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x and y are consecutive positive integers and: x > y ,

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Joined: 12 Feb 2008
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x and y are consecutive positive integers and: x > y , [#permalink]

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29 Oct 2008, 07:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

x and y are consecutive positive integers and:
x > y , x^2 - 1 > y^2 - 4y + x - 1
Which of the following represents all the possible values of Y?

* y >=0
* y > 0
* y > 1
* y > 7
* y > 8

SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 67

Kudos [?]: 707 [0], given: 19

Re: consecutive positive integers [#permalink]

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29 Oct 2008, 08:39
elmagnifico wrote:
x and y are consecutive positive integers and:
x > y , x^2 - 1 > y^2 - 4y + x - 1
Which of the following represents all the possible values of Y?

* y>=0
* y > 0
* y > 1
* y > 7
* y > 8

x^2 - 1 > y^2 - 4y + x - 1
x^2 - x > y^2 - 4y
x (x - 1) > y (y - 4)
since x and y are +ve consecutive integers, y = x - 1 or x = y + 1

(y+1) y > y (y - 4)
y + 1 > y - 4
y > (y - 5)

what value for y work in the above expression? anything but y should be a +ve. so y has to be at least 1 or y > 0.

But I am still missing a more succint way to solve this problem.
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Joined: 05 Jul 2006
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Kudos [?]: 256 [0], given: 39

Re: consecutive positive integers [#permalink]

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29 Oct 2008, 09:45
[quote="elmagnifico"]x and y are consecutive positive integers and:
x > y , x^2 - 1 > y^2 - 4y + x - 1

Which of the following represents all the possible values of Y?

y >=0
y > 0
y > 1
y > 7
y > 8

(x-1)(x+1)-x > (y-1)(y+1)-4y

x is = (y-1)) as x>y AND THEY ARE CONSECS

thus

(y-2)(y)> (y-1)(y+1)-4y

5*3>4*6 - 4*5.....true

0> -1*1 true, BUT GIVEN IS y+VE THUS

I will go with Y>0
VP
Joined: 30 Jun 2008
Posts: 1043
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Re: consecutive positive integers [#permalink]

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29 Oct 2008, 10:01
check this thread http://gmatclub.com/forum/post532313.html#p532313
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Re: consecutive positive integers [#permalink]

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29 Oct 2008, 10:14
elmagnifico wrote:
x and y are consecutive positive integers and:
x > y , x^2 - 1 > y^2 - 4y + x - 1
Which of the following represents all the possible values of Y?
* y >=0
* y > 0
* y > 1
* y > 7
* y > 8

lets see x>y then x=y+1

x^2 - 1= (x-1)(x+1) => (y+1-1)(Y+1+1)=y(Y+2) or y^2 +2y>y^2-4y+y+1-1

5y>0 which means y>0 , however, the stems says x and y are consecutive POSITIVE intgers..therefore y>0 or y=1 ..or 2..
Re: consecutive positive integers   [#permalink] 29 Oct 2008, 10:14
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x and y are consecutive positive integers and: x > y ,

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