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Algebra- DS questions
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Is |x−z−y| > x−z+y?
(1) 0<x<z<y
(2) (x-z-y) is negative
Statement 1:
Case 1: x=1, z=2, y=3
Plugging this combination into |x−z−y| > x−z+y, we get:
|1−2−3| > 1−2+3
4 > 2
YES.
Case 2: x=1, z=1.1, y=1.2
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 1.1 − 1.2| > 1 − 1.1 + 1.2
1.3 > 1.1
YES.
Case 3: x=1, z=5, y=10
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 5 − 10| > 1 − 5 + 10
14 > 6
YES.
In every case -- whether x, y and z are close in value or far apart from each other -- the answer is YES.
SUFFICIENT.
Statement 2:
Cases 1, 2 and 3 also satisfy Statement 2.
In Cases 1, 2 and 3, the answer to the question stem is YES.
Case 4: x=0, z=0, y=1, with the result that x-z-y = 0-0-1 = -1.
Plugging this combination into |x−z−y| > x−z+y, we get:
|0 − 0 − 1| > 0 - 0 + 1
1 > 1
NO.
Since the answer is YES in Cases 1, 2 and 3 but NO in Case 4, INSUFFICIENT.
The correct answer is A.
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Followed here and elsewhere by over 1900 test-takers.
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.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
PERFECT!! Thanks! Plugging the values seemed the best way to solve..
[quote="GMATGuruNY"][quote]Is |x−z−y| > x−z+y?
(1) 0<x<z<y
(2) (x-z-y) is negative [/quote]
[b]
Statement 1:[/b]
Case 1: x=1, z=2, y=3
Plugging this combination into |x−z−y| > x−z+y, we get:
|1−2−3| > 1−2+3
4 > 2
YES.
Case 2: x=1, z=1.1, y=1.2
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 1.1 − 1.2| > 1 − 1.1 + 1.2
1.3 > 1.1
YES.
Case 3: x=1, z=5, y=10
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 5 − 10| > 1 − 5 + 10
14 > 6
YES.
In every case -- whether x, y and z are close in value or far apart from each other -- the answer is YES.
SUFFICIENT.
[b]Statement 2:[/b]
Cases 1, 2 and 3 also satisfy Statement 2.
In Cases 1, 2 and 3, the answer to the question stem is YES.
Case 4: x=0, z=0, y=1, with the result that x-z-y = 0-0-1 = -1.
Plugging this combination into |x−z−y| > x−z+y, we get:
|0 − 0 − 1| > 0 - 0 + 1
1 > 1
NO.
Since the answer is YES in Cases 1, 2 and 3 but NO in Case 4, INSUFFICIENT.
The correct answer is [spoiler]A[/spoiler].[/quote]
[quote="GMATGuruNY"][quote]Is |x−z−y| > x−z+y?
(1) 0<x<z<y
(2) (x-z-y) is negative [/quote]
[b]
Statement 1:[/b]
Case 1: x=1, z=2, y=3
Plugging this combination into |x−z−y| > x−z+y, we get:
|1−2−3| > 1−2+3
4 > 2
YES.
Case 2: x=1, z=1.1, y=1.2
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 1.1 − 1.2| > 1 − 1.1 + 1.2
1.3 > 1.1
YES.
Case 3: x=1, z=5, y=10
Plugging this combination into |x−z−y| > x−z+y, we get:
|1 − 5 − 10| > 1 − 5 + 10
14 > 6
YES.
In every case -- whether x, y and z are close in value or far apart from each other -- the answer is YES.
SUFFICIENT.
[b]Statement 2:[/b]
Cases 1, 2 and 3 also satisfy Statement 2.
In Cases 1, 2 and 3, the answer to the question stem is YES.
Case 4: x=0, z=0, y=1, with the result that x-z-y = 0-0-1 = -1.
Plugging this combination into |x−z−y| > x−z+y, we get:
|0 − 0 − 1| > 0 - 0 + 1
1 > 1
NO.
Since the answer is YES in Cases 1, 2 and 3 but NO in Case 4, INSUFFICIENT.
The correct answer is [spoiler]A[/spoiler].[/quote]
