arithmetic

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arithmetic

by John fran kennedi » Tue Mar 15, 2016 10:44 am
156. Are all of the numbers in a certain list of 15 numbers equal?
(1) The sum of all the numbers in the list is 60
(2) The sum of any 3 numbers in the list is 12

of course the statement 1 is unsufficient, but I don't know why the statement 2 is sufficiet.
I have seen the explanation answer of my OG 2016, it is really confusing tho.
Could you please explain it to me?

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by Brent@GMATPrepNow » Tue Mar 15, 2016 11:06 am
Are all of the numbers in a certain list of 15 numbers equal?

(1) The sum of all the numbers in the list is 60.
(2) The sum of any 3 numbers in the list is 12.
Target question: Are all 15 numbers equal?

Statement 1: The sum of all the numbers in the list is 60.
There are several possible scenarios that satisfy this statement. Here are two.
Case a: numbers are: {4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4}, in which case all of the numbers are equal
Case b: numbers are: {4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 1, 7}, in which case all of the numbers are not equal
Statement 1 is NOT SUFFICIENT

Statement 2: The sum of any 3 numbers in the list is 12.
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,c and d be four of the 15 numbers in the set.
We know that a + b + c = 12
Notice that if I replace ANY of these three values (a,b or c) with d, the sum must still be 12.
This tells us that a, b and c must all equal d.
I can use a similar approach to show that e, f and g must also equal d.
In fact, 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.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

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Brent
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by John fran kennedi » Tue Mar 15, 2016 11:34 am
I thought that the statement 2 was insufficient because it could have been solved by plugging diverse numbers with the same result 12. for example (4,4,4), (1,2,9), ( 2,4,6) etc

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by Brent@GMATPrepNow » Tue Mar 15, 2016 12:41 pm
John fran kennedi wrote:I thought that the statement 2 was insufficient because it could have been solved by plugging diverse numbers with the same result 12. for example (4,4,4), (1,2,9), ( 2,4,6) etc
But the statement says: The sum of any 3 numbers in the list is 12.
Your examples look at very select groups of 3 numbers.

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by [email protected] » Tue Mar 15, 2016 3:18 pm
Hi John fran kennedi,

This DS question is really about considering the "possibilities" and making sure that you're thorough with your thinking.

We're told that there is a group of 15 numbers. We're asked if they're all equal. This is a YES/NO question.

Fact 1: The sum of the numbers is 60

IF.....
We have fifteen 4s, then the answer to the question is YES.

IF....
We have ANY OTHER option (e.g. fourteen 3s and one 18), then the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact 2: The sum of ANY 3 numbers in the list is 12.

With THIS information, we know that all the numbers MUST be 4s. Here's why:

With fifteen 4s, we know that selecting ANY 3 of them will give us a sum of 12. If we change EVEN 1 of those numbers to something else though, then there's no way to GUARANTEE that we get a total of 12 from ANY 3.

For example, if we have fourteen 4s and one 5. It's possible that we could pick 3 numbers and get 4+4+5 = 13, which is NOT a sum of 12. We're told that picking ANY 3 numbers gets us a sum of 12 though, so this serves as proof that no other option exists. Therefore, all fifteen numbers MUST be 4s and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer: B

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by John fran kennedi » Tue Mar 15, 2016 3:18 pm
You are absolutely right. That's why I went wrong. Thank you so much.

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by Matt@VeritasPrep » Thu Mar 17, 2016 10:49 pm
You can also use a little algebra to decode S2.

Suppose five of the numbers in our set are {a, b, c, d, e}. From S2, we know that

a + b + c = 12

and

a + b + d = 12

so c = d.

We also know that

a + c + d = 12

and

b + c + d = 12

so a = b, etc.

Pretty quickly it becomes clear that we can choose any two groups of three to show that any integers we like in our set are equal, forcing ALL the integers in our set to be equal! Algebra to the rescue!

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by ceilidh.erickson » Fri Mar 18, 2016 7:59 am
John fran kennedi wrote:I thought that the statement 2 was insufficient because it could have been solved by plugging diverse numbers with the same result 12. for example (4,4,4), (1,2,9), ( 2,4,6) etc
To add to what Brent said, it might be helpful just to remember that if you ever see a construction like "the sum of ANY three of the numbers," then it must be the case that all of the numbers are equal - that's the only way that you could get the same sum with any possible combination of terms.
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Harvard Graduate School of Education