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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OG 12 <number list>
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vaishalijain7
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S1 is clearly insufficient.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.
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.
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Domnu
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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.
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.
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