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# If x and y are positive, is x < y?

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If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 01:43
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If x and y are positive, is x < y?

(1) $$\sqrt{x} < \sqrt{y}$$
(2) $$(x-3)^2 < (y-3)^2$$
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If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 02:13
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If x and y are positive, is x < y?

(1) $$\sqrt{x} < \sqrt{y}$$. Since both sides of the inequality are positive (the square root from a positive number is positive), then we can safely square: x < y. Directly answers the question. Sufficient.

(2) $$(x-3)^2 < (y-3)^2$$. If $$x=3$$ and $$y\neq 3$$, the inequality will hold true: the left hand side will be 0, while the right hand side will be more than 0. Thus, if $$x=3$$, y can be less than 3, giving a NO answer to the question, as well as more than 3, giving an YES answer to the question. Not sufficient.

Hope it's clear.
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Re: if x and y are positive, is x<y? [#permalink]

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25 Mar 2016, 02:13
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Gurshaans wrote:
if x and y are positive, is x<y?

(1) $$\sqrt{x} < \sqrt{y}$$
(2) $$(x-3)^2 < (y-3)^2$$

Hi,
we know that x and y are positive..
is x<y?

lets see the statements

(1) $$\sqrt{x} < \sqrt{y}$$
we can say that ans is YES,
but lets solve it algebrically too..
is x<y can be written as x-y<0..
$$\sqrt{x}^2-\sqrt{y}^2$$<0..
$$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) <0$$..
$$(\sqrt{x}+\sqrt{y}) >0$$, as x and y are +ive,
so we have to find
if $$(\sqrt{x}-\sqrt{y})<0$$ or $$\sqrt{x}<\sqrt{y}$$..
statement 1 tells us exactly this..
SUFF

(2) $$(x-3)^2 < (y-3)^2$$
It will hold in many cases : two case
a) if x = 1 and y= 7.. y>x
b) if x= 4 and y=1.. y<x.
Insuff

A
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 02:49
Hi guys, thank you for your explanations Bunuel chetan2u . I have a question, I wanted to understand why we cant use an algebraic approach on the second statement? Since we have squares on either side, can't we take take the square root on either side, which would give us x - 3 < y - 3, then by adding 3 on either side we get x < y .. ? This is exactly what I did and marked D as the answer when I saw this question for the first time. From both the posts, my understanding is that we cant do what i mentioned earlier because we wouldn't know after taking the root whether x-3 and y-3 are positive or negative... Is this right? or am I mistaken? I'm asking this because I do not want to be confused in the exam under time pressure.. for example what would have happened had the question not specified that x and y are positive... and the first statement was x^2<y^2... ? would the answer in that case be E... Sorry about the long post, I just wanted clarity on this.. I do NOT want to get a 600-700 level question wrong in the exam.

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If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 03:08
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Gurshaans wrote:
Hi guys, thank you for your explanations Bunuel chetan2u . I have a question, I wanted to understand why we cant use an algebraic approach on the second statement? Since we have squares on either side, can't we take take the square root on either side, which would give us x - 3 < y - 3, then by adding 3 on either side we get x < y .. ? This is exactly what I did and marked D as the answer when I saw this question for the first time. From both the posts, my understanding is that we cant do what i mentioned earlier because we wouldn't know after taking the root whether x-3 and y-3 are positive or negative... Is this right? or am I mistaken? I'm asking this because I do not want to be confused in the exam under time pressure.. for example what would have happened had the question not specified that x and y are positive... and the first statement was x^2<y^2... ? would the answer in that case be E... Sorry about the long post, I just wanted clarity on this.. I do NOT want to get a 600-700 level question wrong in the exam.

MUST KNOW: $$\sqrt{x^2}=|x|$$:

The point here is that since square root function cannot give negative result then $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function: $$|x|=x$$, if $$x\geq{0}$$ and $$|x|=-x$$, if $$x<0$$. That is why $$\sqrt{x^2}=|x|$$.

BACK TO THE QUESTION:

According to the above if you take the square root from $$(x-3)^2 < (y-3)^2$$ you'll get |x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3, which is not sufficient to say whether x < y.

Hope it helps.
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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 03:59
Hi bunuel,
Does this mean given 16<25, can we assume that square root of these will have the same sign?
Bunuel

Does x^2
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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 04:04
rahulkashyap wrote:
Hi bunuel,
Does this mean given 16<25, can we assume that square root of these will have the same sign?
Bunuel

Does x^2
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If we take the square root from 16 < 25 we get 4 < 5.

As for your second question it is addressed in my previous post: x^2
Hope it helps.
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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 04:06
Bunuel, but it can also be that square root of 16 can be 4 and square root of 25 be - 5, in which case 4>-5
Bunuel

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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 04:09
rahulkashyap wrote:
Bunuel, but it can also be that square root of 16 can be 4 and square root of 25 be - 5, in which case 4>-5
Bunuel

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hi,
square root is always positive so $$\sqrt{25}$$ will always be 5
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 04:10
Hi, could u pls explain that statement? According to me -5x-5 gives 25, so does 5x5

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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 04:12
rahulkashyap wrote:
Bunuel, but it can also be that square root of 16 can be 4 and square root of 25 be - 5, in which case 4>-5
Bunuel

Posted from my mobile device

You should go through basics again before attempting the questions.

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.

That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

P.S. You might find the following post useful: All You Need for Quant
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Re: If x and y are positive, is x < y? [#permalink]

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25 Mar 2016, 04:24
Sorry, so only when there's an x² term can we assume it gives positive /negative when we take the root?

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If x and y are positive, is x < y? [#permalink]

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26 Mar 2016, 08:22
The problem can be solved using arbitrary numbers which are positive:

So as per 1) √x<√y
let the values be √4 and √9. Since all positive numbers will always satisfy this relation. Hence 4 is less than 9.
This implies x<y. There cannot be a scenario wherein given the above relation (A) x will be greater than y(√x or y will always give a positive result). Hence Ist statement is sufficient.

2) (x-3)^2<(y-3)^2

Lets again take two arbitrary numbers to satisfy this equation:

Case I) (4-3)^2<(5-3)^2 (taking x and y as 4 & 5)

Hence 1<4.

CASE II - Lets test the relation with x as 4 and y as 1

(4-3)^2 = 1 < (1-3)^2 = 4 here 1<4 again however x>y
we can get multiple cases like these for other numbers as well (try fractions also)
Basically if you try inserting values of x which are bigger in magnitude however closer to 3 then the squared value will be lesser as compared to a number y which although smaller in magnitude but far away from 3.

Hence statement 2 alone cannot be sufficient to answer.

(Give kudos if you like the solution)
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If x and y are positive, is x < y? [#permalink]

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11 Jun 2016, 11:19
If x and y are positive, is x < y?

(1) sqrt (x) < sqrt (y)
(2) (x-3)^2 < (y-2)^2

--------------------------------------------------

(1) Sufficient because you square both sides and since they are positive numbers x < y
(2) Take the sqrt of both sides and get (x-3) <(y-3). Because we are taking the sqrt ourselves, I thought only the positive situation (x-3) < (y-3) was possible? In that case, isn't this sufficient? OA is A.

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Re: If x and y are positive, is x < y? [#permalink]

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11 Jun 2016, 12:24
Sqrt(x^2) = |X|. We are given that x and y are positive, but x-3 and y-2 can result in negative values.
Now for statement 2, assume x = 3 and y = 1. This is true and here x > y.
Now for the 2nd statement x =3 and y =4. This is again true and here x < y. So statement 2 is not sufficient. Hence A.

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Re: If x and y are positive, is x < y? [#permalink]

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19 Jan 2017, 09:48
BACK TO THE QUESTION:

According to the above if you take the square root from $$(x-3)^2 < (y-3)^2$$ you'll get |x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3, which is not sufficient to say whether x < y.

Hope it helps.

Dear Bunuel,

Stem tells x>0 and y>0, so why cannot we take (x-3)<(y-3) as it is since we know the signs for both x and y.

In that case, I thought x was indeed less than y. Pls let me know why this is wrong?

Thnak you

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Re: If x and y are positive, is x < y? [#permalink]

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20 Jan 2017, 07:15
Liza99 wrote:
BACK TO THE QUESTION:

According to the above if you take the square root from $$(x-3)^2 < (y-3)^2$$ you'll get |x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3, which is not sufficient to say whether x < y.

Hope it helps.

Dear Bunuel,

Stem tells x>0 and y>0, so why cannot we take (x-3)<(y-3) as it is since we know the signs for both x and y.

In that case, I thought x was indeed less than y. Pls let me know why this is wrong?

Thnak you

We know that |x| = x, when $$x \geq{0}$$ (so |something| = something, when that something is >=0) and |x| = -x, when $$x \leq{0}$$ (so |something| = -something, when that something is =<0).

Know for positive x, x-3 (expression in modulus) can be positive (when x>3) as well as negative (when x<3), thus |x-3| = x-3, when x>3 and |x-3| = -(x-3), when x<3. Thus knowing that x>0 is not enough to say that |x-3| = x-3.

Hope it's clear.
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Re: If x and y are positive, is x < y? [#permalink]

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20 Jan 2017, 09:35
THANK YOU BUNUEL !!! QUANT GOD!

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Re: If x and y are positive, is x < y? [#permalink]

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09 Apr 2017, 20:44
Bunuel wrote:
Gurshaans wrote:
Hi guys, thank you for your explanations Bunuel chetan2u . I have a question, I wanted to understand why we cant use an algebraic approach on the second statement? Since we have squares on either side, can't we take take the square root on either side, which would give us x - 3 < y - 3, then by adding 3 on either side we get x < y .. ? This is exactly what I did and marked D as the answer when I saw this question for the first time. From both the posts, my understanding is that we cant do what i mentioned earlier because we wouldn't know after taking the root whether x-3 and y-3 are positive or negative... Is this right? or am I mistaken? I'm asking this because I do not want to be confused in the exam under time pressure.. for example what would have happened had the question not specified that x and y are positive... and the first statement was x^2<y^2... ? would the answer in that case be E... Sorry about the long post, I just wanted clarity on this.. I do NOT want to get a 600-700 level question wrong in the exam.

MUST KNOW: $$\sqrt{x^2}=|x|$$:

The point here is that since square root function cannot give negative result then $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function: $$|x|=x$$, if $$x\geq{0}$$ and $$|x|=-x$$, if $$x<0$$. That is why $$\sqrt{x^2}=|x|$$.

BACK TO THE QUESTION:

According to the above if you take the square root from $$(x-3)^2 < (y-3)^2$$ you'll get |x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3, which is not sufficient to say whether x < y.

Hope it helps.

Hi Bunuel, I can't get this statement clear, can you help?

|x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3.

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Re: If x and y are positive, is x < y? [#permalink]

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09 Apr 2017, 22:08
Cez005 wrote:
Bunuel wrote:
Gurshaans wrote:
Hi guys, thank you for your explanations Bunuel chetan2u . I have a question, I wanted to understand why we cant use an algebraic approach on the second statement? Since we have squares on either side, can't we take take the square root on either side, which would give us x - 3 < y - 3, then by adding 3 on either side we get x < y .. ? This is exactly what I did and marked D as the answer when I saw this question for the first time. From both the posts, my understanding is that we cant do what i mentioned earlier because we wouldn't know after taking the root whether x-3 and y-3 are positive or negative... Is this right? or am I mistaken? I'm asking this because I do not want to be confused in the exam under time pressure.. for example what would have happened had the question not specified that x and y are positive... and the first statement was x^2<y^2... ? would the answer in that case be E... Sorry about the long post, I just wanted clarity on this.. I do NOT want to get a 600-700 level question wrong in the exam.

MUST KNOW: $$\sqrt{x^2}=|x|$$:

The point here is that since square root function cannot give negative result then $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function: $$|x|=x$$, if $$x\geq{0}$$ and $$|x|=-x$$, if $$x<0$$. That is why $$\sqrt{x^2}=|x|$$.

BACK TO THE QUESTION:

According to the above if you take the square root from $$(x-3)^2 < (y-3)^2$$ you'll get |x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3, which is not sufficient to say whether x < y.

Hope it helps.

Hi Bunuel, I can't get this statement clear, can you help?

|x - 3| < |y - 3|, which means that the distance between x and 3 is less than the distance between y and 3.

Absolute value a number is the distance between this number and 0. For example, |x| is the distance from 0 to x. Similarly |x - 3| is the distance between x-3 and 0 or between x and 3.
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Re: If x and y are positive, is x < y?   [#permalink] 09 Apr 2017, 22:08

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