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Is x > y?

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Is x > y? [#permalink] New post 24 Jul 2003, 04:23
00:00
A
B
C
D
E

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  25% (medium)

Question Stats:

71% (01:31) correct 29% (00:52) wrong based on 80 sessions
Is x>y?

(1) \sqrt{x}>\sqrt{y}
(2) x^2>y^2 -
[Reveal] Spoiler: OA

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Re: DS by Stolyar [#permalink] New post 24 Jul 2003, 23:09
Lynov Konstantin wrote:

Is X>Y ?

(1) square root X > square root Y
(2) X^2 > Y^2



All we know from (1) is that X and Y are positive (i.e. it could be an integer or a fraction). Consider an example:

X=4, Y=9 => sqrt(X)<sqrt(Y);
X=1/4, Y=1/9 => sqrt(X)>sqrt(Y);

Therefore, (1) alone is clearly not sufficient.

From (2) X and Y could be either positive or negative, thus, (2) alone is also not sufficient.

Combine => X&Y are positive.

Looks like the answer is E.
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Re: DS by Stolyar [#permalink] New post 25 Jul 2003, 00:07
Lynov Konstantin wrote:
Lynov Konstantin wrote:

Is X>Y ?

(1) square root X > square root Y
(2) X^2 > Y^2



All we know from (1) is that X and Y are positive (i.e. it could be an integer or a fraction). Consider an example:

X=4, Y=9 => sqrt(X)<sqrt(Y);
X=1/4, Y=1/9 => sqrt(X)>sqrt(Y);

Therefore, (1) alone is clearly not sufficient.

From (2) X and Y could be either positive or negative, thus, (2) alone is also not sufficient.

Combine => X&Y are positive.

Looks like the answer is E.


IMO, I think you need to re=examine your analysis for condition (1).
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 [#permalink] New post 25 Jul 2003, 00:19
You should start from data sets to answer the question, not vice versa.

(1) square root X > square root Y

X and Y are nonnegative

sqrt 9> sqrt 4 => 3>2 => X>Y
sqrt 1/4 > sqrt 1/9 => 1/2>1/3 => X>Y
sqrt 100 > sqrt 1/9 => 10>1/3 => X>Y
sqrt 1/4 > sqrt 0 => 1/2>0 => X>Y

More than enough

(2) clearly not sufficient because of even powers

9>4 => 3>2 => X>Y OR
9>4 => -3<2 => X<Y

Finally, A.
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Re: An old one: Is X>Y ? (1) square root X > square root Y [#permalink] New post 14 Jan 2012, 21:28
Explaination by stolyar is perfect.

Sqrt of X is greater than X if X < 1 but here the comparison is between 2 different numbers X and Y. So A is sufficient.
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Re: Is x>y? (1) square root x>square root y (2) x^2>y^2 [#permalink] New post 15 Jan 2012, 03:08
Expert's post
Konstantin Lynov wrote:
An old one:

Is X>Y ?

(1) square root X > square root Y
(2) X^2 > Y^2

Through the discussion people agreed that the answer is A, since on the GMAT we are dealing only with arithmetic radicals and, therefore, sqrt(X) and sqrt(Y) are non negative.

Disagree on that answer is:


USEFUL TO KNOW
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
2<4 --> we can square both sides and write: 2^2<4^2;
0\leq{x}<{y} --> we can square both sides and write: x^2<y^2;

But if either of side is negative then raising to even power doesn't always work.
For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
-2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3;
x<y --> we can raise both sides to third power and write: x^3<y^3.

Is x>y?

(1) \sqrt{x}>\sqrt{y} --> as both parts of the inequality are non-negative then according to A we can square them --> x>y. Sufficient.

(2) x^2>y^2 --> |x|>|y| --> x is farther from zero than y, but this info is insufficient to say whether x>y (if x=2 and y=1 - YES but x=-2 and y=1 - NO). Not sufficient.

Answer: A.
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Re: Is x > y? [#permalink] New post 11 Oct 2013, 01:37
pls correct my approach in solving this problem..

1, square root X > square root Y
Consider value: x=9, y= 4 so, suare root of 9 = 3, AND 4 = 2, AND 9>4 so AD...
2, X^2>Y^2
here x= 1,y= 2 so 1 is not greater than 2, so

only A....

can we consider negative values in square root??
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Re: Is x > y? [#permalink] New post 11 Oct 2013, 04:36
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Expert's post
kanusha wrote:
pls correct my approach in solving this problem..

1, square root X > square root Y
Consider value: x=9, y= 4 so, suare root of 9 = 3, AND 4 = 2, AND 9>4 so AD...
2, X^2>Y^2
here x= 1,y= 2 so 1 is not greater than 2, so

only A....

can we consider negative values in square root??


You cannot get sufficiency based only on one set of numbers. Thus theoretically testing only one set of numbers can give you an incorrect answer.

As for your other questions: the even roots from negative numbers are not defined for the GMAT, which means that for \sqrt{x} to be defined, x must be more than or equal to zero.

Hope it's clear.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Is x > y?   [#permalink] 11 Oct 2013, 04:36
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