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Re: Quant Rev v.2, DS # 66: Consecutive Integer Problem [#permalink]
09 Dec 2010, 14:59

19

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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.

Re: Quant Rev v.2, DS # 66: Consecutive Integer Problem [#permalink]
12 Dec 2010, 04:24

7

This post received KUDOS

Expert's post

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. _________________

Re: If the average (arithmetic mean) of n consecutive odd [#permalink]
29 Sep 2012, 22:27

I think solution D is wrong, what is numbers are : -5, -3, -1, 1, 3,5,7, 9 then range is 14 thus least value in set is : -5 However, if we consider numbers from 3 to 11 then least value is 3.

Re: If the average (arithmetic mean) of n consecutive odd [#permalink]
29 Sep 2012, 22:41

1

This post received KUDOS

Expert's post

bandgmat wrote:

I think solution D is wrong, what is numbers are : -5, -3, -1, 1, 3,5,7, 9 then range is 14 thus least value in set is : -5 However, if we consider numbers from 3 to 11 then least value is 3.

Yeah, but is the average of these numbers 10? _________________

Re: Quant Rev v.2, DS # 66: Consecutive Integer Problem [#permalink]
28 Oct 2012, 10:36

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.

Doesn't the highlighted statement actually mean that the highest number in the series is 17??

Re: Quant Rev v.2, DS # 66: Consecutive Integer Problem [#permalink]
29 Oct 2012, 01:26

Expert's post

avaneeshvyas 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.

Doesn't the highlighted statement actually mean that the highest number in the series is 17??

Yes. We generally use the terms greatest/largest. _________________

Re: If the average (arithmetic mean) of n consecutive odd [#permalink]
09 Jul 2013, 06:30

1

This post received KUDOS

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.

Re: If the average (arithmetic mean) of n consecutive odd [#permalink]
09 Jul 2013, 20:57

Or, since this is DS, we can skip the math and use the fact that for a consecutive sequence we only need 2 pieces of information (among mean, smallest number, greatest number, and range) to determine it. So D is correct.

Re: If the average (arithmetic mean) of n consecutive odd [#permalink]
11 Sep 2014, 20:42

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Re: If the average (arithmetic mean) of n consecutive odd [#permalink]
28 Feb 2015, 21:19

As per the question average of n consecutive integers is 10 ;Sum of n consecutive integers =10n or lets say lowest integer is k then k+ k+2+k+4...+ k+2(n-1) =10n Simplifying further nk+2(1+2...+n-1)=10n ----> A

lets go with option I

i) The range of n integers is 14

so we know highest integer - lowest integer =14 highest integer =k+2(n-1) lowest integer =k

Hence we get 2(n-1) =14 or n=8 ,Substituting we get value of k hence I is sufficient

ii) if greatest integer is 17, then sum would be 17 + 17-2 ...+(17 -(n-1)) = 10n Simplifying 17n - 2(1+2...+(n-1))=10n or 7n = 2(1+2+...+(n-1))----> B

From A and B we get K=3 ,this sufficient to get all numbers in series Hence II is sufficient

gmatclubot

Re: If the average (arithmetic mean) of n consecutive odd
[#permalink]
28 Feb 2015, 21:19

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