nverma wrote:
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.
For (a) you are checking scenario x > z > y. But if we have this case then the answer to the question is x>y>z is still NO.
Let's do this in the way you are solving:
Question: is \(z<y<x\)?(1) We have \(x-y = |x-z|+|z-y|\). This statement is true. Now let's check in which cases is this true:
As \(x\geq{y}\) (see my solution to see why this must be true) then there can be 4 scenarios as you've written:
A. \(x=y\) --> \(x-y=0=|x-z|+|z-y|\) --> \(x=y=z\). Which means that \(x-y = |x-z|+|z-y|\) is true when \(x=y=z\).
B. \(z<y<x\) ---z---y---x--
\(x-y=|x-z|+|z-y|\) --> \(x-y=x-z-z+y\) --> \(z=y\). Which means that \(x-y = |x-z|+|z-y|\) is also true when \(y={z}<x\). Two points \(z\) and \(y\) must coincide.
C. \(y<z<x\) ---y---z---x--
\(x-y=|x-z|+|z-y|\) --> \(x-y=x-z+z-y\) --> \(0=0\). Which means that \(x-y = |x-z|+|z-y|\) is always true for any values of x, y, and z, when \(y<z<x\). You can see this on diagram: if x, y, and z are placed on number line as above then 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 \(y\).
D. \(y<x<z\) ---y---x---z---
\(x-y=|x-z|+|z-y|\) --> \(x-y=-x+z+z-y\) --> \(z=x\). Which means that \(x-y = |x-z|+|z-y|\) is true when \(y<z=x\). Two points \(z\) and \(x\) must coincide.
So we've got that statement as \(x-y = |x-z|+|z-y|\) is true, only following 4 scenarios are possible:
A. \(x=y=z\);
B. \(z=y<x\);
C. \(y<z<x\);
D. \(y<z=x\).
Among the above scenarios there is no case when \(z<y<x\). Hence the answer to the question "is \(z<y<x\)" is NO.
Hope it's clear.
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