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For a certain set of n numbers, where n > 1, is the average (arithmeti

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For a certain set of n numbers, where n > 1, is the average (arithmeti  [#permalink]

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New post 26 Feb 2014, 02:24
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For a certain set of n numbers, where n > 1, is the average (arithmetic mean) equal to the median?

(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.
(2) The range of the n numbers in the set is 2(n- 1).


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Re: For a certain set of n numbers, where n > 1, is the average (arithmeti  [#permalink]

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New post 26 Feb 2014, 02:24
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SOLUTION

For a certain set of n numbers, where n > 1, is the average (arithmetic mean) equal to the median?

(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2. This implies that the set is even;y spaced. In any evenly spaced set the mean (average) is equal to the median. Sufficient.

(2) The range of the n numbers in the set is 2(n- 1). This is completely useless. Not sufficient.

Answer: A.

P.S. The range is the difference between the smallest and largest elements of the set.
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Re: For a certain set of n numbers, where n > 1, is the average (arithmeti  [#permalink]

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New post 26 Feb 2014, 04:22
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For a certain set of n numbers, where n > 1, is the average (arithmetic mean) equal to the median?

(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.
(2) The range of the n numbers in the set is 2(n- 1).

Sol:

St 1: The above statement basically means that the terms in the set are in Arithmetic Progression and thus Mean=Median.

Note that if it is given that Mean=Median then it does not mean that the set is in order

St 2: The range of n numbers is 2(n-1)

Consider n=3

Then trange =4

Now if Set consists of elements 1,3 and 5 then Mean=Median
but set consists of 1,4,5 then Mean is not equal to Median

Ans is A

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Re: For a certain set of n numbers, where n > 1, is the average (arithmeti  [#permalink]

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New post 27 Feb 2014, 07:15
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A) x-4,x-2,x,x+2,x+4 (condition is successive difference is 2)
average = (x-4+x+4)/2 = 2x/2 = x
median = x

x-4,x-2,x,x+2,x+4,x+6 (condition is successive difference is 2)
average = (x-4+x+6)/2 = (2x+2)/2 = 2(x+1)/2 = x+1
median = x+1

SUFFICENT

B) What is a range? largest number - smallest number. It doesn't tell us much about the mean or median.
You can try some numbers here.
Constraint here is n > 1 which means you can take only 2 or 3 nos and find their average and that should be sufficient to answer the question.

INSUFFICIENT
Answer A

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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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New post 15 May 2014, 11:50
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ugimba wrote:
For a certain set of n numbers, where n > 1, is the average (arithmetic mean) equal to the median?

(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.
(2) The range of n numbers in the set is 2(n - 1).


I thought that 2(n-1) was the range for both n consecutive even/odd integers
Therefore, we still have an evenly spaced set
Thus, 2 was sufficient

Could someone please advice where am I going wrong here?

Thanks!
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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New post 16 May 2014, 02:26
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jlgdr wrote:
ugimba wrote:
For a certain set of n numbers, where n > 1, is the average (arithmetic mean) equal to the median?

(1) If the n numbers in the set are listed in increasing order, then the difference between any pair of successive numbers in the set is 2.
(2) The range of n numbers in the set is 2(n - 1).


I thought that 2(n-1) was the range for both n consecutive even/odd integers
Therefore, we still have an evenly spaced set
Thus, 2 was sufficient

Could someone please advice where am I going wrong here?

Thanks!
Cheers
J :)


But the reverse is not necessarily true: not all n-element sets which have the range equal to 2(n - 1) are consecutive odd or consecutive even integers.

For example, {2, 3, 6} has the range equal to 2(n - 1) but this set is not of that type.
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Re: For a certain set of n numbers, where n > 1, is the average (arithmeti  [#permalink]

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New post 13 May 2016, 05:29
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If numbers are listed in increasing order (consecutive even numbers, consecutive odd numbers and consecutive multiples), median=mean. Hence, since the condition 1) is always yes, the condition is sufficient. The correct answer choice is A.


l For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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Re: For a certain set of n numbers, where n > 1, is the average (arithmeti  [#permalink]

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New post 28 Jan 2017, 14:56
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A neat little trick to remember is that for any series that is an Arithmetic progression, namely difference between each successive term is constant, the median is always = mean. I'll try to prove it below. Let us say there are n terms. There are two pssibillities n is odd or n is even. Let us say the constant difference is d (2 in the case of this problem).

1st term: a, 2nd term = a+d, 3rd term = a+2*d .... nth term = a+(n-1)*d
Adding all you get Sum = a*n +[ (1+2+3...+(n-1))]*d = a*n +[ n*(n-1)/2]*d ( another interesting result sum of the first n-1 integers is n*(n-1)/2]. Hence the Average = Sum/n = a+(n-1)/2*d

n is odd: Median = (n+1)/2 th term = a+(n-1)/2*d = Average so if n is odd we have proved avg always equal to mean.
n is even: Median = average of n/2 and (n/2+1)th term =[ a+(n/2-1)*d+a+(n/2)*d ]/2 = a+(n-1)*d/2 = Average so if n is even too we have proved avg always equal to mean. Thus (1) is always sufficient to answer such questions of if avg = median.

(2) Since only range is given we cant determine anything about the numbers in between so this informatino is insufficient. Hence answer is A.
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Re: For a certain set of n numbers, where n > 1, is the average (arithmeti  [#permalink]

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