DS Question - Need Explanation

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DS Question - Need Explanation

by ksatish » Tue Jul 30, 2013 4:12 am
Is x > y?
1. x^2 > y^2+1.
2. The product of xy is negative.

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by GMATGuruNY » Tue Jul 30, 2013 5:13 am
ksatish wrote:Is x > y?
1. x^2 > y^2+1.
2. The product of xy is negative.
Always consider the following values:
-2, -1, -1/2, 0, 1/2, 1, 2.

Both statements are satisfied by x=2 and y=-1.
Statement 1: 2² > (-1)² + 1
Statement 2: (2)(-1) < 0.
In this case, x>y.

Both statements are satisfied by x=-2 and y=1.
Statement 1: (-2)² > 1² + 1
Statement 2: (-2)(1) < 0.
In this case, x<y.

Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by Matt@VeritasPrep » Tue Jul 30, 2013 8:50 am
Just to supplement this with an algebraic approach:

From S1, we have

x² - y² > 1, or
(x + y)(x - y) > 1

Now let's divide both sides by (x + y) to isolate (x - y).

If (x + y) is positive, then (x - y) > 1/(x + y).

Since that makes (x - y) > a positive number, we can say x - y > 0, and x > y.

However, if (x + y) is negative, then (x - y) < 1/(x + y), since the sign flips when we divide by a negative. This gives us (x - y) < some negative number, or (x - y) < 0, or x < y. No good!

This does allow us to rephrase the question, however. Given Statement 1, if (x + y) > 0, x > y, and if (x + y) < 0, y > x, so our question becomes "Is (x + y) positive?"

The second statement alone tells us that either x is positive and y is negative or that y is positive and x is negative. In other words, either x > y or y > x. Also no good!

Together, (x + y) could still be either positive or negative, so the answer is E.

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by lunarpower » Tue Aug 06, 2013 1:25 am
ksatish wrote:Is x > y?
1. x^2 > y^2+1.
2. The product of xy is negative.
you can also approach the problem like this:

when you look at statement 1, the important part is that x^2 is bigger than y^2. (the +1 is of no consequence, at least in this problem.)
what does this mean?
remember that squares of positives are positive, but that squares of negatives are also positive. so, if x^2 is bigger than y^2, then we know that the "size" of x is more than the "size" of y -- but we don't know the signs.
this can be written as "|x| > |y|", but to me, at least, that's not a particularly meaningful statement.

here's how i like to put it:
if we know that x^2 > y^2, then
x = ±BIG
y = ±small
(or 0, if the problem allows).

so... now that we have this, let's kill this problem.

(1)
if x = -BIG, then either value of y = ±small will be greater than x.
if x = +BIG, then that's greater than either value of y = ±small.
not sufficient.

(2)
in this case, one of the following is true:
x = positive, y = negative
or
x = negative, y = positive
the first is always a yes, the second always a no. not sufficient.


(together)
here there are still two possibilities:
x = -BIG and y = +small
or
x = +BIG and y = -small
the first is a no, the second a yes. not sufficient.

so, (e).

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mitch's method of testing numbers is also valuable, although you shouldn't memorize a specific set of numbers to test -- too limiting. (for instance, if i change y^2 + 1 to y^2 + 10, then the essential character of the problem doesn't change, but none of those numbers will work anymore.)
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by GMATGuruNY » Tue Aug 06, 2013 10:06 am
lunarpower wrote:you shouldn't memorize a specific set of numbers to test -- too limiting.
A test-taker who cannot see what types of values are relevant has to start somewhere.
The list of values offered in my post above is -- to quote Julie Andrews in The Sound of Music -- a very good place to start.
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by lunarpower » Wed Aug 07, 2013 6:41 am
GMATGuruNY wrote:A test-taker who cannot see what types of values are relevant has to start somewhere.
true.
basically, as long as you didn't mean the word "always" literally, then you and i are thinking along the same lines here.

i would add two more numbers to the list, though: -100 and 100 (or -1000 and 1000, or two other suitably "huge" numbers). a lot of expressions don't reveal their true colors until you get into numbers bigger than 2.
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