227. Is x > y?
(1) sqrt x > y
(2) x^3 > y
Square Root
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 66
- Joined: Wed Jul 13, 2011 2:27 pm
- Followed by:2 members
-
- Master | Next Rank: 500 Posts
- Posts: 423
- Joined: Fri Jun 11, 2010 7:59 am
- Location: Seattle, WA
- Thanked: 86 times
- Followed by:2 members
St1: √x > y
√4 = 2 > 1 Is x>y - Yes
√1/4 = ½ > 1/3 Is x>y - No
Hence Insufficient
St2: x^3 > y
23 = 8 > 5 Is x>y - No
23 = 8 > 1 Is x>y - Yes
Hence Insufficient
Together: x^3 > y AND √x > y (See attached table)
Hence Sufficient
Ans C
√4 = 2 > 1 Is x>y - Yes
√1/4 = ½ > 1/3 Is x>y - No
Hence Insufficient
St2: x^3 > y
23 = 8 > 5 Is x>y - No
23 = 8 > 1 Is x>y - Yes
Hence Insufficient
Together: x^3 > y AND √x > y (See attached table)
Hence Sufficient
Ans C
- Attachments
-
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Statement 1: √x > yIs x > y ?
(1) √x > y
(2) x³ > y
It's possible that x=4 and y=1, in which case x>y.
It's possible that x=1/4 and y=1/3, in which case x<y.
INSUFFICIENT.
Statement 2: x³ > y
It's possible that x=2 and y=1, in which case x>y.
It's possible that x=-1/2 and y=-1/4, in which case x<y.
INSUFFICIENT.
Statements 1 and 2 combined:
Since we can't take the square root of a negative, statement 1 implies that x≥0.
Thus, when we combine the two statements, if y<0, we know that y<x.
Our concern is what happens when y≥0.
One approach is to memorize the shapes of some basic graphs:
Only in the yellow region is y<√x and y<x³.
The entire yellow region is below the graph of y=x, implying that y<x throughout the entire region.
Thus, combining the two statements, we know that y<x.
SUFFICIENT.
The correct answer is C.
An alternate approach would be to use algebra to test the 3 cases: y=x, y>x, and y<x.
Case 1: y=x.
Statement 1: If y=x and y<√x, then x < √x.
Statement 2: If y=x and y<x³, then x < x³.
No value for x will work here: a number cannot be less than both its root and its cube.
Thus, y≠x.
Case 2: y>x.
Statement 1: If x<y and y<√x, then x < √x.
Statement 2: If x<y and y<x³, then x < x³.
No value for x will work here: a number cannot be less than both its root and its cube.
Thus, it is not possible that y>x.
Since it is not possible that y=x or that y>x, we know that y<x.
SUFFICIENT.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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
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