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:

If x, y, and z are integers, and x < y < z, is z y = y [#permalink]

Show Tags

18 Nov 2009, 21:43

4

This post received KUDOS

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

28% (03:51) correct
72% (02:20) wrong based on 407 sessions

HideShow timer Statistics

If x, y, and z are integers, and x < y < z, is z – y = y – x?

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.

If x, y, and z are integers, and x < y < z, is z – y = y – x?

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.

Question Stem : Is 2y = z + x ; x , y , z , are integers such that x < y < z

St. (1) : \(\frac{x+y+z+4}{4} > \frac{x+y+z}{3}\) This simplifies to : \(12 > x + y + z\) Consider the following two sets both of which satisfy all the given conditions: (a) {2, 3, 4} ----> Proves the question stem true. (b) {2, 3, 5} ----> Proves the question stem false. Since we get contradicting solutions, statement is Insufficient.

St. (2) : median of the set {x, y, z, 4} < median of the set {x, y, z} The problem here is that we don't know the relation of 4 with respect to the other numbers. Let us consider all the possible cases and see how St. (2) relates to them : Case 1 : {4, x, y, z} --> \(\frac{x+y}{2} < y\) --> y > x which is already specified. Case 2 : {x, 4, y, z} --> \(\frac{4+y}{2} < y\) --> y > 4 Case 3 : {x, y, 4, z} --> \(\frac{y+4}{2} < y\) --> y > 4 which is false since we assumed y < 4 in the set. Case 4 : {x, y, z, 4} --> \(\frac{y+z}{2} < y\) --> y > z which is false since it violates information in question stem. Thus the only two valid cases are Case 1 and 2. However, they do not give us enough information to answer the question stem. Hence, Insufficient.

St. (1) and (2) together : \(12 > x + y + z\) ; Case 1 and 2 from above. Now let us see how Case 1 and 2 relate to St. (1) : Case 1 : {4, x, y, z} --> y > 4 (assumed for the set) Case 2 : {x, 4, y, z} --> y > 4 (assumed for the set) Note: The important thing to note here is that the only valid cases from St. (2) are those in which we have assumed y > 4. Thus, for the conditions: (a) y > 4 (b) x + y + z < 12 (c) x < y < z (d) x, y, z are integers The only set of values possible are x < ; y >= 5; z >= 6 More importantly, this tells us that x, y and z can never be equidistant from each other (which is a condition necessary for 2y = x + z). Thus the question stem will always be false. Sufficient.

If x, y, and z are integers, and x < y < z, is z – y = y – x?

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.

Question Stem : Is 2y = z + x ; x , y , z , are integers such that x < y < z

St. (1) : \(\frac{x+y+z+4}{4} > \frac{x+y+z}{3}\) This simplifies to : \(12 > x + y + z\) Consider the following two sets both of which satisfy all the given conditions: (a) {2, 3, 4} ----> Proves the question stem true. (b) {2, 3, 5} ----> Proves the question stem false. Since we get contradicting solutions, statement is Insufficient.

St. (2) : median of the set {x, y, z, 4} < median of the set {x, y, z} The problem here is that we don't know the relation of 4 with respect to the other numbers. Let us consider all the possible cases and see how St. (2) relates to them : Case 1 : {4, x, y, z} --> \(\frac{x+y}{2} < y\) --> y > x which is already specified. Case 2 : {x, 4, y, z} --> \(\frac{4+y}{2} < y\) --> y > 4 Case 3 : {x, y, 4, z} --> \(\frac{y+4}{2} < y\) --> y > 4 which is false since we assumed y < 4 in the set. Case 4 : {x, y, z, 4} --> \(\frac{y+z}{2} < y\) --> y > z which is false since it violates information in question stem. Thus the only two valid cases are Case 1 and 2. However, they do not give us enough information to answer the question stem. Hence, Insufficient.

St. (1) and (2) together : \(12 > x + y + z\) ; Case 1 and 2 from above. Now let us see how Case 1 and 2 relate to St. (1) : Case 1 : {4, x, y, z} --> y > 4 (assumed for the set) Case 2 : {x, 4, y, z} --> y > 4 (assumed for the set) Note: The important thing to note here is that the only valid cases from St. (2) are those in which we have assumed y > 4. Thus, for the conditions: (a) y > 4 (b) x + y + z < 12 (c) x < y < z (d) x, y, z are integers The only set of values possible are x < ; y >= 5; z >= 6 More importantly, this tells us that x, y and z can never be equidistant from each other (which is a condition necessary for 2y = x + z). Thus the question stem will always be false. Sufficient.

Answer : C

Good explanation! OA is C +1

I don't think it's possible to finish this problem in 2 minutes. However, who knows....

If x, y, and z are integers, and x < y < z, is z – y = y – x?

(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.

Another way of tackling this problem. (Graphical/Diagramatic Method)

Q: ( is z – y = y – x? ) Question is asking whether x,y, z are equidistant, while y being middle number. Means.. Is Y middle number or integer? Clearly individual statemnets are not sufficient.. they can be equidistance or not .

Combined: Assume that y is middle number. Now check, whether two statements satisfy this or not.

x---------y---------z x---------y---------z -----4 (stmt 1 satisfy, 2 not satisfy) x---------y----4--- z (stmt 1 satisfy, 2 not satisfy) x-------4--y--------z (stmt 1 not satisfy, 2 satisfy)

Clearly, 4 will fall on right hand side of y (For statment 1 to be true) and on left hand side of y (for statement 2 be true).

Clearly this is not possible.. our assumption (Y is middle integer) is not correct C is the answer. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

\(x,y,z,k\) are integers such that \(x<y<z\) and \(k>0\). Is \((y-x) = (z-y)\) ?

(1) The mean of set \(\{x,y,z,k\}\) is greater than the mean of set \(\{x,y,z\}\) (2) The median of set \(\{x,y,z,k\}\) is less than the median of set \(\{x,y,z\}\) _________________

Yeah, I think so. But there is no reason to take the special case of 4, it can be proven for a general k.

I had an alternate solution :

(1) & (2) are clearly not sufficient as k is arbitrary and can be large enough or small enough to force any condition on mean or median seperately

For (1) + (2) : Statement (2) implies y>k. Statement (1) implies x+y+z<3k. Now if you assume x+z=2y the above two inequalities are inconsistent ==> Contradiction ==> The answer to the question is always "No"

Yeah, I think so. But there is no reason to take the special case of 4, it can be proven for a general k.

I had an alternate solution :

(1) & (2) are clearly not sufficient as k is arbitrary and can be large enough or small enough to force any condition on mean or median seperately

For (1) + (2) : Statement (2) implies y>k. Statement (1) implies x+y+z<3k. Now if you assume x+z=2y the above two inequalities are inconsistent ==> Contradiction ==> The answer to the question is always "No"

Hence (C)

short & sweet

Yes you are right: as "4" in original question is purely arbitrary we can replace it with any other number or with unknown k as you did.

Question: is \(2y=x+z\)?

For (1): \(x+y+z<3k\); For (2): \(k\) can not be the the largest term, thus it must be less than \(y\);

(1)+(2) As \(x+y+z<3k\) and \(k<y\) then \(2y=x+z\) can not be true. Sufficient.

Re: If x, y, and z are integers, and x < y < z, is z y = y [#permalink]

Show Tags

22 Feb 2014, 11:30

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 x, y, and z are integers, and x < y < z, is z y = y [#permalink]

Show Tags

17 Apr 2015, 21:44

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 x, y, and z are integers, and x < y < z, is z y = y [#permalink]

Show Tags

21 Apr 2016, 04:04

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. _________________

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...