Is |x - y| > |x + y|?
(1) x^2 - y^2 = 9
(2) x - y = 2
The OA is C.
I don't know how to use both statements together. Experts, may you show me how to do it? Thanks.
Is |x - y| > |x + y|?
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We know from the question stem that we want to find out information about x - y and x + y. We can express Statement 1 as:
(x - y) (x + y) = 9
Using Statement 2, we can substitute 2 in for x - y:
2 (x + y) = 9
x + y = 9/2
So |x - y| = |2| = 2, and |x + y| = |9/2| = 9/2
So |x - y| is NOT greater than |x + y|. Both statements together are sufficient.
(x - y) (x + y) = 9
Using Statement 2, we can substitute 2 in for x - y:
2 (x + y) = 9
x + y = 9/2
So |x - y| = |2| = 2, and |x + y| = |9/2| = 9/2
So |x - y| is NOT greater than |x + y|. Both statements together are sufficient.
Last edited by ErikaPrepScholar on Thu Nov 30, 2017 7:12 am, edited 1 time in total.
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(1) x^2 - y^2 = 9Vincen wrote:Is |x - y| > |x + y|?
(1) x^2 - y^2 = 9
(2) x - y = 2
The OA is C.
I don't know how to use both statements together. Experts, may you show me how to do it? Thanks.
Case 1: Say x = 3 and y = 0, then 3^2 - 0^2 = 9. We see that |x - y| > |x + y| => |3 - 0| > |3 + 0| => 3 = 3. The answer is No.
Case 2: Say x = -4 and y = 1, then (-4)^2 - 1^2 = 9. We see that |x - y| > |x + y| => |-4 - 1| > |-4 + 1| => 5 > 3. The answer is Yes.
No unique answer. Insufficient.
(2) x - y = 2
Case 1: Say x = 2 and y = 0, then 2 - 0 = 2. We see that |x - y| > |x + y| => |2 - 0| > |2 + 0| => 2 = 2. The answer is No.
Case 2: Say x = 1 and y = -1, then 1 + 1 = 2. We see that |x - y| > |x + y| => |1 + 1| > |1 - 1| => 2 > 0. The answer is Yes.
No unique answer. Insufficient.
(1) and (2) combined together
From (2), we get that x = y + 2. Plugging in the value of x in (1), we get x^2 - y^2 = 9 => (y + 2)^2 - y^2 = 9 => (y+2+y).(y+2-y) = 9 => (2y+2).2 = 9 => y = 5/4. Thus, x = 13/4.
At x = 13/4 and y = 5/4, we see that we see that |x - y| > |x + y| => |13/4 - 5/4| > |13/4 + 5/4| => 2 < 9/2 . The answer is No. Sufficient,
The correct answer: C
Hope this helps!
-Jay
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Is |x-y| > |x+y|?Vincen wrote:Is |x - y| > |x + y|?
(1) x^2 - y^2 = 9
(2) x - y = 2
When each side of an inequality is enclosed in absolute value symbols, we can square the inequality:
(x-y)² > (x+y)²
x² - 2xy + y² > x² + 2xy + y²
0 > 4xy
xy < 0.
Question stem, rephrased:
Do x and y have DIFFERENT SIGNS?
Statement 1: x² - y² = 9
Perfect squares: 1, 4, 9, 16, 25...
The difference between the two perfect squares in blue is 9, implying that one solution for Statement 1 is as follows:
x² = 25, with the result that x = ±5.
y² = 16, with the result that y = ±4.
Since x and y could have the same sign or different signs, INSUFFICIENT.
Statement 2: x - y = 2
If x=3 and y=1, then x and y have the same sign.
If x=1 and y=-1, then x and y have different signs.
INSUFFICIENT.
Statements combined:
Statement 1 implies the following:
(x+y)(x-y) = 9.
Substituting x-y=2 into (x+y)(x-y) = 9, we get:
(x+y)(2) = 9
x+y = 9/2.
Since we have two variables (x and y) and two distinct linear equations (x-y = 2 and x+y = 9/2), we can solve for both variables, allowing us to determine whether x and y have the same sign or different signs.
SUFFICIENT.
The correct answer is C.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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