Whether xyz?

Question

Whether x>y>z?

(1) x-y = |x-z|+|z-y|
(2) x > y

Bunuel · Accepted Answer

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.

KarishmaB · Answer

Alternative approach: Stmnt 1: x-y = |x-z|+|z-y| If the RHS has to be equal to x - y, we have to get rid of 'z'. That will happen only when EITHER |x-z| = x - z and |z-y| = z - y i.e. x - z and z - y are both positive which implies x > z and z > y OR |x-z| = -(x - z) and |z-y| = -(z - y) i.e. x - z and z - y are both negative which implies x < z and z < y. (Anyway, in this case, RHS becomes y - x instead of x - y) In both the cases, the relation between x, y and z is not x>y>z. Hence statement 1 is sufficient to say "No, x>y>z does not hold." Statement 2 doesn't say anything about z so is obviously not sufficient alone. Hence A.

nverma · Answer

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.

nverma · Answer

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.

Bunuel · Answer

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 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 x-y=x-z-z+y --> z=y . Which means that x-y = |x-z|+|z-y| is also true when y={z} 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 x-y=-x+z+z-y --> z=x . Which means that x-y = |x-z|+|z-y| is true when y

abhi758 · Answer

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.

Look forward to your reply.

Bunuel · Answer

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.

abhi758 · Answer

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

FQ · Answer

great explanation.

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

MHIKER · Answer

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.

amit2k9 · Answer

good question indeed.

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

brandy96 · Answer

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

shameekv · Answer

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 ; 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 :-D

ongy · Answer

what about the scenario when X=Y=Z

Bunuel · Answer

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.

ssshyam1995 · Answer

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

CEdward · Answer

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.

Bunuel · Answer

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.

TarunKumar1234 · Answer

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.

Whether x>y>z?

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