 21/3 = 7
Since , the above is valid for any three terms in set J , the average is 7 for the entire set .
Each number in this set will be equal to 7 .
Sufficient

2. number of terms=12
Average= sum/ number of terms
Not sufficient

Answer A
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Re: What is the average of the terms in set J? [#permalink]

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New post 08 Oct 2015, 07:36
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Bunuel wrote:
What is the average of the terms in set J?

(1) The sum of any three terms in set J is 21.
(2) Set J consists of 12 total terms.

Kudos for a correct solution.


Target question: What is the average of the terms in set J?

Statement 1: The sum of any three terms in set J is 21.
This is a very powerful statement, because it tells us that all of the numbers in the set are equal.
Let's let a,b and c be three of the numbers in set J.
We know that a + b + c = 21
Notice that if I replace ANY of these three values (a,b or c) with d, the sum must still be 21.
This tells us that a, b and c must all equal d.
Using similar logic, I can show that ALL of the numbers in the set must equal d, which means all of the numbers in the set must be equal.
If all of the numbers are equal, then EVERY number must equal 7, which means the average of set J MUST equal 7
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Set J consists of 12 total terms
There are several possible scenarios that satisfy this statement. Here are two.
Case a: J = {2,2,2,2,2,2,2,2,2,2,2,2}, in which case the average of set J = 2
Case b: J = {1,1,1,1,1,1,1,1,1,1,1,1}, in which case the average of set J = 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

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Re: What is the average of the terms in set J? [#permalink]

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Bunuel wrote:
What is the average of the terms in set J?

(1) The sum of any three terms in set J is 21.
(2) Set J consists of 12 total terms.

Kudos for a correct solution.



Question : What is the average of the terms in set J?

Statement 1: The sum of any three terms in set J is 21.
this is possible only when all the terms in the set are 7 or when there are only three terms with sum of them as 21
in both cases the average of terms will be 7. hence
SUFFICIENT

Statement 2: Set J consists of 12 total terms
There is no information about sum of the terms. Hence,
NOT SUFFICIENT

Answer: option A
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Re: What is the average of the terms in set J? [#permalink]

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New post 08 Oct 2015, 20:57
Bunuel wrote:
What is the average of the terms in set J?

(1) The sum of any three terms in set J is 21.
(2) Set J consists of 12 total terms.

Kudos for a correct solution.


Statement 1:

suppose set has 4 terms a,b, c and d

a+b+c=21
b+c+d=21
a+b+d=21

above situation is only possible when all the numbers are same.
hence Mean=7
Sufficient

Sstatement 2: Doesn't speak about the numbers in set.

Insufficient.

Ans :A
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Re: What is the average of the terms in set J? [#permalink]

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New post 08 Oct 2015, 21:18
Average of the set J will be sum of all the digits by the number of digits.

1) The sum of any three terms in set J is 21....
Number of digits not specified...So the statement is insufficient.

(2) Set J consists of 12 total terms.
Sum of the digits not specified....So the statement is insufficient.

Considering both, we have 12 digits and we have the sum of the any 3 digits....So we cannot calculate the sum of all the 12 digits.

So the answer is 'E'.

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Re: What is the average of the terms in set J? [#permalink]

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Bunuel wrote:
What is the average of the terms in set J?

(1) The sum of any three terms in set J is 21.
(2) Set J consists of 12 total terms.

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Question Type: What Is the Value? This question asks for the average of the terms in set J.

Given information in the question stem or diagram: No important information is given in the question stem.

Statement 1: The sum of any three terms in the set is 21. This is a very difficult statement to consider. Clearly this WILL be true if all the terms are 7, but you need to make sure there are not any other possibilities. Will it work with any other sets, such as “6, 7, 8, 6, 7, 8”? Adding certain numbers in this set will give you 21, but not “any three terms.” The only ways to ensure that any three randomly chosen terms will sum to 21 are either to have only three terms (say, 6 + 7 + 8) or to have all terms in the set equal 7. In either case, the average must be 7, so this is sufficient information. The answer is A or D.

Statement 2: Set J consists of 12 terms. This statement is not sufficient on its own since it does not give any values for the terms in the set. However, as is the case with any clearly insufficient statement such as this, you must consider whether it is important with the other statement. This statement was clearly designed to make choice C a more attractive option, but you do not need to know anything about the number of terms for the first statement to be sufficient. The correct answer is A.
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Re: What is the average of the terms in set J? [#permalink]

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New post 08 Apr 2016, 11:50
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I often find set question difficult to contemplate. Here is how I reasoned to correct choice:

S1:
Of course, it was easy to think of a set such as {7,7,7,7,..}. What if there is another set which satisfies given condition? Not that easy! I moved onto next statement without rejecting this. Let's come back rather than wasting 30secs.

S2: Clearly insufficient.

S1/2: Now I go back to S1 and add this extra info from S2. Nah! Doesn't make any sense at all because I can have any of this set - {7,7,7} or {7,7,7,7,7,7}. At that point, I have that extra inkling about S1 which should somehow give us required result. Now is the time I spend extra few secs to disapprove extra sets. Even if you end up taking longer, a right guess would be A.

In right state of mind, this whole process will take much less time. My point is - Don't rush to answers like C or E in such questions but to make some sense out of combined statements. Bunnel's explanation rightly points out that subtle but important point.

Cheers!

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Re: What is the average of the terms in set J? [#permalink]

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New post 08 Jul 2017, 10:47
Bunuel wrote:
Bunuel wrote:
What is the average of the terms in set J?

(1) The sum of any three terms in set J is 21.
(2) Set J consists of 12 total terms.

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Question Type: What Is the Value? This question asks for the average of the terms in set J.

Given information in the question stem or diagram: No important information is given in the question stem.

Statement 1: The sum of any three terms in the set is 21. This is a very difficult statement to consider. Clearly this WILL be true if all the terms are 7, but you need to make sure there are not any other possibilities. Will it work with any other sets, such as “6, 7, 8, 6, 7, 8”? Adding certain numbers in this set will give you 21, but not “any three terms.” The only ways to ensure that any three randomly chosen terms will sum to 21 are either to have only three terms (say, 6 + 7 + 8) or to have all terms in the set equal 7. In either case, the average must be 7, so this is sufficient information. The answer is A or D.

Statement 2: Set J consists of 12 terms. This statement is not sufficient on its own since it does not give any values for the terms in the set. However, as is the case with any clearly insufficient statement such as this, you must consider whether it is important with the other statement. This statement was clearly designed to make choice C a more attractive option, but you do not need to know anything about the number of terms for the first statement to be sufficient. The correct answer is A.


Hi Bunuel,

Can we generalize statement 1 for any number of terms? i.e. "Sum of ANY X terms in set J is Y" --> all numbers in the set = X/Y?

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Re: What is the average of the terms in set J? [#permalink]

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New post 08 Jul 2017, 11:29
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bianalyst wrote:
Bunuel wrote:

Can we generalize statement 1 for any number of terms? i.e. "Sum of ANY X terms in set J is Y" --> all numbers in the set = X/Y?


Almost always, with the one exception I mention below, though I think you meant to write: "all numbers in the set are Y/X" (not X/Y).

The one exception is if your list contains exactly X terms. For example, if I tell you "the sum of any 2 terms in the list J is equal to 0", then we can make all kinds of lists that meet that condition which contain only 2 terms:

-10, 10
0, 0
π, -π
etc

but if the list is going to have more than 2 things in it, they would all need to be zeros.

And an unrelated thing - I object to the use of the word "set" in the original question. In math, a set cannot have repeated elements. There's a reason the real GMAT always talks about "data sets" or "lists" when it describes collections of values that can have repeated elements.
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Re: What is the average of the terms in set J? [#permalink]

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New post 21 Nov 2017, 11:26
Bunuel wrote:
What is the average of the terms in set J?

(1) The sum of any three terms in set J is 21.
(2) Set J consists of 12 total terms.

Kudos for a correct solution.


Average = \(\frac{Sum}{# of terms}\)

S1) Sum of any 3 terms is 21
=> All terms in Set J are 7
=> Average is always 7, regardless of the number of terms in the set.
Sufficient.

S2) No. of Terms = 12
Sum = ?
Insufficient.

A is the answer.
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Re: What is the average of the terms in set J?   [#permalink] 21 Nov 2017, 11:26
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