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# x and y are consecutive positive integers and: \left{

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Senior Manager
Joined: 05 Oct 2008
Posts: 272
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Kudos [?]: 117 [0], given: 22

x and y are consecutive positive integers and: \left{ [#permalink]  25 Oct 2008, 20:16
$x$ and $y$ are consecutive positive integers and:
$\left{ \begin{eqnarray*} x &>& y\\ x^2 - 1 &>& y^2 - 4y + x - 1\\ \end{eqnarray*}$

Which of the following represents all the possible values of $y$ ?

* $y \ge 0$
* $y > 0$
* $y > 1$
* $y > 7$
* $y > 8$
Senior Manager
Joined: 05 Oct 2008
Posts: 272
Followers: 3

Kudos [?]: 117 [0], given: 22

Re: Math Question [#permalink]  25 Oct 2008, 20:50
Mostly, everyone who posts questions on this website has an answer to the question, unless specified otherwise. Please provide an explanation to the answer as a mere Answer Choice in the reply is not of much help.

Thanks!!
Director
Joined: 23 May 2008
Posts: 840
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Kudos [?]: 37 [0], given: 0

Re: Math Question [#permalink]  25 Oct 2008, 21:41
study wrote:
Mostly, everyone who posts questions on this website has an answer to the question, unless specified otherwise. Please provide an explanation to the answer as a mere Answer Choice in the reply is not of much help.

Thanks!!

x and y are positive and consecutive integers. I just plugged in values that satisfied those conditions. lowest value for Y is 1
Manager
Joined: 23 Aug 2008
Posts: 64
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Kudos [?]: 49 [0], given: 0

Re: Math Question [#permalink]  26 Oct 2008, 05:42
study wrote:
$x$ and $y$ are consecutive positive integers and:
$\left{ \begin{eqnarray*} x &>& y\\ x^2 - 1 &>& y^2 - 4y + x - 1\\ \end{eqnarray*}$

Which of the following represents all the possible values of $y$ ?

* $y \ge 0$
* $y > 0$
* $y > 1$
* $y > 7$
* $y > 8$

No creative solution from me I'm sorry, I'm also going for B.

Explanation:
By definition of condition 1, x>y, therefore x=y+1 (since they are consecutive)

I then just plugged in values, starting from y=1, x=2; y=1 seems to work, therefore y>0 is my answer. Note I start from y=1, as y=0 is not possible due to question conditions (y is positive integer)

Seems a bit too straight forward though, what did I miss?
VP
Joined: 30 Jun 2008
Posts: 1048
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Kudos [?]: 345 [1] , given: 1

Re: Math Question [#permalink]  26 Oct 2008, 06:32
1
KUDOS
study wrote:
$x$ and $y$ are consecutive positive integers and:
$\left{ \begin{eqnarray*} x &>& y\\ x^2 - 1 &>& y^2 - 4y + x - 1\\ \end{eqnarray*}$

Which of the following represents all the possible values of $y$ ?

* $y \ge 0$
* $y > 0$
* $y > 1$
* $y > 7$
* $y > 8$

here we do not need to plug in numbers

Ok.... now we know x>0 and y>0 and x,y are consecutive

we know x>y ..... this means x=y+1 ( x,y consecutive)

x² -1 > y²-4y+x-1
(y+1)² -1 > y²-4y+ (y+1) -1 -------- [ here i have substituted x=y+1]
y²+2y+1 -1 > y²-4y +y+1-1
y²+2y > y²-4y +y
y²+2y > y²-3y
2y > -3y
5y >0
y>0

SO B

Hope this helps
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Manager
Joined: 23 Aug 2008
Posts: 64
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Kudos [?]: 49 [0], given: 0

Re: Math Question [#permalink]  26 Oct 2008, 07:33
amitdgr wrote:
here we do not need to plug in numbers

Ok.... now we know x>0 and y>0 and x,y are consecutive

we know x>y ..... this means x=y+1 ( x,y consecutive)

x² -1 > y²-4y+x-1
(y+1)² -1 > y²-4y+ (y+1) -1 -------- [ here i have substituted x=y+1]
y²+2y+1 -1 > y²-4y +y+1-1
y²+2y > y²-4y +y
y²+2y > y²-3y
2y > -3y
5y >0
y>0

SO B

Hope this helps

Thanks, that solution seems more logical.
Manager
Joined: 15 Oct 2008
Posts: 54
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Kudos [?]: 2 [0], given: 0

Re: Math Question [#permalink]  28 Oct 2008, 05:28
WOW !! thank you amitdgr..
Senior Manager
Joined: 05 Oct 2008
Posts: 272
Followers: 3

Kudos [?]: 117 [0], given: 22

Re: Math Question [#permalink]  29 Oct 2008, 07:54
Amitgdr very good explanation - kudos given. thanks.
Re: Math Question   [#permalink] 29 Oct 2008, 07:54
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