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

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Senior Manager
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This problem has been recently posted by halle i believe, I [#permalink]  26 Jun 2004, 17:33
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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

Senior Manager
Joined: 21 Mar 2004
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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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ash
________________________
I'm crossing the bridge.........

Manager
Joined: 07 May 2004
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Location: Ukraine, Russia(part-time)
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Kudos [?]: 9 [0], given: 0

Re: DS101 [#permalink]  27 Jun 2004, 23:15
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

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.
Manager
Joined: 07 May 2004
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Location: Ukraine, Russia(part-time)
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Re: DS101 [#permalink]  27 Jun 2004, 23:22
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.
Senior Manager
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Posts: 358
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Re: DS101 [#permalink]  28 Jun 2004, 05:31
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

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
Senior Manager
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

Kudos [?]: 9 [0], given: 0

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.
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 9 [0], given: 0

Re: DS101 [#permalink]  28 Jun 2004, 05:39
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

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...
Manager
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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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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.
_________________

ash
________________________
I'm crossing the bridge.........

Senior Manager
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

Kudos [?]: 9 [0], given: 0

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.
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 9 [0], given: 0

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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Ash,

I have a question based on your explaination.Why do you presume for (2) that x<y will not hold true because |x|<|y| always?What is the relationship b/w both sets of inequality?How did you draw the // that x>y 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 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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