Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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]

Show Tags

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.

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?
_________________

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??

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]

Show Tags

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]

Show Tags

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]

Show Tags

11 Sep 2014, 20:42

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

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

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

Show Tags

09 Oct 2015, 04:43

Below is a very simple logical approach to the problem.

Set={consecutive odd integers} for eg:{3,5,7}---Avg=5(an odd number;this is because number of integers in set= odd) Avg given=10 (even number) Thus, obviously the number of terms are even. For eg: {9;11} or {7,9,11,13} Avg:10 Thus; possible entries in set={1,3,5,7,9,11,13,15,17}

1) range=Greatest-least= 14 Check set above ;only 17-3=14 ; thus highest number is 17,lowest 3-- Sufficient

2) Greatest =17 in consecutive integer set greatest+lowest/2= mean 17+Low/2=10 Low= 3 ~~ sufficient.

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

Show Tags

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

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} ...
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...