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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'

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.

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

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.

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.

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???!!!

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

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

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. _________________

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!

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"

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) _________________