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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 integers is 10, [#permalink]
Bunuel wrote:
tonebeeze wrote:
Hello All,

I got this problem correct. I just would like to see a technical explanation of how to arrive at both occasions of sufficiency.

Thanks!

"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"


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.



Bunuel ,

My question is from the problem statement itself only one solution is possible. [ 3,5,7,9,11,13,15,17]. Are there any chances to encounter such a question on actual exam
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
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rt1601 wrote:
Bunuel wrote:
tonebeeze wrote:
Hello All,

I got this problem correct. I just would like to see a technical explanation of how to arrive at both occasions of sufficiency.

Thanks!

"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"


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.



Bunuel ,

My question is from the problem statement itself only one solution is possible. [ 3,5,7,9,11,13,15,17]. Are there any chances to encounter such a question on actual exam


Unfiniftley many sets are possible:
{9, 11}
{7, 9, 11, 13}
{5, 7, 9, 11, 13, 15}
...
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
Bunuel wrote:
tonebeeze wrote:
Hello All,

I got this problem correct. I just would like to see a technical explanation of how to arrive at both occasions of sufficiency.

Thanks!

"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"


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.


Hi Bunuel

Could you please explain what evenly spaced set means?
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
Expert Reply
zanaik89 wrote:
Bunuel wrote:
tonebeeze wrote:
Hello All,

I got this problem correct. I just would like to see a technical explanation of how to arrive at both occasions of sufficiency.

Thanks!

"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"


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.


Hi Bunuel

Could you please explain what evenly spaced set means?


Evenly spaced set (aka arithmetic progression) is a special type of sequence in which the difference between successive terms is constant. Fore example, 1, 4, 7, 10, ... is an evenly spaced set. Check for more here: https://gmatclub.com/forum/math-sequenc ... 01891.html

Hope it helps.
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
Expert Reply
ArupRS wrote:
Bunuel wrote:
ArupRS wrote:
What if we consider negative odd integers?
-1,1,3,5,7,9,11,13,15,17,19,21 -- here also our avg is 10. In this way we can include 289 for 17^2.

chetan2u, VeritasKarishma, Bunuel, gmatbusters : Please provide your suggestion.


Sorry but your question is not clear at all. Also, your set does not satisfy any of the statements given:

(1) The range of the n integers is 14. The range of your set is 21 - (-1) = 22, NOT 14

(2) The greatest of the n integers is 17. Te greatest integer in your set is 21, not 17.



Hello Bunuel

Statement 2 says The greatest integer is 17 or \(17^n\) ?

Quote:
The greatest of the n integers is 17"


(2) The greatest of the n integers is 17
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
Whoa, what has happened to me. I read 14 as 4 unfortunately
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
Can someone please explain me why we haven't taken negative odd integers into consideration ?
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
Bunuel wrote:
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.


Bunuel why did you assume from the beginning that there are two odd consecutive integers ? and not more
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
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dave13 wrote:
Bunuel wrote:
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.


Bunuel why did you assume from the beginning that there are two odd consecutive integers ? and not more


I think you missed the highlighted part. x1 is the first term of the sequence and xn is the nth term of the sequence.
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
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For any evenly spaced set:
average = median

tonebeeze wrote:
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



Since the average of the consecutive odd integers = 10, the median of the set must also be 10.

Statement 1:
For a range of 14 to be yielded:
The smallest integer in the set must be 7 places to the left of 10 --> 10-7 = 3
The greatest integer in the set must be 7 places to the right of 10 --> 10+7 = 17
Thus, the smallest integer = 3
SUFFICIENT.

Statement 2:
Since the greatest integer is 7 places to the right of 10, the smallest integer must be 7 places to the left of 10:
10-7 = 3
SUFFICIENT.

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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
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If the average (arithmetic mean) of n consecutive odd integers is 10, what is the least of the integers?

The average of an evenly spaced set must be the median. Since 10 is even, we must have an even number of integers in the set.

9 + 11 = average 10
3 + 5 + 7 + 9 + 11 + 13 + 15 + 17= average 10

To determine the least of the integers, we need to know the number of integers, the greatest number in the set, or the range.

(1) The range of the n integers is 14

SUFFICIENT.

(2) The greatest of the n integers is 17

SUFFICIENT.

Answer is D.
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
If the average is 10, then 2 middle numbers are 9 and 11
(1) Range of the sequence is 14. The smallest and the largest number of the sequence is symmetric though 10.
Largest - Smallest = 14 --> (10+a) - (10-a) = 14 --> a = 7
--> Smallest number is 3, largest is 17, then we can count how many number
--> Suff

(2) We know the largest number --> Suff (similar to (1))
--> Answer D
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
Hi KarishmaB

With only D in our hand we can also assume the following; -1,1,3,5,7,9,11,13,15,17
why didn't we consider negative numbers and instead cut it at 3?
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
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gloomybison wrote:
Hi KarishmaB

With only D in our hand we can also assume the following; -1,1,3,5,7,9,11,13,15,17
why didn't we consider negative numbers and instead cut it at 3?


Because the average of this list is 8, not 10.

The average will be the middle term. So here average will be middle of 7 and 9.

When you added 1 and -1 to the list, the sum did not change but the average did because number of terms increased.
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, [#permalink]
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tonebeeze wrote:
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


Question Stem Analysis:

We are told that the average of n consecutive odd integers is 10. Remember that in an evenly spaced set, mean = median, so the median of this set is also 10. Since the median is even, n must also be even, so the median is the average of the two numbers in the middle. Thus, the possibilities for the set are:

{9, 11}

{7, 9, 11, 13}

{5, 7, 9, 11, 13, 15}

and so on.

Statement One Alone:

\(\Rightarrow\) The range of the n integers is 14

Let's calculate the ranges for the pattern we obtained above:

{9, 11}, range = 2

{7, 9, 11, 13}, range = 6

{5, 7, 9, 11, 13, 15}, range = 10

We see that the range increases by 4 each time. Following the pattern, we can determine the set with a range of 14 and determine the smallest integer in that set. Actually, the next set in our pattern is the set we are looking for: {3, 5, 7, 9, 11, 13, 15, 17}. Statement one alone is sufficient.

Eliminate answer choices B, C, and E.

Statement Two Alone:

\(\Rightarrow\) The greatest of the n integers is 17

Following the same process above, we can keep listing the possibilities until we hit the set where the largest element is 17. Once we determine the set, we can determine its smallest element. Statement two alone is sufficient.

Answer: D
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