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

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

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26 Feb 2014, 01:24
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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).

Data Sufficiency
Question: 112
Category: Arithmetic Statistics
Page: 161
Difficulty: 650

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

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26 Feb 2014, 01:24
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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.

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  [#permalink]

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26 Feb 2014, 03:22
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2
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  [#permalink]

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26 Feb 2014, 15:22
I am not clear with the solution...

From Stm 1- if we have , 3, 5, 7,9 , then mean is not equal to median..

but if wehave 3,5,7,9,11- then mean - median..

then how stm1 is suff??
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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26 Feb 2014, 15:32
Mountain14 wrote:
I am not clear with the solution...

From Stm 1- if we have , 3, 5, 7,9 , then mean is not equal to median..

but if wehave 3,5,7,9,11- then mean - median..

then how stm1 is suff??

Hi Mountain14,

when the numbers are 3, 5, 7, 9 mean = (9+3)/2 = 6 and median = (5+7)/2 = 6 so both are equal...

Hope this clarifies...

However, can someone let me know the concept of range??? I did not understand the 2nd statement so unable to answer the question....
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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27 Feb 2014, 05:22
gayam wrote:
Mountain14 wrote:
I am not clear with the solution...

From Stm 1- if we have , 3, 5, 7,9 , then mean is not equal to median..

but if wehave 3,5,7,9,11- then mean - median..

then how stm1 is suff??

Hi Mountain14,

when the numbers are 3, 5, 7, 9 mean = (9+3)/2 = 6 and median = (5+7)/2 = 6 so both are equal...

Hope this clarifies...

However, can someone let me know the concept of range??? I did not understand the 2nd statement so unable to answer the question....

Solutions will be published on weekend.
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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27 Feb 2014, 06:15
1
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

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

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27 Feb 2014, 22:15
gayam wrote:
Mountain14 wrote:
I am not clear with the solution...

From Stm 1- if we have , 3, 5, 7,9 , then mean is not equal to median..

but if wehave 3,5,7,9,11- then mean - median..

then how stm1 is suff??

Hi Mountain14,

when the numbers are 3, 5, 7, 9 mean = (9+3)/2 = 6 and median = (5+7)/2 = 6 so both are equal...

Hope this clarifies...

However, can someone let me know the concept of range??? I did not understand the 2nd statement so unable to answer the question....

Range is simple - all it means is the difference between the highest and the lowest number in a set(arranged in assending order)

For Eg - 1,3,5,7,9,11
Range = 11-1 = 10. So we say the range for this set is 10

Eg 2 - 10,20,30,40,50
Range = 50-10 = 40. Range for the above set is 40.

Thats all . Hope u got it.
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Posts: 52294
Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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01 Mar 2014, 04:59
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.

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  [#permalink]

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18 May 2014, 09:06
1

if n=3, then numbers are 2,4,6....mean=median
if n=4, then numbers are 2,4,6,8....mean equals median

insuff

2
number of elements in the set not clear
insuff

combined 1 and 2
suff
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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19 May 2014, 01:45
gmatkum wrote:
1

if n=3, then numbers are 2,4,6....mean=median
if n=4, then numbers are 2,4,6,8....mean equals median

insuff

2
number of elements in the set not clear
insuff

combined 1 and 2
suff

Please note that the correct answer is A, not C. Check here: for-a-certain-set-of-n-numbers-where-n-1-is-the-average-167948.html#p1336840
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Re: DS - Word Translation Problem  [#permalink]

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28 Jan 2015, 21:09
keshav11 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 the n numbers in the set is 2( n -1).

I do have the answer for this question , but am not able to understand. If anybody could help me understand this problem in a different/simpler way. Thanks in advance

I will go with A.
1) If the numbers are listed in increasing order and the difference in any 2 consecutive numbers is 2, then it is an increasing set of consecutive odd or even numbers. So the mean and median should be same. For e.g. (2,4,6) or (2,4,6,8). Same mean and median. SUFFICIENT.
2) This has nothing to do with mean or median. For e.g for n=3, range = 4 and this is true for both sets (1, 1, 5) and (2, 4, 6) but the statement is not same in both cases. INSUFFICIENT.

Let me know if I am right and give kudos!
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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29 Mar 2015, 05:47
keshav11 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 the n numbers in the set is 2( n -1).

I do have the answer for this question , but am not able to understand. If anybody could help me understand this problem in a different/simpler way. Thanks in advance

Just remember this rule:

If a set of numbers is evenly spaced (consecutive, odds, multiples of 5, etc.), the median is equal to the mean.

I'd say that's as simple as we can get
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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30 Mar 2015, 02:36
1) any successive pair's diff = 2 => all elements are in AP with d = 2.
Thus mean = median (prop of AP) SUFF
2) Range is given, but no info about the elements is known. Thus Not SUFF.
Ans A.
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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13 May 2016, 04:29
1
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  [#permalink]

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28 Jan 2017, 13:56
1
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  [#permalink]

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28 Dec 2017, 09:55
Bunuel wrote:
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.

P.S. The range is the difference between the smallest and largest elements of the set.

Let's take a set {2,4,6,8,10}, then we apply Range=2(n-1) then 2(5-1)=8 [10-2]. The formula for range applies to all consecutive even integers. Why is it not sufficient?
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Re: For a certain set of n numbers, where n > 1, is the average  [#permalink]

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13 Jan 2019, 06:54
Hi Bunuel,

The question asks is mean equal to median. In other words, basically it asks is this set evenly spaced or arithmetic progression?

I think we have to check the stmnt 2 with a couple of sets. Only then we can conclude that is it insufficient. For example, the range 2(n-1) is true for all consecutive odd and consecutive even sets. However, it doesn't work with consecutive integers. That's why it is insufficient. Am I right?

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Re: For a certain set of n numbers, where n > 1, is the average &nbs [#permalink] 13 Jan 2019, 06:54
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