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Bunuel
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M03-14 [#permalink] New post  15 Sep 2014, 23:20
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Is \(x \lt y\)?


(1) \(x \gt y^2\)

(2) \(x\) and \(y\) are positive integers
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Re M03-14 [#permalink] New post  15 Sep 2014, 23:20
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Official Solution:


(1) \(x \gt y^2\). If \(x=2\) and \(y=1\) then the answer is NO but if \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then the answer is YES. Not sufficient.

(2) \(x\) and \(y\) are positive integers. Not sufficient on its own.

(1)+(2) Since from (2) \(x\) and \(y\) are positive integers then from (1) \(x \gt y^2 \ge 1\), which means that \(x \gt y\), so the answer to the question is NO. Sufficient.


Answer: C
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Re: M03-14 [#permalink] New post  12 Oct 2014, 17:34
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Bunuel wrote:
Official Solution:


(1) \(x \gt y^2\). If \(x=2\) and \(y=1\) then the answer is NO but if \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then the answer is YES. Not sufficient.

(2) \(x\) and \(y\) are positive integers. Not sufficient on its own.

(1)+(2) Since from (2) \(x\) and \(y\) are positive integers then from (1) \(x \gt y^2 \ge 1\), which means that \(x \gt y\), so the answer to the question is NO. Sufficient.


Answer: C



If x and y are positive numbers then from 1 and 2 x>y2>=0 why you are considering it as '>=1'. In this case answer would be E. Please explain. Thanks.
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Re: M03-14 [#permalink] New post  12 Oct 2014, 23:07
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kd1989 wrote:
Bunuel wrote:
Official Solution:


(1) \(x \gt y^2\). If \(x=2\) and \(y=1\) then the answer is NO but if \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then the answer is YES. Not sufficient.

(2) \(x\) and \(y\) are positive integers. Not sufficient on its own.

(1)+(2) Since from (2) \(x\) and \(y\) are positive integers then from (1) \(x \gt y^2 \ge 1\), which means that \(x \gt y\), so the answer to the question is NO. Sufficient.


Answer: C



If x and y are positive numbers then from 1 and 2 x>y2>=0 why you are considering it as '>=1'. In this case answer would be E. Please explain. Thanks.


y is a positive integer means that y >=1, so y^2>=1, so \(x \gt y^2 \ge 1\).
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M03-14 [#permalink] New post  29 Oct 2014, 06:12
Please let me know if this reasoning is correct:

(1) insuff: x>y^2 -->|√x|=y; since we only know that the abs val of the x term is greater than y, it could still be possible that the x term is neg, such that its less than y.

(2) insuff: we don't know the values of x and y.

(1&2) from stmt 1, we know that |√x|>y, which means the pos.√x is greater than y. Stmt 2 tells us that we are dealing with the pos √x only, therefore, we know x>y if the pos √x >y. Suff.
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M03-14 [#permalink] New post  20 May 2015, 03:24
kd1989 wrote:
Bunuel wrote:
Official Solution:


(1) \(x \gt y^2\). If \(x=2\) and \(y=1\) then the answer is NO but if \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then the answer is YES. Not sufficient.

(2) \(x\) and \(y\) are positive integers. Not sufficient on its own.

(1)+(2) Since from (2) \(x\) and \(y\) are positive integers then from (1) \(x \gt y^2 \ge 1\), which means that \(x \gt y\), so the answer to the question is NO. Sufficient.


Answer: C



If x and y are positive numbers then from 1 and 2 x>y2>=0 why you are considering it as '>=1'. In this case answer would be E. Please explain. Thanks.


This question is simple. Only think tricky and which shouldn't be is the term positive. I think you are confused with it or you overlooked that term.
The integer 0 cannot be classified as either positive or negative.
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Re: M03-14 [#permalink] New post  20 May 2015, 03:33
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bimalpandeyr9 wrote:
kd1989 wrote:
Bunuel wrote:
Official Solution:


(1) \(x \gt y^2\). If \(x=2\) and \(y=1\) then the answer is NO but if \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then the answer is YES. Not sufficient.

(2) \(x\) and \(y\) are positive integers. Not sufficient on its own.

(1)+(2) Since from (2) \(x\) and \(y\) are positive integers then from (1) \(x \gt y^2 \ge 1\), which means that \(x \gt y\), so the answer to the question is NO. Sufficient.


Answer: C



If x and y are positive numbers then from 1 and 2 x>y2>=0 why you are considering it as '>=1'. In this case answer would be E. Please explain. Thanks.


This question is simple. Only think tricky and which shouldn't be is the term positive. I think you are confused with it or you overlooked that term.
The integer 0 cannot be classified as either positive or negative.


y can be 1, no?
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Re: M03-14 [#permalink] New post  20 May 2015, 04:01
Bunuel wrote:
bimalpandeyr9 wrote:
Bunuel wrote:
Official Solution:


(1) \(x \gt y^2\). If \(x=2\) and \(y=1\) then the answer is NO but if \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then the answer is YES. Not sufficient.

(2) \(x\) and \(y\) are positive integers. Not sufficient on its own.

(1)+(2) Since from (2) \(x\) and \(y\) are positive integers then from (1) \(x \gt y^2 \ge 1\), which means that \(x \gt y\), so the answer to the question is NO. Sufficient.


Answer: C



If x and y are positive numbers then from 1 and 2 x>y2>=0 why you are considering it as '>=1'. In this case answer would be E. Please explain. Thanks.


This question is simple. Only think tricky and which shouldn't be is the term positive. I think you are confused with it or you overlooked that term.
The integer 0 cannot be classified as either positive or negative.


y can be 1, no?[/quote]

I don't know if you are asking it to me or replying to kd1989. I'm new here and getting used to this probably needs time. Since you have already explained to kd1989 why the answer is C, I assume that you are still asking to him. If I have to comment on this, then yes y can be 1 since, 1 is integer an positive.
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Re: M03-14 [#permalink] New post  20 May 2015, 16:42
JackSparr0w wrote:
Please let me know if this reasoning is correct:

(1) insuff: x>y^2 -->|√x|=y; since we only know that the abs val of the x term is greater than y, it could still be possible that the x term is neg, such that its less than y.

(2) insuff: we don't know the values of x and y.

(1&2) from stmt 1, we know that |√x|>y, which means the pos.√x is greater than y. Stmt 2 tells us that we are dealing with the pos √x only, therefore, we know x>y if the pos √x >y. Suff.


Hi JackSparr0w,

I tried a similar approach...
However, for condition (1)...I took a Sqrt on LHS and RHS:
x > y^2
√x > √(y)^2
√x > |y|
Now, √x can only be +ve but y can take either +ve or -ve values
So, for +ve y: √x > y
-ve y: √x < y
Therefore, (1) is not sufficient. (2) is clearly insufficient, but 1&2 combined preclude a negative value for y...hence answer is C

What do you think?
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Re: M03-14 [#permalink] New post  20 May 2015, 16:43
rohitd80 wrote:
JackSparr0w wrote:
Please let me know if this reasoning is correct:

(1) insuff: x>y^2 -->|√x|=y; since we only know that the abs val of the x term is greater than y, it could still be possible that the x term is neg, such that its less than y.

(2) insuff: we don't know the values of x and y.

(1&2) from stmt 1, we know that |√x|>y, which means the pos.√x is greater than y. Stmt 2 tells us that we are dealing with the pos √x only, therefore, we know x>y if the pos √x >y. Suff.


Hi JackSparr0w,

I tried a similar approach...
However, for condition (1)...I took a Sqrt on LHS and RHS:
x > y^2
√x > √(y)^2
√x > |y|
Now, √x can only be +ve but y can take either +ve or -ve values
So, for +ve y: √x > y
-ve y: √x < y
Therefore, (1) is not sufficient. (2) is clearly insufficient, but 1&2 combined preclude a negative value for y...hence answer is C

What do you think?


Hi Bunuel,
Can you please comment on my approach for (1) above?

Thanks
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Re: M03-14 [#permalink] New post  21 May 2015, 02:43
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rohitd80 wrote:
JackSparr0w wrote:
Please let me know if this reasoning is correct:

(1) insuff: x>y^2 -->|√x|=y; since we only know that the abs val of the x term is greater than y, it could still be possible that the x term is neg, such that its less than y.

(2) insuff: we don't know the values of x and y.

(1&2) from stmt 1, we know that |√x|>y, which means the pos.√x is greater than y. Stmt 2 tells us that we are dealing with the pos √x only, therefore, we know x>y if the pos √x >y. Suff.


Hi JackSparr0w,

I tried a similar approach...
However, for condition (1)...I took a Sqrt on LHS and RHS:
x > y^2
√x > √(y)^2
√x > |y|
Now, √x can only be +ve but y can take either +ve or -ve values
So, for +ve y: √x > y
-ve y: √x < y
Therefore, (1) is not sufficient. (2) is clearly insufficient, but 1&2 combined preclude a negative value for y...hence answer is C

What do you think?


How did you get the red part???
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Re M03-14 [#permalink] New post  28 Jun 2017, 00:18
I agree with explanation.
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Re: M03-14 [#permalink] New post  17 Jun 2019, 05:55
STATEMENT 1:
X>Y^2.
When a number(FIRST NUMBER) is greater than square of another number (SECOND), THREE situations are possible:

(1) Both numbers are EQUAL in case of proper fraction (example, x=y=1/2).

(2) First number is LESS than Second Number if both numbers are proper fraction.
(Example: (x=1/3, y=1/2)

(3) First number is GREATER than Second Number if both numbers are INTEGERS .
(Example: (x=2, y=1)

so, from statement 1, we can't definitely say whether y is greater than x or not.

NOT SUFFICIENT.

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Re: M03-14 [#permalink] New post  17 Jun 2019, 05:59
Statement 2:
If X=2, y=1 (x>y: NO to main question)

If x=1, y=2 (x<y: Yes to main question).

So, NOT SUFFICIENT.

COMBINING:
Statement 2 rules out first two cases of my previous previous post. Thus, only third scenario (both are INTEGERS) remains, stating that X>Y (No to main question).

SUFFICIENT.

My answer: C

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Re: M03-14 [#permalink] New post  17 Jun 2019, 06:33
Silly mistake... I assumed X as an integer :(

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Re: M03-14 [#permalink] New post  17 Jun 2019, 08:03
x < y? A yes/no type of DS question and we need a definitive yes or definitive no.

Statement 1) x > y^2

If we assume x = 1/4 and y = 1/3 then yes satisfies the statement and provides a yes answer.

However, if x = 1/2 and y = 1/3 then 1/2 > 1/9

But 1/2 < 1/3 No so insufficient.


Statement 2) provides that both are positive integers without any further information so insufficient.

Since both are positive integers then say x = 2 and y = 1 then 2 > 1

If x = 5 and y = 2 then 5 > 4

x < y (no)

Hence, answer choice C

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Re: M03-14 [#permalink] New post  17 Jun 2019, 09:13
Bunuel

Do you have any similar questions?

Thanks in advance.
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Re: M03-14 [#permalink] New post  17 Jun 2019, 22:01
Expert Reply
bhChicago wrote:
Bunuel

Do you have any similar questions?

Thanks in advance.


9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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M03-14 [#permalink] New post  30 Nov 2020, 07:55
Bunuel wrote:
Is \(x \lt y\)?


(1) \(x \gt y^2\)

(2) \(x\) and \(y\) are positive integers


Is \(x - y < 0?\)

(1) \(x > y^2\)

\(x - y^2 > 0\)

\(y^2\) = non-negative. Therefore \(x > 0\). We don't know if \(x - y < 0\), however.

(2) x and y are positive integers

Clearly insufficient.

(1&2) If x and y are positive integers, then \(x - y^2 ≥ 0\) or \(x ≥ y^2 ≥ 1\)

Therefore \(x - y > 0\)

SUFFICIENT.

Answer is C.
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Re: M03-14 [#permalink] New post  30 Nov 2020, 09:09
This is E. Check x=1, y=1
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Re: M03-14 [#permalink]
30 Nov 2020, 09:09
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