Is xy |x| − |y| (2) x y

Question

Is xy < 0?

(1) |x − y| > |x| − |y|
(2) x > y

amanvermagmat · Accepted Answer

(1) we can try with plugging in values which satisfy this statement. Lets take x=4, y=3 (both +ve and x >y). Here LHS = 1 and RHS = 1. Not true
Now lets take x=3, y=4 (both +ve, x<y). Here LHS = 1, RHS = -1. True, and in this case x*y is positive
Now lets take x=3, y=-3 (one +ve, one -ve). Here LHS = 6, RHS = 0. True, and in this case x*y is negative. 
So we cannot determine with surety whether x*y is positive or negative. Not sufficient.

(2) Not sufficient obviously.

Combining, x has to be greater than y. So in this case x-y will always be positive but there could be various cases:
If x is positive and y is 0, then statement 1 will not be true. 
If x is positive and y is negative, then statement 1 will be true.
If x is 0 and y is negative, then again statement 1 will be true.

So both statement 1 and 2 are coming true to be in two cases: one where x is positive, y is negative (here x*y < 0) OR second where x is zero, y is negative (here x*y = 0). So we cant be sure whether x*y will be 0 or less than 0. Not sufficient.

Hence  E answer

jayantbakshi · Answer

Dear  chetan2u ,
Is there a way to solve this without plugging in numbers?
Thanks a lot, always.

chetan2u · Answer

Hi Jayant,

If you can imagine these numbers on a number line it will help you..

Points which can help you reduce number of plugging...
1) |X|-|y| will never be greater than |x-y|
2) whenever |y|>|X|, LHS, |x-y|, will be POSITIVE while RHS, |X|-|y| will be negative..
So when both X and y are positive and y>X, 
Or both negative and x>y
3) whenever the signs are different, again statement I will be true..

Knowing above points, you would know that statement I is true for both cases- one when both are same sign and second when both are different...
 but we are looking for both X and y of different sign..

Combined..
Even if x>0....
Both can be negative
Or both can be of different sign..
So insufficient..

Even rules for |X+y|<|X|+|y| can be learnt

GyMrAT · Answer

Question: is xy < 0 ? Statement 1: |x − y| > |x| − |y| for x = 0, y = -1, we get 1 > -1....hence xy = 0...therefore NO for x = 1, y = -1, we get 2 > 0....hence xy < 0...therefore YES Statement 1 is not sufficient. Statement 2: x > y for x = 0, y = -1, we get 0 > -1...hence xy = 0....therefore NO for x = 1, y = -1, we get 1 > -1...hence xy < 0....therefore YES Statement 2 is not sufficient. Combining, we can use x = 0, y = -1, to satisfy both statements & we get xy = 0...NO x = 1, y = -1, to satisfy both statements & we get xy < 0...YES Combining is not Sufficient. Answer E. Thanks, GyM

kaladin123 · Answer

Let me attempt to solve this using Avi's number line approach. avigutman - request you to share your expert advice!

The question is asking us whether x & y are on the opposite sides of 0 on the number line. https://i.ibb.co/VDHgwPp/nl1.jpg From S1: |x-y| > |x|-|y| |x-y| represents the distance between x & y on the number line. https://i.ibb.co/W5bY9Qk/nl2.jpg S1 will only be true when x and y are only opposite sides of 0. Now, we need to check this statement for boundary values. What if x or y are 0. If x=0 , then S1 is always true. If y=0 , then S1 doesn't hold. There is a danger of making a mistake here. I got this step wrong while solving the problem the first time. To summarize: S1 tells us that xy < 0 as long as y e 0 . But, x could be 0. So, xy \le 0 . S1 is insufficient. From S2: x > y https://i.ibb.co/xmFzhNq/nl3.jpg S2 tells us the relative position of x and y on the number line. It doesn't tell us where 0 is on the number line. Insufficient. Hence E.

Fdambro294 · Answer

Is: X < Y?

Absolute Value Rule:
For any real values of X and Y, it will always hold true that:

[X - Y] >/= [X] - [Y]

The right side can NEVER be greater than the left side for any real values of X and Y

(1)the 2 sides can equal in a couple of cases

Case 1:

X * Y > 0

And

[X] > [Y] ————> if the absolute value of Y is GREATER than the absolute value of X, then you will end up with a negative value on the right hand side, while the left hand side will remain positive

OR

Case 2: Y = 0

Then the inequality changes and becomes: [X] = [X]

We are given in the first statement:

Statement (1): [X - Y] > [X] - [Y]

remembering that the right hand side can take a (-)negative value, while the right hand side can never be negative can help you decide whether the statement is sufficient or not

Finding a YES: XY < 0

Case 1: if the signs of X and Y are different, the subtraction inside the absolute value Modulus on the left side will 1st INCREASE the magnitude of the 2 values when combined

However, on the right side, because we are taking the absolute value of each first and THEN subtracting, the value on the right hand side will still decrease

Case 1: X = +4 And Y = -2

[4 - (-2)] > [4] - [-2]

[4 + 2] > 4 - 2

6 > 2

So we can have the signs of X and Y be opposite (XY < 0) and still satisfy statement 1.

Furthermore: X > Y ——- +4 > -2
(satisfies statement 2)

YES to the question stem

Finding a NO ——— XY > 0

Case 2: as long as the MAGNITUDE of the Y value is greater than the Magnitude of the X value, we will always end up with a (-)negative value on the right hand side of statement 1’s inequality.

This is true even if X and Y both have the same signs (XY > 0) —— X’s value can be greater than (but magnitude less than) Y’s value when both variables are (-)negative values

Case 2: X = -2 and Y = -4

When we multiply two negative values we get a positive value ———> XY > 0 ——-NO to the question stem

Satisfy s1: [-2 - (-4)] > [-2] - [-4]

[-2 + 4] > 2 - 4

+2 > -2 ——- s1 satisfied

S2 satisfied: X > Y ———-> -2 > -4

NO

Since we can get a YES and a NO when both statements are applied, the answer is

E

Posted from my mobile device

ShubhadeepB · Answer

As per my understanding, the question actually asks:  Can x be -ve AND y +ve OR x +ve AND y -ve?

Statement 1: |x-y| > |x| - |y|

Statement 1 satisfies the following conditions:
Condition 1: x -ve AND y +ve e.g. x = -2 and y = 3 ----> |-2-3| > |-2| - |3| so, 5 > -1

Condition 2: x +ve AND y -ve e.g. x = 3 and y = -2 -----> |3-(-2)| > |3| - |-2| so, 5 > 1

Condition 3: x and y BOTH -ve e.g. x = -2 and y = -3 ----> |-2-(-3)| > |-2| - |-3| so, 1 > -1

Only statement 1 is not sufficient to answer the question

Statement 2: x > y

Here, both x and y could be positive OR both x and y could be negative OR x is +ve AND y is -ve

Only statement 2 is not sufficient to answer the question

Combining statements 1 and 2, we get two conditions:
1. Both x AND y are -ve ------> xy > 0
2. x is +ve AND y is -ve ------> xy < 0

Even by combining two statements, we get two different answers to the questions stem. Therefore,  E   is the correct answer choice.

avigutman · Answer

Nice work,  kaladin123 !
Just a bit of feedback here on your analysis of statement (1):

No need to check boundary values. Rather, observe that statement (1) allows for any case in which y is farther away from zero than x is, as such cases would mean a positive left hand side with a negative right hand side.
Whenever you're dealing with absolute values, you must take note that they're incapable of being negative. This has a great impact on the analysis of  the absolute value of a difference  vs  the difference of absolute values  (the former can't be negative, but the latter can be negative).

So, statement (1) implies that x and y are on opposite sides of zero (leading to a YES)  or  that y is farther away from zero than x is (leading to either a YES or a NO).

avigutman · Answer

Video solution from Quant Reasoning starts at 0:26: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=9X6Y96oDACI

sonalshenoy · Answer

But there is this rule we've learnt:
|x| - |y| <= | x-y|
a. If xy > 0, |x| - |y| = | x-y |
b. If xy < 0, |x| - |y| < | x-y|

Isn't statement 1 of the question just the same as point b here?

Bunuel · Answer

|x|-|y|\leq{|x-y|} . Note that "=" sign holds when x \leq y \leq 0 or when 0 \leq y \leq x . So, essentially when xy \geq 0 AND |x| \geq |y| Hence, |x|-|y| < |x - y| , given in (1), does not necessarily mean that xy < 0. It means that we can have any of the following four cases: y < x \leq 0 . In this case xy \geq 0 . 0 < x < y . In this case xy > 0 . x \leq 0 < y . In this case xy \leq 0 . y<0

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Is xy < 0? (1) |x − y| > |x| − |y| (2) x > y

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