Is lXl>lYl?

[dianapaolasanchez]

Is lXl>lYl?
1) X^2>Y^2
2) X>Y
Rephrase:
lXl>lXl
x> y or -x> y or x> -y or -x> -y ?
1) x> y or -x> y or x> -y or -x> -y then Suff.
2) x>y Suff ?

The answer is A or D? I am confused
Thank you

[parthatayi]

The OA is A.
Reason is :
x2>y2
=> x2-y2>0
this is possible only if x>y
or |x| >|y|

[Ben Ku]

Is lXl>lYl?
1) X^2>Y^2
2) X>Y
Rephrase:
lXl>lXl
x> y or -x> y or x> -y or -x> -y ?
1) x> y or -x> y or x> -y or -x> -y then Suff.
2) x>y Suff ?

The answer is A or D? I am confused
Thank you

The rephrase would not really be all of the inequalities you suggest; instead, it really means "is X further away from 0 than Y?"

Statement (1) is sufficient because we can take the square root of both sides:
x^2 > y^2 => | x | > | y |

Statement (2) is insufficient. Just because x > y doesn't mean |x| > |y|. While statement (2) works if x = 5 and y = 3, it doesn't work if x = -3 and y = -5. Therefore statement (2) is insufficient.

[vivekcall81]

This can be checked as follows take value
1st statement:
X=3,Y=2
X^3>Y^2 YES
9>4
then
-3^2>2^2
yes lXl > lYl

[editor: this is only one plug-in, which is nowhere near enough information to establish the sufficiency of this statement. see below.]

2nd Statement:
2>-3
but
l2l is not > l-3l NO
3>2

l3l > l2l yes
[editor: in case it's unclear what is going on, the poster plugged in x = 2 and y = -3 for the first plug-in, and x = 3 and y = 2 for the second. this is a valid approach.]

hence only Ist statement is okay.

[RonPurewal]

@ vivekcall

in statement 1, you appear to be implying that you can evaluate the sufficiency of this statement by plugging in only one value for each of x and y.
important:
it is impossible to evaluate a data sufficiency problem using only one set of plug-ins.

it's possible to investigate data sufficiency problems by plugging, but you must obey the following protocol:
* plug in MULTIPLE values
* if any of these values give you different answers to the question, you can stop and select "insufficient".
* if ALL of those values give the same result, you should suspect that the statement is sufficient -- although you are still not 100% sure (unless your list is exhaustive or exhibits an extremely reliable pattern).

[sprparvathy]

Hi,

Though its clear that x^2 > y^2 only when |x| > |Y|,
I have a doubt reg. st. 1.

x^2 > y^2 can be simplified as

(x+y) (x-y) > 0

Can this not be read as X > Y or X > -Y?

Both are inconclusive in establishing whether |x| > |Y|.

Is this interpretation not correct?

[mschwrtz]

That's an interesting approach sprparvathy. You are correct that S1 can be rewritten as, (x+y) (x-y) > 0. I don't know that I'd describe that move as simplifying, though. In fact, I'll chase down the answer based on your rephrasing, and I think that you'll find it pretty subtle. It will turn out that (x+y) (x-y) > 0 is sufficient to determine that |x|>|y|.

If (x+y) (x-y) > 0, then (x+y) and (x-y) have the same sign; they're either both positive or both negative.

There are three ways that this could be true: x and y could both be positive, and x greater than y; x and y could both be negative, and x less than y (that is further from 0 than y); x could be a positive and y negative, and x further from 0 than y. In every one of these cases, |x|>|y|.

Here's a way to justify that claim with algebra. Keep in mind the definition of absolute value,
If n>0, then |n|=n
If n<0, then |n|=-n

If both expressions are positive, then it's possible that x and y are both positive: x-y>0, so x>y, so |x|>|y|.

If both expressions are positive, then it's possible that x is positive and y is negative: x+y>0, so x>-y, so |x|>|y|.

If they're both negative, then x+y<0, so x<y, so |x|>|y|.

The best take-away here is that you should probably not approach this algebraically. Instead, think about absolute value as "distance from 0," at least for some problems. And if you do test values, explicitly consider various combinations of positive and negative values for x and y.

When you're testing values in an absolute value question, the relevantly different values are positive vs. negative, and big vs. small. Choose values accordingly.