HOT Competition: Is |x+y| > |x-y| ? (1) |x| > |y| (2) |x - y| < |x|

Is |x + y| > |x - y| ?

Is \(|x+y| > |x-y|\)? --> square it: is \(x^2+2xy+y^2>x^2-2xy+y^2\) --> is \(xy>0\)?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: \(x^2-2xy+y^2<x^2\) --> \(xy>\frac{y^2}{2}\) --> since \(y^2\geq{0}\) (the square of any number is more than or equal to 0), then we have that \(xy>\frac{y^2}{2}\geq{0}\). Sufficient.

Answer: B.
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Statement 1: |X| > |Y|
X>±Y
Case 1: X>Y
X-Y > 0

Case 2: X>-Y
X+Y > 0

No sufficient constraint to determine answer.
(Insufficient)

Statement 2: |x-y|<|X|
If |x-y|<|X|
|x-y| will definitely be less than |X+Y|
(Sufficient)

Answer is B

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Answer: Option B

Please check the video for the step-by-step solution.

GMATinsight's Solution

Target
Is |x+y|>|x−y| ?

#1
|x|>|y|
check at x ,y pairs of ( 2,1) & (-2,-1) we get yes and
at x,y pairs of (-1,1/2) we get no to our target ; insufficient
#2
|x−y|<|x|
possible at x,y pairs of same sign eg ( -2,-1 ) & ( 2,1)
we get yes for our target value ; sufficient
OPTION B


Is |x+y|>|x−y| ?


(1) |x|>|y||x|>|y|

(2) |x−y|<|x|

Squaring the stem
---> 2xy > -2xy
===> 4xy >0 --- rephrasing is xy >0 or do x and y have opposite signs ?

1. Absolute values are given. We can't comment on polarity of the variables. Insufficient
2. Squaring --> y^2 - 2x < 0 ---> 2x > y^2 ----> minnimum value of Y^2 can be zero .... SO R.H.S. is at least Zero
Hence, LHS > 0 .... 2x > 0 ---> x > 0 . But we dont know the polarity of Y. Insufficient

Combining ,

We know that absolute value of X is more than the absolute value of Y
and x>0

These are still not suffcient as we dont know the polairty of Y. For example,
consider x = 3 and y =-1/2 . Our answer is NO
and x = 3 and y =+1/2 . Our answer is Yes.
Hence, E
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Is |x+y|>|x−y| ?

(St 1) |x|>|y|
If, |2|>|1|; then |2+1|>|2−1|; |3|>|1|? Yes
If, |2|>|-1|; then |2-1|>|2+1|; |1|>|3|? No

St. 1 has more than 1 value, so it is insufficient

(St 2) |x−y|<|x|
If x=-2 and y=-1; then |-2-1|>|-2+1|; |-3|>|-1|? Yes
If x=2 and y=1; then |2+1|>|2−1|; |3|>|1|? Yes

St. 2 is sufficient as it has one value

ANS. B
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|x−y|>|x+y| square both sides to get

x2+y2−2xy>x2+y2+2xy=>4xy<0

so the question Is xy<0 (means are they of opposite signs)

1) |x|>|y| --> x can be 2 or y can be 1 (same sign) or x can be 2 and y can be -1. (Insufficient)
2) |x-y|>|y| --> using the above example, if x=2, y=1, the statement is null. However, if x and y are opposite signs, the statement is always true. Plug in another example (x = -2, y = 1). Using same sign x=-2 and y=-3 also does not work. (Sufficient)

Is |x+y|>|x−y| ?


(1) |x|>|y|
Case1: x=2; y=1
3>1 Yes
Case2: x=-2 y=1
1>3 No

Cancel AD

(2) |x−y|<|x|
Here notice we cant take a negative value to test because this equation wont be true, so
x=2 y =1
Yes it is sufficient to answer the question

Answer - B

Is |x+y| > |x-y| ?

Squaring both sides:
Is (x+y)^2 > (x-y)^2
Is xy > 0 ?

Statement 1:
|x| > |y|
(i) x = 10, y = 5
=> xy > 0
(ii) x = 10, y = -5
=> xy < 0

Statement 1 is insufficient


Statement 2:
|x-y| < |x|
=> (x-y)^2 < x^2
=> y^2 - 2xy < 0
=> y(y-2x) < 0

Either y < 0 and y - 2x > 0
Or
y - 2x < 0 and y > 0

In either case, xy > 0

Statement 2 is sufficient

Answer: B

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Number line approach:

|x-y| = the distance between x and y
|x+y| = |x - (-y)| = the distance between x and -y

Is |x-y| < |x+y| ?
In words:
Is the distance between x and y less than the distance between x and -y?

..........-y..........0..........y..........
...........y..........0..........-y..........
In each number line:
For x to be closer to y than to -y, x and y must be together in a green area.
In other words, x and y must have the SAME SIGN.

Question stem, rephrased:
Do x and y have the same sign?

Statement 1:
No way to determine whether x and y have the same sign.
INSUFFICIENT.

Statement 2, put into words:
The distance between x and y is less than the distance between x and 0.
........................0.....|.....y..........
...........y.....|.....0........................
In each number line:
The vertical line lies halfway between y and 0.
For x to be closer to y than to 0, x and y must be together in a green area.
Implication:
x and y have the SAME SIGN.
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.


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|x+y|>|x−y| only true if x and y are same sign

(1) insufic
x,y could be any sign

(2) sufic
only true if x and y are different signs
x,y=1,-1: 1--1=|2|>|1|
x,y=-2,2: -2-2=|-4|>|-2|

(B)

Hi Bunuel,

Great explanation. Just one query - what if x = y = 0? Since the same is not stated explicitly in the question, is the equality implied?

Thanks

This case is not possible because x = y = 0 does not satisfy |x - y| < |x|.
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it's a quite simple question A being clearly insuff i am gonna just concentrate on B
B it states that y>0 is y was negative then it would have added while subtracting and would have given a different result
imo B

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