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If the average (arithmetic mean) of n consecutive odd intege

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If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

(1) The range of the n integers is 14.
(2) The greatest of the n integers is 17.

Data Sufficiency
Question: 66
Category: Arithmetic Statistics
Page: 157
Difficulty: 600


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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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SOLUTION

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

Odd consecutive integers is an evenly spaced set. For any evenly spaced set the mean equals to the average of the first and the last terms, so in our case \(mean=10=\frac{x_1+x_{n}}{2}\) --> \(x_1+x_{n}=20\). Question: \(x_1=?\)

(1) The range of the n integers is 14 --> the range of a set is the difference between the largest and smallest elements of a set, so \(x_{n}-x_1=14\) --> solving for \(x_1\) --> \(x_1=3\). Sufficient.

(2) The greatest of the n integers is 17 --> \(x_n=17\) --> \(x_1+17=20\) --> \(x_1=3\). Sufficient.

Answer: D.
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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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What we know:
If n consecutive integers have arithmetic mean of 10; those numbers are distributed either side of 10 perfectly symmetrically.


Statement (1)


Range is the difference between the greatest and the least numbers in the series.
If the range is known the least of these integers must be less than the mean, i.e. 10 - (14/2) = 3 (Note we don't really need to calculate this... to know we can is sufficient.)


Statement (2)


Similarly, the distance from the greatest number to mean equals to the distance from mean to the least number. So knowing the greatest number also sufficient to calculate the least number. (in this case 10 - (17 - 10) = 3)


The answer is D : Both Statement 1 and Statement 2 are sufficient ALONE .

Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

(1) The range of the n integers is 14.
(2) The greatest of the n integers is 17.

Data Sufficiency
Question: 66
Category: Arithmetic Statistics
Page: 157
Difficulty: 600


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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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Ans. D
Let the series start from x=>x,(x+2),(x+4),...(x+k) id the series.
Average=[x+(x+k)]/2=10(1)
x+(x+k)=20=>2x+k=20

Given in S1, x+k-x=14=>k=14.We can find out x,the smallest integer.Sufficient

From S2: x+k=17
By putting this value in (1),we can get x,the answer.
Sufficient.

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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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SOLUTION

If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

Odd consecutive integers is an evenly spaced set. For any evenly spaced set the mean equals to the average of the first and the last terms, so in our case \(mean=10=\frac{x_1+x_{n}}{2}\) --> \(x_1+x_{n}=20\). Question: \(x_1=?\)

(1) The range of the n integers is 14 --> the range of a set is the difference between the largest and smallest elements of a set, so \(x_{n}-x_1=14\) --> solving for \(x_1\) --> \(x_1=3\). Sufficient.

(2) The greatest of the n integers is 17 --> \(x_n=17\) --> \(x_1+17=20\) --> \(x_1=3\). Sufficient.

Answer: D.
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If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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New post 15 Aug 2014, 08:52
I did it this way.

Since consecutive odd integers are evenly spaced then avg = median = 10

Since median is even (10) for a set of consecutive odd integers, set must have even number of integers
Ex: if set has 3,5,7 then median 5 (odd)
if set has 3,5,7,9 then median = 5+7/2 = 6 (even)

so we can say (x+ (x+2))/2 = 10 => 2x+2=20 => x=9.
So middle terms of the set are 9 & 11. If I know one more thing like range or first/last term I can create full list.

Hence both statements are sufficient. Ans: D

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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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New post 13 Jun 2015, 05:47
(Min + Max)/2 = 10 --> Min + Max=20
1)Range means Max - Min=14, sow we have two equations and two unknowns --> can be solved Min = 3
2)Max=17, Plug it here Min+Max=20 --> Min=3
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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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New post 20 Dec 2016, 17:05
Great Official Question.
Here is what i did in this one =>


Given data =>
N consecutive odds
Mean =Median (For any evenly spaced set)=10
AS medan 10 => N must be even
Sum of deviations around the mean is always zero.
E.g => 9,11 OR 7,9,11,13

Hence we are adding elements in pair around the mean 10.
We need to get the least element now

Statement 1-->
Range =14
Series =>
3,5,7,9,11,13,15,17=> Least Integer will be 3
Hence Sufficient

Statement 2-->
Greasers integer =17

Number of terms tot he right of the median = Number of terms to the left of median
Median =10
11,13,15,17 => 4 terms to the right of it=> 4 terms must be there t the left of it.
Series => 3,5,7,9..
Hence least term =3
Hence Sufficient


Hence D

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Re: If the average (arithmetic mean) of n consecutive odd intege [#permalink]

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New post 23 Feb 2017, 04:53
PROMPT ANALYSIS
Let a, a+2, a+4, a+6…...n terms have average as 10
Sum of all values = na +n(n-1) = 10n or a +n-1 =10

SUPERSET
a can be any odd number.

TRANSLATION
In order to find a:
1# we need exact value of a
2# another equation with a and n as variable.

STATEMENT ANALYSIS

St 1: a +2n-2 - a =14 or n=8 therefore a =3 ANSWER. Option B, C, E eliminated.
St 2: a+2n-2 = 17.Solving with the equation in PROMPT, we get a =3 and n= 8. ANSWER. OptionA eliminated.

Option D.

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Re: If the average (arithmetic mean) of n consecutive odd intege   [#permalink] 23 Feb 2017, 04:53
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