singhpreet1 wrote:Is |x-z| >|x-y|

1.|z|> |y|
2. 0>x

OA: E

from Kaplan practice test...i thought it should be A, but the OA disagrees. Would someone explain stepwise please.

Preet

Try rephrasing the stem:

It asks Is Z>y or Is Z<Y??


St 1: .|z|> |y|


Z can be positive or negative but the Absolute value of Z will always be greater than absolute value of Y.

So you can't say of Z is greater than or lesser than Y

Example :

|3| >|2|
But it can be like 3> 2 or it can also be like -3<-2

Inconsistent..so Insufficient

St2 : X>0

Really of No use. Insufficient

Combining st 1 & St2

Not really, we are still stuck with the inconsistencies

Pick E

Try rephrasing the stem:

It asks Is Z>y or Is Z<Y??


St 1: .|z|> |y|


Z can be positive or negative but the Absolute value of Z will always be greater than absolute value of Y.

So you can't say of Z is greater than or lesser than Y

The statement in RED seems contrdictory to me, please explain.

gmatmachoman wrote:

Z can be positive or negative but the Absolute value of Z will always be greater than absolute value of Y.

So you can't say of Z is greater than or lesser than Y

Example :

|3| >|2|

But it can be like 3> 2 or it can also be like -3<-2

Inconsistent..so Insufficient

St2 : X>0

Really of No use. Insufficient

Combining st 1 & St2

Not really, we are still stuck with the inconsistencies

Pick E

Is |x-z| >|x-y|

1.|z|> |y|
2. 0>x


lets assume X=2, Y=3 and Z=4, for 1st set of equations and plug these values

|2-4|> |2-3|
as we have absolute values 2>1 correct?

lets assume X=-2, Y=-3 and Z=-4, for 2nd set of equations and plug these values

|-2+4| > |-2+3|

2>1 ; as we have absolute values 2>1 this is also correct?

lets assume X=-2, Y=3 and Z=-4, for 2nd set of equations and plug these values

|-2-4|> |-2+3|

6>1; as we have absolute values this is also correct?


lets assume X=2, Y=-3 and Z=4, for 2nd set of equations and plug these values

|2-4|> |2+3|

2>5?? no so we have a yes + no condition. therefore A is not sufficient..i just answered my own query lol!

hope it helps everyone else.

Preet

The Jock wrote:

Try rephrasing the stem:

It asks Is Z>y or Is Z<Y??


St 1: .|z|> |y|


Z can be positive or negative but the Absolute value of Z will always be greater than absolute value of Y.

So you can't say of Z is greater than or lesser than Y

The statement in RED seems contrdictory to me, please explain.

I tried to mean that even though the absolute value of Z is greater than absolute value of Y, it does NOT mean that value of Z is greater than value of Y.

Let see a example
case 1: Z>Y
Z= 3, y= 2

Mod Z >Mod Y

YES

case 2 Z<Y
Z= -3, Y= -2

Mod Z>Mod Y

but Z is < Y...

Hope I am clearing the smoke..

singhpreet1 wrote:Is |x-z| >|x-y|

1.|z|> |y|
2. 0>x

OA: E

from Kaplan practice test...i thought it should be A, but the OA disagrees. Would someone explain stepwise please.

Preet

The first step of the Kaplan Method for Data Sufficiency is to ask three questions about the question stem. First, what type of Data Sufficiency question is this? Here, it's a Yes/No question--that means answering yes or no is sufficient, but answering maybe is insufficient. Second, can we simplify this question? Not really; Absolute Value questions tend to resist simplification. Third, is there anything obviously missing that would lead us directly to an answer? Not really; we may not literally need all three variables here, but we can't just ignore x for a straightforward solution.

So now we'll check the statements. In most cases, I stat with statement (1) because it's first. However, a Kaplan trained test-taker will see that (2) is a whole lot easier to evaluate--a quick glance makes that clear. So let's start there.

(2) gives us no information about y or z. Insufficient, so we can eliminate choices (B) and (D)

Now let's look at statement (1). This one is pretty vague; Absolute Values once again make things difficult. But when in doubt, we can Pick Numbers. Plugging in sample sets of numbers will help us get a much better feel for this problem. However before we begin, remember two key factors when picking numbers for picking numbers on DS questions: First, since one set of numbers will always give us one solution, regardless of the statement's sufficiency, we need to pick at least two distinct sets of values to test things out. Second, our goal when picking numbers should be insufficiency. A thousand examples of a 'yes' answer doesn't actually prove that 'no' is impossible (though it makes a strong case), but one 'yes' and one 'no' is all we need to establish insufficiency.


So we'll start by picking x = -1. Note that we do not need to consider the rule in statement (2) while we evaluate statement (1). however, by picking a number consistent with statement (2) now, we may save ourselves some work down the line if we do end up having to combine statements.

Then z and y. The easiest set of numbers to pick here would be something like z = 2 and y = 0. We plugging these values into the statement confirms their validity (but does NOT give us a yes/no answer; if we broke the rule, we have to toss these numbers out and pick new ones): Is |z| > |y|? |1| > |0|. Great, we're following all the rules we have to. Now we ask the question in the question stem: is |x - z| > |x - y|? Plugging is, we can rephrase the question as whether |-1 - 2| > |-1 -0| ? Yes, | -3| > |-1|.

So now that we've picked a set of numbers that give us a 'Yes' solution, we WANT a 'No' solution. This is absolute values, so we can predict that negatives might throw things for a loop--as will zeroes. So let's keep our zero and use a negative: let z = -2 and y = 0. Confirming with the rules, |-2| > |0|, so we're golden; asking the question, |x - z| = |-1 - (-2)| = |1|= 1 and |x - y| = |-1 - 0| = |-1| = 1. 1 is NOT greater than 1, so we have a 'no' answer. Statement 1 can answer 'yes' or 'no', so it's insufficient and we eliminate choice (A).

Finally, when both statements are insufficient, we need to combine them. Fortunately, we're one step ahead. When evaluating statement (1), we picked numbers consistent with statement (2)--although we didn't have to, and could have changed those numbers if it ended up necessary. Since we already have a 'yes' answer AND a 'no' answer that follow both rules (1) and (2), we know that even together rules (1) and (2) cannot answer the question: we must pick (E) as the answer, because the answers are not sufficient together.