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This problem has been recently posted by halle i believe, I

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lastochka
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This problem has been recently posted by halle i believe, I [#permalink] New post   26 Jun 2004, 18:33
00:00
A
B
C
D
E

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(N/A)

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This problem has been recently posted by halle i believe, I couldn't trace the orginal post, so I'm reposting it in effort to clarify an issue:

[x]>=[x-y]+[y], is y>x?

1. x>0
2. y>0

explain your solution



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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   27 Jun 2004, 23:56
Is this a greatest integer function problem or a mod problem ?

for mod problems I would just take appropriate values and use. However not all times we can use only values.......we might have to use a trick here and there.

- let me know b4 i can give it a try

- ash
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 00:15
lastochka wrote:
This problem has been recently posted by halle i believe, I couldn't trace the orginal post, so I'm reposting it in effort to clarify an issue:

[x]>=[x-y]+[y], is y>x?

1. x>0
2. y>0

explain your solution


From |x| >= |x-y| + |y| we get that |x| = |x-y| + |y|, because actually for every x and y it is true that |x| <= |x-y| + |y|(this can be proven...)!

Then x, y, and x - y have the same sign (from =).

1 is sufficient: if x > 0, then x - y >= 0, then y < x is not true.

2 is sufficient too because the same is true for x - y >= 0.

C.
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 00:22
You may ask, why from |x| = |x-y| + |y| => x, x-y and y have the same sign?

It follows from:

(|x|-|y|)^2 = |x-y|^2 => -2*|x|*|y| = -2*x*y => x*y >= 0. And if x > 0, then y >= 0.

etc.
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 06:31
Emmanuel wrote:
lastochka wrote:
This problem has been recently posted by halle i believe, I couldn't trace the orginal post, so I'm reposting it in effort to clarify an issue:

[x]>=[x-y]+[y], is y>x?

1. x>0
2. y>0

explain your solution


From |x| >= |x-y| + |y| we get that |x| = |x-y| + |y|, because actually for every x and y it is true that |x| <= |x-y| + |y|(this can be proven...)!

Then x, y, and x - y have the same sign (from =).

1 is sufficient: if x > 0, then x - y >= 0, then y < x is not true.

2 is sufficient too because the same is true for x - y >= 0.

C.


since both are sufficient you mean the answer is D
other than that, I agree with your solution
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 06:33
ashkg wrote:
Is this a greatest integer function problem or a mod problem ?

for mod problems I would just take appropriate values and use. However not all times we can use only values.......we might have to use a trick here and there.

- let me know b4 i can give it a try

- ash


not sure I understand the difference in your question ash. This is an absolute values problem.
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 06:39
lastochka wrote:
Emmanuel wrote:
lastochka wrote:
This problem has been recently posted by halle i believe, I couldn't trace the orginal post, so I'm reposting it in effort to clarify an issue:

[x]>=[x-y]+[y], is y>x?

1. x>0
2. y>0

explain your solution


From |x| >= |x-y| + |y| we get that |x| = |x-y| + |y|, because actually for every x and y it is true that |x| <= |x-y| + |y|(this can be proven...)!

Then x, y, and x - y have the same sign (from =).

1 is sufficient: if x > 0, then x - y >= 0, then y < x is not true.

2 is sufficient too because the same is true for x - y >= 0.

C.


since both are sufficient you mean the answer is D
other than that, I agree with your solution


Yes, lastochka, I don't remember exact definitions for A,B,C,D,E, but I know the solution...
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 06:40
lastochka wrote:
not sure I understand the difference in your question ash. This is an absolute values problem.


lastochka, ashkg want to say that [x] nay mean greatest integer, which is less than x.
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 09:42
lastochka wrote:
ashkg wrote:
Is this a greatest integer function problem or a mod problem ?

for mod problems I would just take appropriate values and use. However not all times we can use only values.......we might have to use a trick here and there.

- let me know b4 i can give it a try

- ash


not sure I understand the difference in your question ash. This is an absolute values problem.


The notation used for absolute values(modulus func) is |x|
The notation used for greatest integer value of x is [x]

Thats my understanding which I hope is correct. I didnt want to solve the problem before knowing that :)

Here's my attempt to solve the problem.
Let E => |x| >= |x-y| + |y|

1. given x > 0

for all values of y>x, E will not hold true.

So y>x cannot be true.
So A is sufficient.


2. given y > 0

for all x, where x<y E wont hold true because |x|<|y| always.

for x>y, E will hold true.

Sor for E to hold true, x>y must be true. So y>x is not true. B is sufficient.

MY ans is D.
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 14:19
ashkg wrote:
lastochka wrote:
ashkg wrote:
Is this a greatest integer function problem or a mod problem ?

for mod problems I would just take appropriate values and use. However not all times we can use only values.......we might have to use a trick here and there.

- let me know b4 i can give it a try

- ash


not sure I understand the difference in your question ash. This is an absolute values problem.


The notation used for absolute values(modulus func) is |x|
The notation used for greatest integer value of x is [x]

Thats my understanding which I hope is correct. I didnt want to solve the problem before knowing that :)

Here's my attempt to solve the problem.
Let E => |x| >= |x-y| + |y|

1. given x > 0

for all values of y>x, E will not hold true.

So y>x cannot be true.
So A is sufficient.


2. given y > 0

for all x, where x<y E wont hold true because |x|<|y| always.

for x>y, E will hold true.

Sor for E to hold true, x>y must be true. So y>x is not true. B is sufficient.

MY ans is D.


I couldn't figure out what key will produce "|" symbol (absolute value symbol). Which key is it on a keyboard? Thanks in advance.
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Jun 2004, 15:16
lastochka wrote:
I couldn't figure out what key will produce "|" symbol (absolute value symbol). Which key is it on a keyboard? Thanks in advance.


lastochka, this symbol appears when you press Shift and backslash (it is to the right from "backspace" key).
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Re: This problem has been recently posted by halle i believe, I [#permalink] New post   28 Dec 2004, 05:33
Ash,

I have a question based on your explaination.Why do you presume for (2) that xy will hold in this case?Tx.

Anna

ashkg wrote:
lastochka wrote:
ashkg wrote:
Is this a greatest integer function problem or a mod problem ?

for mod problems I would just take appropriate values and use. However not all times we can use only values.......we might have to use a trick here and there.

- let me know b4 i can give it a try

- ash


not sure I understand the difference in your question ash. This is an absolute values problem.


The notation used for absolute values(modulus func) is |x|
The notation used for greatest integer value of x is [x]

Thats my understanding which I hope is correct. I didnt want to solve the problem before knowing that :)

Here's my attempt to solve the problem.
Let E => |x| >= |x-y| + |y|

1. given x > 0

for all values of y>x, E will not hold true.

So y>x cannot be true.
So A is sufficient.


2. given y > 0

for all x, where xy, E will hold true.

Sor for E to hold true, x>y must be true. So y>x is not true. B is sufficient.

MY ans is D.




Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: This problem has been recently posted by halle i believe, I [#permalink]
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