OG 12 <number list>

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vaishalijain7: Master | Next Rank: 500 Posts; Posts: 102; Joined: Sun Mar 29, 2009 7:21 pm; Location: toronto; Thanked: 5 times

by vaishalijain7 » Mon Jun 29, 2009 5:08 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.

Please explain how could the answer be [spoiler]'B'[/spoiler]. I understand [spoiler]'A'[/spoiler] is not sufficient.

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Source: Beat The GMAT — Data Sufficiency | Mon Jun 29, 2009 5:08 pm

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Re: OG 12 <number list>

by Ian Stewart » Mon Jun 29, 2009 6:02 pm

vaishalijain7 wrote: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.

Please explain how could the answer be [spoiler]'B'[/spoiler]. I understand [spoiler]'A'[/spoiler] is not sufficient.

S1 is clearly insufficient.

S2 is sufficient. Take a smaller set as an example: {a, b, c, d}. Suppose the sum of any 3 numbers in this set is 12. Then,

a+b+c = 12

or c = 12 - a - b. But it must also be true that

a+b+d=12

or d = 12 - a - b. So c = d.

Similarly:

c + d + a = 12 and c + d + b = 12, so a = b;
a + d + b = 12 and a + d + c = 12, so b = c.

So we've shown that a = b = c = d. Of course you could continue with this logic for any set containing four or more elements.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Domnu: Master | Next Rank: 500 Posts; Posts: 385; Joined: Sun May 24, 2009 3:55 pm; Thanked: 11 times; GMAT Score:740

by Domnu » Mon Jun 29, 2009 6:03 pm

Okay, so A isn't sufficient. Now, here's why B is sufficient.

Let us go about constructing such a set. Now, let the first element of the set be x, and the second element be y (where x and y could or couldn't be equal). Now, all the other elements absolutely have to be 12 - x - y, since the sum of the first and second element with any of the other elements has to be 12. Now, by the same argument, the third, fourth, and fifth elements all have to sum to 12. Because of this,

3 * (12 - x - y) = 12, so 12 - x - y = 4, so x + y = 8.

So, the sum of the first two elements is 8. Now, the sum of the first three elements is 12, so the third element is thus 4. But there's nothing special about the third element... we can let x be the tenth element and y be the eleventh element and repeat the above. By this procedure, we get all the elements to equal 4.

Therefore, the answer has to be B.

Have you wondered how you could have found such a treasure? -T

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