Bunuel wrote:
xyztroy wrote:
Whether x>y>z?
1. x-y = |x-z|+|z-y|
2. x > y
Q: is \(x>y>z\)?
(1) \(x-y=|x-z|+|z-y|\)
First of all as RHS is the sum of two non-negative values LHS also must be non-negative. So \(x-y\geq{0}\).
Now, we are told that the distance between two points \(x\) and \(y\), on the number line, equals to the sum of the distances between \(x\) and \(z\) AND \(z\) and \(y\).
The question is: can the points placed on the number line as follows ---z---y---x---. If you look at the number line you'll see that it's just not possible. Sufficient.
OR algebraic approach:
If \(x>y>z\) is true, then \(x-y=|x-z|+|z-y|\), will become \(x-y=x-z-z+y\) --> \(z=y\), which contradicts our assumption \(x>y>z\). So \(x>y>z\) is not possible. Sufficient.
(2) \(x>y\) --> no info about \(z\). Not sufficient.
Answer: A.
hello Brunel..
Can u please help me with my approach..where i have gone wrong..!!
IMHO E
from the given equation.
1. x-y = |x-z|+|z-y|
we can have four cases,
a) x > z > y . The given equations gives no result.
b) x > z and z < y. We get y=z
c) x < z and z > y. We get x=z
d) y > z > x. We get x=y
Though we have 3 (b,c and d)results which implies that x > y > z cannot be true.
But the result from a is not clear. So we cannot confidently state that x > y > z is true. INSUFFICIENT.
2. x > y. INSUFFICIENT.
From 1 and 2.
given x > y.(Statement 2).We can only consider cases a, b and c.(From statement 1).
Again from a, b and c. NOT SUFFICIENT.