## Tricky DS ep2 if x and y are positive , is x< y?

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### Tricky DS ep2 if x and y are positive , is x< y?

by Neilsheth2 » Fri May 20, 2016 8:50 am
Hi can any one help me how is Statement 2 inssuff?

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by OptimusPrep » Sat May 21, 2016 7:00 pm
Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?

I will focus just on the statement 2: (x - 3)^2 < (y-3)^2
Hence |x - 3| < |y - 3|

Case 1: x = 4, y = 1, x > y
Hence |x - 3| < |y - 3| => |1| < |-2|
Or 1 < 2, True

Case 2: x = 1, y = 7, x < y
Hence |-2| < |4| or 2 < 4, True

Therefore we cannot say that x < y
INSUFFICIENT.

Does this help?
Last edited by OptimusPrep on Thu May 26, 2016 4:47 am, edited 1 time in total.

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by Neilsheth2 » Sat May 21, 2016 7:09 pm
OptimusPrep wrote:
Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?

I will focus just on the statement 2: (x - 3)^2 < (y-3)^2
Hence |x - 3| < |y - 3|

Case 1: x = 1, y = -1, x > y
Hence, |-2| < |-4| or 2 < 4, True

Case 2: x = 1, y = 7, x < y
Hence |-2| < |4| or 2 < 4, True

Therefore we cannot say that x < y
INSUFFICIENT.

Does this help?
Hey thank you for your reply. I would just like to clear that whenever we have a Square in a variable we make take a modulus right? since the square could be negative or positive.

so for statement 1 we can not since the root always has to be positive? Correct?

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by OptimusPrep » Sat May 21, 2016 7:47 pm
Neilsheth2 wrote: Hey thank you for your reply. I would just like to clear that whenever we have a Square in a variable we make take a modulus right? since the square could be negative or positive.

so for statement 1 we can not since the root always has to be positive? Correct?
Yes, absolutely Correct

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by jain2016 » Wed May 25, 2016 7:21 am
Case 1: x = 1, y = -1, x > y
Hi Optimus,

It is given that x and y are positive, then how come y= -1?

SJ

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by OptimusPrep » Thu May 26, 2016 4:45 am
jain2016 wrote:
Case 1: x = 1, y = -1, x > y
Hi Optimus,

It is given that x and y are positive, then how come y= -1?

SJ
Hi SJ,

Thanks a lot for pointing, i think I overlooked that condition.
We can take the case 1 as x = 4, y = 1, x > y
Hence |x - 3| < |y - 3| => |1| < |-2|
Or 1 < 2, True

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by [email protected] » Thu May 26, 2016 2:43 pm
(y - 3)Â² > (x - 3)Â²

(y - 3)Â² - (x - 3)Â² > 0

Now use the difference of squares:

((y - 3) + (x - 3)) * ((y - 3) - (x - 3)) > 0

(x + y - 6) * (y - x) > 0

We have two sets of solutions here: either both terms are positive, or both terms are negative.

If both terms are positive, we have

(x + y - 6) > 0 and (y - x) > 0, which gives y > x and x + y > 6.

If both terms are negative, we have

0 > (x + y - 6) and 0 > (y - x), which gives 6 > (x + y) and x > y.

Since we get conflicting results, we can't answer.

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by a_new_start » Fri Aug 26, 2016 12:12 pm
[email protected] wrote:(y - 3)Â² > (x - 3)Â²

(y - 3)Â² - (x - 3)Â² > 0

Now use the difference of squares:

((y - 3) + (x - 3)) * ((y - 3) - (x - 3)) > 0

(x + y - 6) * (y - x) > 0

We have two sets of solutions here: either both terms are positive, or both terms are negative.

If both terms are positive, we have

(x + y - 6) > 0 and (y - x) > 0, which gives y > x and x + y > 6.

If both terms are negative, we have

0 > (x + y - 6) and 0 > (y - x), which gives 6 > (x + y) and x > y.

Since we get conflicting results, we can't answer.
Thanks Matt

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by GMATGuruNY » Fri Aug 26, 2016 12:29 pm
|a-b| = the DISTANCE between a and b.

Statement 2: (x-3)Â² < (y-3)Â²

The inequality above implies the following:
|x-3| < |y-3|.
In words:
The distance x and 3 is less than the distance between y and 3.
It's possible that x=2 and y=5, in which case x<y.
It's possible that x=4 and y=1, in which case x>y.
INSUFFICIENT.
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by [email protected] » Sat Aug 27, 2016 3:25 pm
Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?

In case you need it, here is a full solution:

We are given that x and y are positive and we need to determine whether x is less than y.

Statement One Alone:

âˆšx < âˆšy

Using the information from statement one, we can determine that x is less than y. Since x and y are both positive and the square root of x is less than the square root of y, we know that x must be less than y. Statement one alone is sufficient. Eliminate answer choices B, C and E.

Statement Two Alone:

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

Using the information in statement two, we cannot determine whether x is less than y.

For example, if x = 2 and y = 5, we see that (2 - 3)^2 = (-1)^2 = 1 is less than (5 - 3)^2 = (2)^2 = 4 and x is less than y.

However, if x = 2 and y = 1, we see that (2 - 3)^2 = (-1)^2 = 1 is also less than (1 - 3)^2 = (-2)^2 = 4 but x is greater than y. Statement two alone is not sufficient.

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by [email protected] » Thu Sep 01, 2016 6:20 pm
a_new_start wrote:Thanks Matt
No prob! I could've been more concise at the end, though. Once we reach this step:
If both terms are positive, we have

(x + y - 6) > 0 and (y - x) > 0, which gives y > x and x + y > 6.

If both terms are negative, we have

0 > (x + y - 6) and 0 > (y - x), which gives 6 > (x + y) and x > y.

Since we get conflicting results, we can't answer.
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