Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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?

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?

Posted from my mobile device

x = 2 and y = 2 does not satisfy the first statement, which says that x > y.
_________________

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??

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.
_________________

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.

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

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.

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 ?

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 \(\sqrt{x} \gt y\).
_________________

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 !

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 \((\sqrt{y} \approx 1.4) \lt (x=2)\)
_________________

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

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...

(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?

(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?

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
_________________