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If x>0 on the number line, is x>y?

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If x>0 on the number line, is x>y? [#permalink]

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If x>0 on the number line, is x>y?

1) |x-y|>|y|
2) |x-y|=3
[Reveal] Spoiler: OA

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Re: If x>0 on the number line, is x>y? [#permalink]

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New post 21 Oct 2016, 09:55
idontknowwhy94 wrote:
If x>0 on the number line, is x>y?

1) |x-y|>|y|
2) |x-y|=3


1) square both sides
x^2+y^2-2xy>y^2
x^2-2xy>0
x(x-2y)>0
As x>0 then x-2y>0 or x>2y -------->x>y --------->suff..

2) if x=4,y=1------>x>y------>Yes

if x=-4, y=-1------->y>x ------>No

insuff....

Ans A

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If x>0 on the number line, is x>y? [#permalink]

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New post 24 Oct 2016, 22:49
A Sufficient

B Not sufficient
|x-y|= 3
x=3 y= 6 x<y
x=4 y=1 x>y

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Re: If x>0 on the number line, is x>y? [#permalink]

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New post 24 Oct 2016, 22:56
anshuldsharma wrote:
Why is B not sufficient?

x>0 so x cannot be negative. Hence , |x-y|= 3 will always result in x>y.
Any pointers?

Ans: D



|x-y| is always >=0

if x=2 & y=5
|x-y| =|2-5|=3-------->y>x

but if x=5 & y=2 then
|x-y|=|5-2|=3----------->x>y

considering two cases Option B is not sufficient......

Hope it helps
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Re: If x>0 on the number line, is x>y? [#permalink]

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New post 07 Dec 2017, 22:36
idontknowwhy94 wrote:
If x>0 on the number line, is x>y?

1) |x-y|>|y|
2) |x-y|=3


So x is positive, hence it lies to the right of zero on the number line. We need to know whether x > y or not (whether the distance between x and 0 is greater than the distance between y and 0 on the number line or not). Now on the number line, if y lies to the right of x, then y > x but if y lies to the left of x, then x > y.

(1) |x-y|>|y|
This states that the distance between x and y on the number line is more than the distance between y and 0. Now we are already given that x lies to the right of 0. If y further lies to the right of x, then distance between x and y can never be more than distance between y and 0. Illustration:

...0.....x....y . as we can see, since y lies to the right of x, distance between y and 0 will automatically become more than distance between x and y. BUT if y lies to the left of x, then distance between x and y CAN be greater than distance between y and 0. Illustration:

...y....0....x OR ....0....y......x. If y is negative, then automatically distance between x and y will always be greater than that between y and 0. And if y is also positive but less than x, its still possible for distance between y and x to be greater than distance between 0 and y.

so if x > 0, then the only way for |x-y|>|y| to be true if when x lies to the right of y, or when x > y. So first statement is sufficient.


(2) |x-y|=3
This states that distance between x and y on the number line is '3'. But this could happen both where x > y (eg, x=6, y=3) or where y > x (eg, x=3, y=6). So we cant say. Insufficient.


Hence A answer

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Re: If x>0 on the number line, is x>y?   [#permalink] 07 Dec 2017, 22:36
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