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Re: For any 3 given numbers, which of the following is always equivalent [#permalink]
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Bunuel wrote:
For any 3 given numbers, which of the following is always equivalent to adding the 3 numbers together and then dividing the sum by 3?

I. Ordering the 3 numbers numerically, from highest to lowest, and then selecting the middle number.
II. Dividing each of the numbers by 3 and then adding the results together.
III. Multiplying each number by 6, adding the resulting products together, and then dividing the sum by 9.

A. I only
B. II only
C. I and II only
D. II and III only
E. None of the above


I was wondering by highest to lowest does it mean that they have to be consecutive numbers? e.g. 6, 5, 4 or 4, 3, 2 ?
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Re: For any 3 given numbers, which of the following is always equivalent [#permalink]
AtifS wrote:
Bunuel wrote:
For any 3 given numbers, which of the following is always equivalent to adding the 3 numbers together and then dividing the sum by 3?

I. Ordering the 3 numbers numerically, from highest to lowest, and then selecting the middle number.
II. Dividing each of the numbers by 3 and then adding the results together.
III. Multiplying each number by 6, adding the resulting products together, and then dividing the sum by 9.

A. I only
B. II only
C. I and II only
D. II and III only
E. None of the above


I was wondering by highest to lowest does it mean that they have to be consecutive numbers? e.g. 6, 5, 4 or 4, 3, 2 ?


Hi,

It does not mean that it has to be consecutive. It can be anything.
For example: Consider 1, 4 and 1 --> (1 + 4 + 1)/3 = 2 --> Sort the numbers. Middle number is 1 and its not equal to 2
Consider one more example in which all 3 numbers are equal (x, x, x). For this case middle number will always be equal to (x + x + x)/3
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Re: For any 3 given numbers, which of the following is always equivalent [#permalink]
Bunuel wrote:
For any 3 given numbers, which of the following is always equivalent to adding the 3 numbers together and then dividing the sum by 3?

I. Ordering the 3 numbers numerically, from highest to lowest, and then selecting the middle number.
II. Dividing each of the numbers by 3 and then adding the results together.
III. Multiplying each number by 6, adding the resulting products together, and then dividing the sum by 9.

A. I only
B. II only
C. I and II only
D. II and III only
E. None of the above


I. Ordering the 3 numbers numerically, from highest to lowest, and then selecting the middle number.
This will only be true if the numbers are consecutive integers.

II. Dividing each of the numbers by 3 and then adding the results together.
Correct.
a+b+c/3= a/3+b/3+c/3

III. Multiplying each number by 6, adding the resulting products together, and then dividing the sum by 9.
6a+6b+6c/9= 2a+2b+2c/3 and this is not equal to a+b+c/3

Only option 2 satisfies the condition.
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Re: For any 3 given numbers, which of the following is always equivalent [#permalink]
I did the mistake of taking consecutive numbers because of which option A was always correct.
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Re: For any 3 given numbers, which of the following is always equivalent [#permalink]
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