Whether x>y>z?

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.

OA plz.

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.

Bunnel, can you please explain why in the algebraic approach for st1 the signs have not been changed for solving x-y = lx-zl + lz-yl , as your solution shows that this equation becomes (after removing the modulus signs) x-y = x-z-z+y --> z=y

Look forward to your reply.

We know that for \(|x|\):
When \(x\leq{0}\), then \(|x|=-x\);
When \(x\geq{0}\), then \(|x|=x\).

So if \(x>y>z\) is true, then:
As \(x>z\) (\(x-z>0\)) --> \(|x-z|=x-z\) AND as \(y>z\) (\(z-y<0\)) --> \(|z-y|=-(z-y)=-z+y\).

So \(x-y=|x-z|+|z-y|\), will become \(x-y=x-z-z+y\) --> \(z=y\).

Hope it's clear.

Bunnel, thanks as always for the explanation! was considering (x-z) as a negative number..It's clear now..

great explanation.

i think the algebraic method is easier to understand in this one.

1) x-y = -x+z+z-y
x = z
so x>y>z is not true. Sufficient
2) no information is given about z. Insufficient.

good question indeed.

assuming the stem to be correct and then solving A is a cool approach.

I agree with slightly different approach for solving. You rock Bunuel!

We need to answer this question in YES or NO.

Statement 2 gives us no information about z, so Insufficient.

Statement 1 :-
x-y >= 0 ; since the RHS term will always be 0 or positive.

Hence, there can be 3 scenarios with 1st statement.

Imagining a number line :-
1) y-------z---------x :- x>z>y and x-y (Distance between x and y) is equal to the distance between x and z & z and y. So False. since y<z
2) z,y ------------- x :- x > z ; x > y ; y = z , Therefore distance between y and z is 0. Still gives us false since y = z and not greater than that.
3) y----------------x,z :- x = z ; x > y ; y < z , Therefore distance between x and z is 0. Still gives us false since y < z.

In all the 2 scenarios we get FALSE. Hence sufficient.

So only A is sufficient.

Consider Kudos if the post helps anyone. Its a good way to motivate

what about the scenario when X=Y=Z

The answer will remain the same. The question asks whether x>y>z. If x=y=z, then the answer to the questions "is x>y>z" is still NO.

Bunuel Can we assume what the question is asking to be true and solve for each statement as done here for statement 1??

Posted from my mobile device

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Bunuel, How come you only took the negative of the second mod for statement 1? In general, when we have two mods, how many of the following should we do?

1. Open both (taking the positive of both)
2./3. Open both (taking positive of only one)
4. Taking the negative of both

I get confused about that.

The question asks whether x > y > z. For (1) we assumed that x > y > z and showed that IF x > y > z then x - y = |x - z| + |z - y| does not hold true. So, our assumption that x > y > z was wrong.

Now about the ranges. When we assumed that x > y > z, we considered only this case, which gave us x > z (x - z > 0), so |x - z| = x - z and y > z (z - y < 0), so |z - y| = -(z - y).

Hope it's clear.

Whether x>y>z?

Stat1: x-y = |x-z|+|z-y|
So, x-y>0 or, x>y, now, z can have 3 cases,
case-1: z>x>y, then, x-y = -x+z+z-y or, x= z, which is not possible.
case-2: x>z>y, then, x-y = x-z+z-y or, which is possible.
case-3: x>y>z, then, x-y = x-z-z+y or, y= z, which is not possible.

So, we can say, x>z>y is only possible solution. Sufficient to answer, if x>y>z?

Stat2: x > y
we don't know, if position of z between x and y. Not Sufficient.

So, I think A.