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If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?

Notice that we are told that both \(x\) and \(y\) are greater than 1.

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

Notice that if we were told that \(x \gt 0\) and \(y \gt 0\), instead of \(x \gt 1\) and \(y \gt 1\), then this statement wouldn't be sufficient. For example: if \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then \(x=\frac{1}{3} \gt \frac{1}{4}=y^2\) but in this case \(x=\frac{1}{3} \lt \frac{1}{2}=y\).

(2) \(\sqrt{y} \lt x\). Square both sides: \(y \lt x^2\). If \(y=2\) and \(x=3\) then the answer to the question is YES, \(x \gt y\) but if \(y=2\) and \(x=2\) then the answer to the question is NO, since in this case \(x=y\). Not sufficient.


Answer: A
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New post 27 Jan 2015, 06:07
I don't understand why statement 1 can't be deemed insufficient for the same reason as statement 2, i.e. when x=2, and y=2. Why is it possible to use these values when testing statement 2 and not 1?

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I don't understand why statement 1 can't be deemed insufficient for the same reason as statement 2, i.e. when x=2, and y=2. Why is it possible to use these values when testing statement 2 and not 1?

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x = 2 and y = 2 does not satisfy the first statement, which says that x > y.
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Yes, It does not satisfy the first condition that it √x > y . cos √2 is not greater than 2.

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New post 28 Jan 2015, 10:53
Hi Bunnuel

I got the correct answer. But in the process, I have got one genuine doubt.

If in any questions we have variables like p, q, r and certain relation is given, is it safe to assume everytime that p can be equal to q can be equal to r??

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New post 29 Jan 2015, 04:18
Satyarath wrote:
Hi Bunnuel

I got the correct answer. But in the process, I have got one genuine doubt.

If in any questions we have variables like p, q, r and certain relation is given, is it safe to assume everytime that p can be equal to q can be equal to r??


Unless it is explicitly stated otherwise, different variables CAN represent the same number.
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Re: M01-05 [#permalink]

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New post 27 Dec 2016, 07:33
Hello,
I believe that statement 2 is insufficient because of the following:

Statement 2: sqrt(y) < x
Re-arrangement: y < x^2

Due to this we can determine that Y is less that x^2, but not x alone, for any value of x. This would imply that there are multiple value scenarios where both sufficient and insufficient outcomes could occur.

Additionally when reading the explanation for statement 2, which states x = 2, y = 2 is insufficient, I believe this is meant that y is meant to actually equal four with the resulting sqrt(y) equaling 2. Thus when you use the re-arrangement detailed above and plug it in, the result is 4 < 4, which is insufficient.

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New post 16 May 2017, 05:53
The way I came to a conclusion was different from what is given here. But I get this too. I see it is confusing for some people so I'm posting my thought process here. Correct me if I'm wrong.

I just plugged in values and saw the difference.

Statement (1) - \sqrt{x}>y => x>y^2
When you apply statement 1, you always get a situation where x > y let let y = 1.1 or 2, we get x > 1.21 or 4 hence x >y always.
SUFFICIENT

Statement (2) - \sqrt{y}<x => y<x^2
When we apply this we don't always get x>y, since x^2 changes drastically
if we apply y= 3, x=2 : we see that y < x^2 and y>x but if we apply y=2, x = 3, we get y<x^2 but y<x
iNSUFFICIENT

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New post 10 Aug 2017, 03:34
In this question, we are supposed to assume the 2 statements as it is. Also the question is asking for is X > Y? i.e. Yes or No

so Statement 1: \sqrt{x} > y
it means only those values of x and y which make above statement true. so we can have x and y as follows:
X = 3 will have all those values of Y for which Y^2 is less than 3 (so we can have values like Y= 1.41, Y^2 = 2 which answers X > Y as Yes. There is no value of X and Y which satisfies statement1 which will not be able to ans X > Y as Yes or No)

Statement 2: \sqrt{Y} < X
i.e. Y^2 < X
X = 3 will have Y as 1.73, 1.6, .. >1 as its values.
X = 9 will h ave Y as 2, 2.5, 2.6
In this case also we can answer whether X > Y as Yes or No. So answer should be D each statement alone is sufficient.

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Bunuel wrote:
If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?


(1) \(\sqrt{x} \gt y\)

(2) \(\sqrt{y} \lt x\)


hi Bunuel ,

I don't understand why statement 1 can't be deemed insufficient for the same reason as statement 2, i.e. when x=2, and y=2. Why is it possible to use these values when testing statement 2 and not 1 ? can you explain please ?

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dave13 wrote:
Bunuel wrote:
If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?


(1) \(\sqrt{x} \gt y\)

(2) \(\sqrt{y} \lt x\)


hi Bunuel ,

I don't understand why statement 1 can't be deemed insufficient for the same reason as statement 2, i.e. when x=2, and y=2. Why is it possible to use these values when testing statement 2 and not 1 ? can you explain please ?


Because x=2, and y=2 do not satisfy \(\sqrt{x} \gt y\).
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Bunuel wrote:
dave13 wrote:
Bunuel wrote:
If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?


(1) \(\sqrt{x} \gt y\)

(2) \(\sqrt{y} \lt x\)


hi Bunuel ,

I don't understand why statement 1 can't be deemed insufficient for the same reason as statement 2, i.e. when x=2, and y=2. Why is it possible to use these values when testing statement 2 and not 1 ? can you explain please ?


Because x=2, and y=2 do not satisfy \(\sqrt{x} \gt y\).



Bunuel thanks for your reply I am still confused :? .... if x=2, and y=2 dont satisfy \(\sqrt{x} \gt y\), than how can x=2, and y=2 satisfy \(\sqrt{y} \lt x\)[/quote] ? can you please explain in details ? many thanks ! :-)

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Bunuel thanks for your reply I am still confused :? .... if x=2, and y=2 dont satisfy \(\sqrt{x} \gt y\), than how can x=2, and y=2 satisfy \(\sqrt{y} \lt x\) ? can you please explain in details ? many thanks ! :-)


Frankly, I don't know what to explain...

If x = y = 2, then \((\sqrt{y} \approx 1.4) \lt (x=2)\)
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New post 10 Sep 2017, 04:38
Bunuel wrote:
If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?


(1) \(\sqrt{x} \gt y\)

(2) \(\sqrt{y} \lt x\)



Tricky question if we assumed that x & y only integers

(1) \(\sqrt{x} \gt y\)

Both x & y are positive according to stem. So we can square both sides

x > y^2 & by default y^2 > y ...then

x > y^2 > y

Sufficient

(2) \(\sqrt{y} \lt x\)

Both x & y are positive according to stem. So we can square both sides

y < x^2

Let y =3 & x = 2.........Answer is No

Let y =3 & x = 4.........Answer is Yes

Insufficient

Answer: A

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Given : x>1 and y>1, so here we don't have to worry about fractions, negative numbers and value 0 and 1
Now find if x>y

Condition 1:
\(\sqrt{x}\) >y
=> x> \(y^2\)
Now as we know \(y^2\) >y (if y is not equal to 0 and 1)
so => x>\(y^2\) > y
=> x>y . Answer. Sufficient

by putting values:
\(\sqrt{x}\) >y
So we have to find value of \(\sqrt{x}\) is "greater" than y.
Assume y=2 .. => 2.1 > 2
So \(\sqrt{x}\) = 2.1 => x = 4.41 => 4.41 >2 => x>y Sufficient


Condition 2:
\(\sqrt{y}\) < x
=> y <\(x^2\)
=> 3<(3)^2 => but here x=y so Answer x> y answer to main question is No
=> 2< (4)^2 =>Here x> y answer to our main question is yes
=> 9 < (4)^2 => Here x<y so answer to our main question is No
Not sufficient


Answer: A

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Hello dave13,

Let me try and explain how I solved this. It MAY look a bit complex, but once you get a hang of it - this approach is more intuitive (Atleast for me).

Stem : Both X and Y are positive. Both of them are not Proper fractions. Question asks, is X > Y? Another way of Paraphrasing this is - On the number line is X on the right of Y?

S1: \(\sqrt{X} > Y\)

On a Number line, this means \(\sqrt{X}\) is on the right of Y. This is how the Number line would look.

Attachment:
Case 1.png


From this we can very easily infer that X is on the right of Y. Mathematically this means X > Y. Clearly sufficient.


S2: \(\sqrt{Y}< X\)

What this statement means is, on the number line \(\sqrt{Y}\) is on the left of X. The cases that make this conditions true are

Case 1 : Here Y < X

Attachment:
Case 2.png



Case2: : Here Y > X

Attachment:
Case 3.png


S2 gives two Answers for one equation. Hence Insuff.

S1 is Sufficient and S2 is Insufficient. The Answer is hence A

Hope this helps.

dave13 wrote:
What squaring? Which statement are you talking about? What is unclear? Please be more specific...


Bunuel - I am talking about second statement, after squaring both sides to get rid of radical sign why 2 is still 2 and not 2^2 which is 4 http://www.shelovesmath.com/algebra/int ... qualities/ please explain :-)

>> !!!

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New post 02 Dec 2017, 18:25
Bunuel

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


Can you please explain how combining this statement ie. \(x \gt y^2\) along with info in Q stem ie x>1 and y>1 helps us to infer x>y?
Did you try to plug in numbers as in St 2?

niks18

Is Nikkb correct in quoting:

Quote:
Given : x>1 and y>1, so here we don't have to worry about fractions, negative numbers and value 0 and 1


We can still have x and y as improper fractions, right? say 3/2 which is greater than 1.
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New post 02 Dec 2017, 22:25
adkikani wrote:
Bunuel

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


Can you please explain how combining this statement ie. \(x \gt y^2\) along with info in Q stem ie x>1 and y>1 helps us to infer x>y?
Did you try to plug in numbers as in St 2?

niks18

Is Nikkb correct in quoting:

Quote:
Given : x>1 and y>1, so here we don't have to worry about fractions, negative numbers and value 0 and 1


We can still have x and y as improper fractions, right? say 3/2 which is greater than 1.


When 0 < x < 1, then x^2 < x. So, squaring decreases the value. For example, (1/2)^2 < 1/2
When x > 1, then x^2 > x. So, squaring increases the value. For example, (3/2)^2 > 3/2
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