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

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If the average (arithmetic mean) of n consecutive odd integers is 10,  [#permalink]

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New post 09 Dec 2010, 14:43
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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"
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10,  [#permalink]

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New post 09 Dec 2010, 14:59
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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.
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10,  [#permalink]

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New post 12 Dec 2010, 04:24
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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"



If mean of consecutive odd integers is 10, the sequence of numbers will be something like this:
9, 11 or
7, 9, 11, 13 or
5, 7, 9, 11, 13, 15 or
3, 5, 7, 9, 11, 13, 15, 17 or
1, 3, 5, 7, 9, 11, 13, 15, 17, 19
etc
Every time you add a number to the left, you need to add one to the right to keep the mean 10. The smallest sequence will have 2 numbers 9 and 11, the largest will have infinite numbers.

Stmnt 1: Only one possible sequence: 3, 5, 7, 9, 11, 13, 15, 17 will have range 14. Least of the integers is 3. Sufficient.
Stmnt 2: Only one possible sequence:3, 5, 7, 9, 11, 13, 15, 17
Least of the integers is 3. Sufficient.
Answer (D).

Note: You don't actually have to do all this. All such sequences will have distinct number of elements, greatest number, smallest number and range. So each statement alone will be sufficient.
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10,  [#permalink]

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New post 06 Apr 2011, 07:28
3
(1)

so (a + a + 14)/2 = 10

=> 2a = 20 - 14 = 6

=> a =3

(2)

(a+17)/2 = 10

=> a = 3

Answer - D

(a+17)
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10,  [#permalink]

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New post 14 Apr 2012, 10:11
Awesome explanation Bunuel!
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Re: If the average (arithmetic mean) of n consecutive odd integers is 10,  [#permalink]

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

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New post 09 Jul 2013, 06:30
1
Let a be the first term. every term in this sequence can be expressed as a+ (i-1) where i ranges from 1 to n. Thus sum of these terms is a*n +1+2+3+..+n-1= an +n(n-1)/2 = 10 n.

(1) We are given that a+n-1 -a =14. We have two eqns for the unkowns (a and n ) and thus (1) is sufficient. No need to actually solve for and and n.

(2) is also sufficient since it is given a+(n-1) =17.

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

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New post 23 Oct 2016, 02:48
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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New post 23 Oct 2016, 05:32
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]

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New post 07 Sep 2017, 18:19
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]

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New post 07 Sep 2017, 20:39
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]

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Re: If the average (arithmetic mean) of n consecutive odd integers is 10, &nbs [#permalink] 09 Sep 2018, 17:08
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