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# m01 Q5

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27 Oct 2008, 07:08
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If $$x \gt 1$$ and $$y \gt 1$$ , is $$x \gt y$$ ?

1. $$\sqrt{x} \gt y$$
2. $$\sqrt{y} \lt x$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I was thinking that we have to come up either 'Yes' or 'No' to the question so I picked choice 4 (each alone can be sufficient) but answer is different.

by using choice (1) I can say 'Yes'
and by using choice(2) I can say 'No'

Can you guys explain this?
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30 Nov 2012, 05:09
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ugimba wrote:
If $$x \gt 1$$ and $$y \gt 1$$ , is $$x \gt y$$ ?

1. $$\sqrt{x} \gt y$$
2. $$\sqrt{y} \lt x$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I was thinking that we have to come up either 'Yes' or 'No' to the question so I picked choice 4 (each alone can be sufficient) but answer is different.

by using choice (1) I can say 'Yes'
and by using choice(2) I can say 'No'

Can you guys explain this?

If $$x>1$$ and $$y>1$$, is $$x> y$$?

Notice that we are told that both $$x$$ and $$y$$ are greater than 1.

(1) $$\sqrt{x}>y$$ --> since both parts of the inequality are non-negative we can safely apply squaring: $$x>y^2$$. Now, as $$x$$ and $$y$$ are greater than 1 then $$x>y$$. Sufficient.

Notice that if we were told that $$x>0$$ and $$y>0$$, instead of $$x>1$$ and $$y>1$$, then this statement wouldn't be sufficient. For example consider: $$x=\frac{1}{3}$$ and $$y=\frac{1}{2}$$ --> $$x=\frac{1}{3}>\frac{1}{4}=y^2$$ but $$x=\frac{1}{3}<\frac{1}{2}=y$$.

(2) $$\sqrt{y}<x$$ --> $$y<x^2$$. If $$y=2$$ and $$x=3$$ then the answer to the question is YES, $$x>y$$ but if $$y=2$$ and $$x=2$$ then the answer to the question is NO, since in this case $$x=y$$. Not sufficient.

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12 Mar 2013, 03:56
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Condition 1 = sufficient since even if the square root of X is greater than the Y then X will be greater then Y. Plug in for fractions too the answer remains the same.

condition 2 = not sufficient

Take X= 16 Y =100 and X=16 and Y =4

The answers come out to be opposite. So not sufficient

A is the ans
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27 Oct 2008, 07:25
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I'm getting choice A.
B is not sufficient, answer could be "yes" or "no".

I plugged in numbers, maybe there's a more scientific way though.

just working on condition (2) [since there's no issue on condition 1 for both of us, right?]

Example 1:
(root y)=2, x=9
according to the condition, root y < x TRUE, and x>y = TRUE

Example 2:
(root y)=2, x=4
according to the condition, root y < x TRUE, and x>y = FALSE

So (2) is insufficient
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27 Oct 2008, 08:18
[quote="ugimba"]if x >1 and y > 1 , is x>y ?

1) root x > y
2) root y < x

X>Y^2 ..........SUFF

FROM 2

Y<X^2........INSUFF

a
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27 Oct 2008, 09:18
QS clearly says X & Y are >1.

1) root x > y

There will be values >1 such that Root x> Y and x is not greater than Y. Thus A is the answer.

2) root y < x

Root y< x ... we don't know whether x is greater or lesser than y.
Y=4 x=3
root(y)=root(4)=2 <3

The key is, we need to consider each statement separately and see where we stand per that statement before combining both. Hope this helps.
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28 Oct 2008, 20:45
ugimba wrote:
if x >1 and y > 1 , is x>y ?
1) root x > y
2) root y < x
I was thinking that we have to come up either 'Yes' or 'No' to the question so I picked choice 4 (each alone can be sufficient) but answer is different.

by using choice (1) I can say 'Yes' and by using choice(2) I can say 'No'

Can you guys explain this?

1: yes we can say that cuz x and y both are > 1. so if sqrt(x) > y, x >1.
2: yes or no both i.e we cannot say anything certainly because y could be greater or smaller than x. suppose if y is 4 and x is 3, x is not > y. If y is 2 and x is 3, yes x>y. so st. 2 is nsf.

So it is A or 1.
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30 Aug 2009, 10:06
if x >1 and y > 1 , is x>y ?

1) \sqrt{x} > y

SUFFICIENT. Since \sqrt{x} > y. This can be re-written as x > (y^2). And in the range y > 1, (y^2) > y always . Hence x > (y^2) > y.

2) \sqrt{y} < x
INSUFFICIENT. Since \sqrt{y} < x. This can be re-written as y < (x^2). For the range x > 1 and y > 1, by choosing values say

x = 2, y = 3. We have,
x^2 = 2^2 = 4 which is greater than y
but x = 2 is lesser than y.
hence (x^2) > y but x < y.

another set of values say
x = 4, y = 3.We have,
x^2 = 4^2 = 16 which is greater than y
and x = 4 is also greater than y.
hence (x^2) > y and x > y.

Hence A.
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06 Jan 2010, 11:42
If (x > 1) and (y > 1) are true then which of the statement(s) would or wouldn't make (x > y) correct?

1) (root x > y) SUFFICIENT, the square root is always smaller than the original # so if the square is greater than the original number has to be greater as well. Basically, if root x (which is smaller than actual x) is greater than y.... than actual x is going to be greater as well

2) (root y < x) INSUFFICIENT, this would not work in every situation. One situation where it would work is if y = 2 and x = 5. One situation where it wouldn't is if y = 144 (root y = 12) and x = 16. So, since it doesn't work in every situation it can't make (x > y) correct.

thx!
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06 Jan 2010, 23:31
guys, did you try fractions for x and y > 1 . example: x=3/2 and y=2
if you try first condition: root(3/2)=1.22>2 but 3/2 is not > 2
while root(9)=3>2 and it's true that x=9>y=2
so we have true and false answers==>A is insufficient???!!!
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29 Jan 2010, 19:26
maheritani wrote:
guys, did you try fractions for x and y > 1 . example: x=3/2 and y=2
if you try first condition: root(3/2)=1.22>2 but 3/2 is not > 2
while root(9)=3>2 and it's true that x=9>y=2
so we have true and false answers==>A is insufficient???!!!

Statement 1 has more to do with the number property:

If the sq root of a certain number(X) is more than the other number ( Y) ,
thus X will be > than Y

Hope it helps
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26 Nov 2010, 05:06
A.

S1 states \sqrt{x} > y
--> sqrt of x is positive and is greater than 1
--> squaring both sides
x > y^2
for y> 1 y^2 is always greater than y
hence x > y

S2
let y = 1.21 sqrt of y = 1.1
x can be either 1.5 or 1.20
if x is 1.5 the condition holds true however it does not hold true if x is 1.20
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29 Nov 2010, 14:32
ugimba wrote:
If $$x \gt 1$$ and $$y \gt 1$$ , is $$x \gt y$$ ?

1. $$\sqrt{x} \gt y$$
2. $$\sqrt{y} \lt x$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I was thinking that we have to come up either 'Yes' or 'No' to the question so I picked choice 4 (each alone can be sufficient) but answer is different.

by using choice (1) I can say 'Yes'
and by using choice(2) I can say 'No'

Can you guys explain this?

I did the SAME thing ugimba.

Here is how I solved it the second time around...

We knows statement 1 = YES all the way through so it is sufficient

Let's look at statement 2.

If y = 4 the square root of y will be 2. x has to be bigger than 2 so we can choose 3 to be x. Going up to the top we plug the numbers into the equation x>y (3>4) NO

However, to disprove this we need to think of an instance where x is larger than y. Let's try the same setup with y=4. The square root of 4 is 2. X has to be bigger than 2. BUT this time instead of choosing 3 let's choose 7 (really any number greater than 4 will do).

Taking this information we plug into the top equation x>y (7>4) YES

Making statement 2 INSUFFICIENT. Therefore A is the answer.

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I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

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01 Dec 2010, 02:11
maheritani wrote:
guys, did you try fractions for x and y > 1 . example: x=3/2 and y=2
if you try first condition: root(3/2)=1.22>2 but 3/2 is not > 2
while root(9)=3>2 and it's true that x=9>y=2
so we have true and false answers==>A is insufficient???!!!

There is a mistake in your reasoning: root(3/2)<2 and not >2. 3/2 and 2 is not a good example.
You can try any variable >1, it works. So A is sufficient.

For statement 2, we just need to take x=2 and Y=2 / x=4 and y=2 to see that this statement is not sufficient.
=> root(2)<4 and 2<4 so "yes"
=> root(2)<2 but 2=2 so "no"

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10 Jan 2011, 08:29
A!

Example stmt2: y=36; x=7
sqrt36=6<7; But y>x, therefore insufficient.
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11 Jun 2011, 14:50
1. Sufficient

As \sqrt{x} > y , x must be greater than y.

2. Not sufficient

\sqrt{y}<x

=> x >\sqrt{y}

lets say y=4
=>x>2
if x=3 and y=4 x<y no
if x=5 and y=4 x>y yes

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01 Dec 2011, 23:27
I tried to work with easier numbers, so I squared both sides and plugged in two numbers. A was only one that had same results.
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05 Dec 2011, 00:52
I am getting answer A, I solved the problem by picking numbers. Is there any other way to solve. On exam day if we get problem like this what should be strategy to solve such problems (sometimes picking numbers may lead to wrong answer)
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01 Dec 2013, 10:22
(1) is sufficient.
(2) is not sufficient.

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