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

[RonPurewal]

Q. 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}.

The correct answer is given to be:

Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.

i have trouble understanding the logic, please elaborate.

Thanks

hi - since this is one of our problems, please reproduce the answer explanation, and point out the specific parts with which you're having trouble. otherwise, we'll just wind up repeating what's already in the answer explanation. thanks.

i.e., where in the answer explanation did you get lost?

[chitrangada.maitra]

While I get the explanation to this problem, I cant come up with one set of numbers that satisfy both conditions.

For instance, if I pick 0,1,3 as x, y & z, it satisfies the first condition i.e mean should be greater than the median. However,in this case, the second condition is not satisfied.


Again, if I pick 5,6,7 as x, y & z, it satisfies the 2nd condition i.e median should be greater than the mean. However,in this case, the first condition is not satisfied.

Is there a set that satisfies both conditions?

[gokul_nair1984]

Is there a set that satisfies both conditions?

Only values with x<=0,y>4,z>5 will satisfy both criterion.Here's proof:

Stem:If x, y, and z are integers, and x < y < z, is z - y = y - x?
We can simplify this stem to: Is 2y=x+z?

Statement 1: The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}.

Statement 2:The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.

Statement 1 can be simplified to:x+y+z<12
This can be disproved by choosing
Case 1:x=0,y=2,z=3 ----No
Case 2:x=1,y=2,z=3 ----Yes

Statement 2:
you can have 4 cases here:
1. {4,x,y,z}----This means (x+y)/2<y or y>x
2.{x,4,y,z}----Similarly, 4+y<2y=> y>4
3. {x,y,4,z}----Same as above. This cannot be true as in this set y cannot be greater than 4
4. {x,y,z,4}----Finally, y+z<2y =>y>z. This is not true because y<z according to the stem[/b]

So here ideally we have only 2 conditions,
y>x and y>4

So plug in numbers x=1,y=5,z=6 for a No
and x=4,y=5,,z=6 for a Yes

Combining both statements,
x+y+z<12 and y>4 ; Also x<y<z.
x+5+z<12. The least possible value for y in this case will be 5 and z will be 6.Thus x has to be <or = 0.
So x<=0,y>4,z>5.
This is the only valid plug in possible that satisfies both Statements. The answer to the Stem is NO, which makes C sufficient.

Time Taken:5 minutes.
Time Allotted: 2 minutes :)

[chitrangada.maitra]

Thanks, Gokul!

[tim]

:)