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Are all of the numbers in a certain list of 15 numbers equal

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Are all of the numbers in a certain list of 15 numbers equal

by abcdefg » Wed Jul 15, 2009 10:27 pm
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.

OA is [spoiler] B. Basically the solution is that every number has to be 4 4 4 4 4 4 4 etc. But how about the sequence 12-0-0-12-0-0-12-0-0-12-0-0-12-0-0? Wouldn't this also work? That's why I choose E [/spoiler]
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by zenithexe » Wed Jul 15, 2009 10:32 pm
it says "any 3 numbers in the list"

so with your alternative, 12-0-0-12-0-0-12-0-0-12-0-0-12-0-0
12-12-12 would be a valid pick but it won't equal 12
Last edited by zenithexe on Wed Jul 15, 2009 10:32 pm, edited 1 time in total.
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by tohellandback » Wed Jul 15, 2009 10:32 pm
it has to be B
1) INSUFFICIENT-tells nothing except the sum and a lot of combinations can give you 60

2) let's just answer your doubt..
is the sum of 12,12,and 12 equal to 12?? or is the sum of 0,0 and 0 equal to 12? or 12,12,0 etc etc
No. thats what the 2nd statement says/ any three numbers
only option is 4,4,4,4,...
SUFFICIENT
The powers of two are bloody impolite!!
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by abcdefg » Thu Jul 16, 2009 7:24 am
ok i see I got it mixed up with any 3 consecutive numbers. damn tricky tricky. thanks!
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by GMATGuruNY » Thu Jul 15, 2010 4:01 pm
Hi,

Could someone please explain me the second statement? OG has a long explanation and it is really confusing.

Thanks for your help in advance!

Cheers,
Plug in real numbers in order to see the situation more clearly.

Here's statement 2:

The sum of any 3 numbers in the list is 12.

Let's start by seeing whether all the numbers in the list can be equal:

4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 works because no matter which 3 we choose, the sum will be 4 + 4 + 4 = 12.

Now let's see whether we can change one of the numbers:

4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 1? Doesn't work, because we could pick out 4 + 4 + 1 = 9, and the sum of any 3 numbers must be 12.
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2? Doesn't work, because we could pick out 4 + 4 + 2 = 10, and the sum of any 3 numbers must be 12.
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3? Doesn't work, because we could pick out 4 + 4 + 3 = 11, and the sum of any 3 numbers must be 12.

Do you see the situation? If we change any of the numbers, we can't satisfy statement 2. This means the only way to satisfy statement 2 is if each number is 4. So all the numbers must be equal. SUFFICIENT.

Hope this helps!
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by outreach » Thu Jul 15, 2010 9:53 pm
also 0-0-0 will be valid but it won't equal 12

B
zenithexe wrote:it says "any 3 numbers in the list"

so with your alternative, 12-0-0-12-0-0-12-0-0-12-0-0-12-0-0
12-12-12 would be a valid pick but it won't equal 12
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by vardhankirti » Tue Apr 28, 2015 1:44 am
Hi,

The numbers could be 8+2+2 or 4+5+3 etc.How can we be sure whether the numbers would be the same?

Please assist

Regards
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by DavidG@VeritasPrep » Tue Apr 28, 2015 4:11 am
Take another look at Mitch's explanation above. Remember that there are 15 numbers in this set, and it must be the case that any three of them will sum to 12. To simplify, imagine the examples you gave with a fourth element, '1.' (Any number will work to illustrate the point.)

8, 2, 2, 1
Now it's not the case that any three of the elements will sum to 12, because 2 + 2 + 1 = 5, so this set no longer satisfies the condition.

The same would be true for the set:
4, 5, 3, 1
5 + 3 + 1 = 9, so it's no longer true that any three elements sum to 12. The only way that can be the case is if every term in the set is a 4.

Make sense?
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