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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I will focus just on the statement 2: (x - 3)^2 < (y-3)^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 < (y-3)^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?
- OptimusPrep
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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)² > (x - 3)²
(y - 3)² - (x - 3)² > 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)² - (x - 3)² > 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 MattMatt@VeritasPrep wrote:(y - 3)² > (x - 3)²
(y - 3)² - (x - 3)² > 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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|a-b| = the DISTANCE between a and b.
Statement 2: (x-3)² < (y-3)²
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.
Statement 2: (x-3)² < (y-3)²
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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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:
√x < √y
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.