Hi guys, can you help me with this one?
If x<y<z and u<v, is x<u<y<v<z?
(1) y-u = u-x
(2) z-v = v-y
If x<y<z and u<v, is x<u<y<v<z?
(1) y-u = u-x
(2) z-v = v-y
Statement 1: y-u = u-xRJCN wrote:Hi guys, can you help me with this one?
If x<y<z and u<v, is x<u<y<v<z?
(1) y-u = u-x
(2) z-v = v-y
Combining like terms, we get:
x+y = 2u
u = (x+y)/2.
Since u is the AVERAGE of x and y, u is HALFWAY BETWEEN x and y.
Case 1: x=1, u=2, y=3, v=4, z=5.
In this case, x<u<y<v<z.
Case 2: x=1, u=2, y=3, z=4, v=5.
In this case, it is not true that x<u<y<v<z.
INSUFFICIENT.
Statement 2: z-v = v-y
Combining like terms we get:
y+z = 2v
v = (y+z)/2.
Since v is the AVERAGE of y and z, v is HALFWAY BETWEEN y and z.
Case 1: x=1, u=2, y=3, v=4, z=5.
In this case, x<u<y<v<z.
Case 3: u=1, x=2, y=3, v=4, z=5.
In this case, it is not true that x<u<y<v<z.
INSUFFICIENT.
Statements combined:
Since u is hallway between x and y, and v is halfway between y and z -- and x<y<z -- it must be true that x<u<y<v<z.
SUFFICIENT.
The correct answer is C.
