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:

Z is the set of the first n positive odd numbers, where n is [#permalink]
14 Jun 2012, 03:08

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

64% (03:21) correct
36% (02:12) wrong based on 133 sessions

Z is the set of the first n positive odd numbers, where n is a positive integer. Given that n > k, where k is also a positive integer, x is the maximum value of the sum of k distinct members of Z, and y is the minimum value of the sum of k distinct members of Z, what is x + y?

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
14 Jun 2012, 03:21

6

This post received KUDOS

Expert's post

Smita04 wrote:

Z is the set of the first n positive odd numbers, where n is a positive integer. Given that n > k, where k is also a positive integer, x is the maximum value of the sum of k distinct members of Z, and y is the minimum value of the sum of k distinct members of Z, what is x + y?

Probably the easiest way to solve this question would be to assume some values for n and k.

Say n=3, so Z, the set of the first n positive odd numbers would be: Z={1, 3, 5}; Say k=1, so X, the maximum value of the sum of K distinct members of Z would simply be 5. Similarly, Y, the minimum value of the sum of K distinct members of Z would simply be 1.

X+Y=5+1=6.

Now, substitute n=3 and k=1 in the options provided to see which one yields 6. Only asnwer choice E fits: 2kn=2*3*1=6.

Answer: E.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
19 Jan 2013, 09:03

3

This post received KUDOS

1

This post was BOOKMARKED

In the given case, sum of all n odd numbers is n^2.

If the numbers are arranged in increasing order, the max sum is when k is taken from end. So the sum of k odd numbers taken from the end = n^2 - (n-k)^2 (Sum of all numbers) - (Sum of remaining initial numbers) which is equal to 2nk - k^2 -- (1)

Now for minimum sum, take k numbers from the starting and their sum is k^2 -- (2)

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
23 Jan 2013, 11:36

Let us name Zi the general term of the set Z. z(i) = 2*i - 1 for i = 1 to n. For instance z(1) = 2*1 - 1 = 1 z(2) = 2*2 - 1 = 3

z(n) = 2*n - 1

y is the sum of the first k terms of the set Z. x is the sum of the terms from order (n-k+1) to n.

Let us express the sums x and y. Each of the sums contains k terms. y = Sum (2*i - 1) for i = 1, ... k x = Sum (2*j - 1) for j = (n-k+1), ,n

By developing y, we obtain y = 2*Sum (i) - k for i=1,....k y = 2*(1/2) [k*(k+1)]/2 - k = k^2 + k - k = k^2 (we can also remember that the sum of the first k odd numbers equals k^2).

By developing x, we obtain x = 2*Sum (j) - k for j=n-k+1, n We can notice that x is the difference of - the sum of the first "n" odd numbers - and the sum of the first (n-k) odd numbers. We deduce x = n^2 - (n-k)^2=2*n*k - k^2

Re: Z is the set of the first n positive odd numbers, where n is [#permalink]
09 Oct 2014, 08:08

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. _________________

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

Every student has a predefined notion about a MBA degree:- hefty packages, good job opportunities, improvement in position and salaries but how many really know the journey of becoming...