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

Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?


Reply needed ASAP!

Thank you in advance.

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?

OptimusPrep wrote:

Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?


Reply needed ASAP!

Thank you in advance.

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?

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

Case 1: x = 1, y = -1, x > y

Hi Optimus,

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

Please explain.

Many thanks in advance.

SJ

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?

Please explain.

Many thanks in advance.

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

(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.

Matt@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.

Thanks Matt

|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.

Neilsheth2 wrote:Hi can any one help me how is Statement 2 inssuff?


Reply needed ASAP!

Thank you in advance.

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:

√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

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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