Hi can any one help me how is Statement 2 inssuff?
Reply needed ASAP!
Thank you in advance.
Tricky DS ep2 if x and y are positive , is x< y?
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 Neilsheth2
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 OptimusPrep
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I will focus just on the statement 2: (x  3)^2 < (y3)^2Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?
Reply needed ASAP!
Thank you in advance.
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.
 Neilsheth2
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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.OptimusPrep wrote:I will focus just on the statement 2: (x  3)^2 < (y3)^2Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?
Reply needed ASAP!
Thank you in advance.
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?
so for statement 1 we can not since the root always has to be positive? Correct?
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Yes, absolutely CorrectNeilsheth2 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?
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Hi SJ,jain2016 wrote:Hi Optimus,Case 1: x = 1, y = 1, x > y
It is given that x and y are positive, then how come y= 1?
Please explain.
Many thanks in advance.
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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(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.
(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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Thanks Matt[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.
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ab = the DISTANCE between a and b.
Statement 2: (x3)Â² < (y3)Â²
The inequality above implies the following:
x3 < y3.
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.
Statement 2: (x3)Â² < (y3)Â²
The inequality above implies the following:
x3 < y3.
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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In case you need it, here is a full solution:Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?
Reply needed ASAP!
Thank you in advance.
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.
Answer:A
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No prob! I could've been more concise at the end, though. Once we reach this step:a_new_start wrote:Thanks Matt
Everything that isn't in bold is irrelevant clutter.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.