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C
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67%
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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}.
(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}.
19 Nov 2009, 06:38
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kairoshan
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
19 Nov 2009, 09:13
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kairoshan
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.
