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

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

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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}.

Originally posted by kairoshan on 18 Nov 2009, 21:43.
Last edited by Bunuel on 04 Jul 2013, 23:56, edited 1 time in total.
Added the OA.
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 19 Nov 2009, 06:38
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kairoshan wrote:
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
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 19 Nov 2009, 07:40
kairoshan wrote:
sriharimurthy wrote:
kairoshan wrote:
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....:)
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 19 Nov 2009, 08:35
lugeruo wrote:
I don't think it's possible to finish this problem in 2 minutes. However, who knows....:)


agree it will take more than 2mins (atleast I took around 5mins to solve) but does GMAC always intend to give questions which can be solved in 2mins?? :?
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 19 Nov 2009, 09:13
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kairoshan wrote:
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.
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 19 Nov 2009, 09:21
kp1811 wrote:
lugeruo wrote:
I don't think it's possible to finish this problem in 2 minutes. However, who knows....:)


agree it will take more than 2mins (atleast I took around 5mins to solve) but does GMAC always intend to give questions which can be solved in 2mins?? :?



Yes, I agree that this takes more than 2 min.
I guess a few problems take more than 2 min on Gmat, so we should save time in easy ones.
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Means & Medians [#permalink]

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New post 09 Sep 2010, 00:16
Another fairly tough question :

\(x,y,z,k\) 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 \(\{x,y,z\}\)
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Re: Means & Medians [#permalink]

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New post 09 Sep 2010, 01:04
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 09 Sep 2010, 01:12
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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
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 09 Sep 2010, 01:36
shrouded1 wrote:
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.

Answer: C.
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Re: x < y < z, is z – y = y – x? [#permalink]

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New post 10 Sep 2010, 12:13
(1)+(2) As x+y+z<3k and k<y then 2y=x+z can not be true.

Please elaborate on this...
thanks!
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Re: x < y < z, is z – y = y – x? [#permalink]

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

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New post 04 Jul 2016, 04:43
Hi, I have a doubt if x<y<z and x+y+z<12 and y>4 why can I not take x to be a -ve value ? For example if i assume x=-10 and y=4 and z=6 it satisfies all three but contradicts the question stem i.e. z-y=y-x or z+x=2y
z+x= 6-10 = -4
and 2y=8
therefore E
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Re: If x, y, and z are integers, and x < y < z, is z y = y [#permalink]

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New post 04 Jul 2016, 11:55
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The question essentially asks if the integers x,y and z are in progression (equally spaced) or not?

Stmt1: on solving we get, x+y+z<12. - clearly not sufficient.
Stmt2: considering a set is un-ordered, the middle numbers to consider for median can be 4,y or y,4. Therefore, (y+4)/2 < y.
This gives us y>4. The next integer which Y can be is 5, but it does not gives us anything about x and z. Hence Not Sufficient.

Combining, for x+y+z<12 , if Y=5 then Z=6, which gives us x<1. Clearly shows X,Y and Z are not going to be equally spaced out. Hence the answer 'C'
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Re: If x, y, and z are integers, and x < y < z, is z y = y [#permalink]

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New post 12 Feb 2018, 13:26
Question: z - y = y - x => x + z = 2y?

Statement 1: The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}.
=> (x + y + z + 4)/4 > (x + y + z)/3
=> x + y + z < 12
case 1: {0,3,6} => answer to question : Yes
case 2: {1,3,6} => answer to question : No => insuff

Statement 2: The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.
case 1: say 4 > z, then median (y+z)/2 < y => z < y (which is not possible, as per question, x < y < z) - ruled out
case 2: say 4 < x, then median (x+y)/2 < y => x < y (which is possible)
case 3, say x < 4 < z, then median (4 + y)/2 < y => y > 4 (which is possible)
but still insuff to conclude if x + z = 2y

1 + 2
if case 2 ( x > 4) where true, then x + y + z > 12 , as minimum value of x will be 5 -> this case is ruled out

only case possibe is x < 4 < z and y > 4
so set looks like this, {x,4,y,z} => now minimum value of y is 5
so given x + y + z < 12 from statement 1, if y = 5, z has to be 6 and x has to be 0, {0,5,6} => not evenly spaced.

if y = 5, z = 7, x = 3, x + y + z < 12 is not satisfied
if y = 5, z = 9, x = 1, x + y + z < 12 is not satisfied.
so we can conclude that x,y,z are not evenly placed, satisfying y > 4 and x + y + z < 12
Re: If x, y, and z are integers, and x < y < z, is z y = y   [#permalink] 12 Feb 2018, 13:26
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