x, y, and z are integers

[reply2spg]

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}.

Ans. given as C. However, I am getting E.

Both statements together show that (mean) < 4 < (median). This means that the mean and median are not equal, and so the set is not in arithmetic progression, which means that z - y can NOT equal y - x.

However, if I put values x = 10, y = 20 and z = 30

then,

{4, 10, 20, 30} - mean = 32
{10, 20, 30} - mean = 20

{4, 10, 20, 30} - median = 15
{10, 20, 30} - median = 20

As per above, for statement 1 mean {4,x,y,z} > mean {x, y, z}
and as per st II median {4,x,y,z} < {x,y,z}

and z - y can equal y - x

Please let me know where am I wrong? and why answer is C?

[vaibhav.tripathi]

{4, 10, 20, 30}
mean = (4 + 10 + 20 + 30)/4 = 64/4 = 16


correct me if i am wrong

[esledge]

{4, 10, 20, 30}
mean = (4 + 10 + 20 + 30)/4 = 64/4 = 16


correct me if i am wrong

You are right.

The thing the set of (10,20,30) didn't consider:
From (1), (x+y+z+4)/4 > (x+y+z)/3
Cross-multiply to get 3x+3y+3z+12 > 4x+4y+4z
Subtract 3x+3y+3z from each side to get 12 > x+y+z.

reply2spg, I think you may have lost track of the constraint by writing it this way (bold emphasis mine)...

Both statements together show that (mean) < 4 < (median).

Also, let's think more about that "4 < median" part from (2).
The median of (x,y,z) is y because x<y<z.
The median of (x,y,z,4) depends on where 4 is on the number line relative to the variables. The median will be the average of the middle two terms, so that will only be less than y if 4 itself is less than y. That is, the x and 4 are below the median (either one can be the minimum term) and y and z are above the median.

Putting the statements together, we have two scenarios from (2):
(a) 4 < x < y < z
Since the variables are integers, the minimum values are x = 5, y = 6, z = 7. But then x+y+z > 12, violating statement (1). We throw this scenario out.

(b) x < 4 < y < z
The minimum y and z are 5 and 6, respectively. As long as x is an integer less than or equal to 0, the sum x+y+z<12. The question is essentially asking "are x, y, and z evenly spaced?" The answer is no for (0,5,6) and certainly also for any set with (negative, 5, 6).