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# x and y are consecutive positive integers and x>y. X^2 -

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CEO
Joined: 21 Jan 2007
Posts: 2760
Location: New York City
Followers: 9

Kudos [?]: 387 [0], given: 4

x and y are consecutive positive integers and x>y. X^2 - [#permalink]  06 Nov 2007, 21:49
x and y are consecutive positive integers and x>y.

X^2 - 1 > Y^2 - 4y + x - 1

Which represents all the possible values of Y?

y>=0
y>0
y>1
y>7
y>8
CEO
Joined: 21 Jan 2007
Posts: 2760
Location: New York City
Followers: 9

Kudos [?]: 387 [0], given: 4

Re: inequalities [#permalink]  06 Nov 2007, 22:01
bmwhype2 wrote:
x and y are consecutive positive integers and x>y

X^2 - 1 > Y^2 - 4y + x - 1

Which represents all the possible values of Y?

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

simplifying the equation we get
X(x+1) > y(y+4)
(y+1) (y+1-1)> y(y-4)
y+1 > y-4
1 > 4

can someone elaborate on why y+1 > y-4 becomes 1 > 4?????
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

Re: inequalities [#permalink]  06 Nov 2007, 23:15
bmwhype2 wrote:
bmwhype2 wrote:
x and y are consecutive positive integers and x>y

X^2 - 1 > Y^2 - 4y + x - 1

Which represents all the possible values of Y?

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

simplifying the equation we get
X(x+1) > y(y+4)
(y+1) (y+1-1)> y(y-4)
y+1 > y-4
1 > 4

can someone elaborate on why y+1 > y-4 becomes 1 > 4?????

We have 1 > -4 not 1 > 4.... That means it's always true for y > 0
SVP
Joined: 29 Aug 2007
Posts: 2493
Followers: 59

Kudos [?]: 576 [0], given: 19

Re: inequalities [#permalink]  07 Nov 2007, 00:34
bmwhype2 wrote:
bmwhype2 wrote:
x and y are consecutive positive integers and x>y

X^2 - 1 > Y^2 - 4y + x - 1

Which represents all the possible values of Y?

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

simplifying the equation we get
X(x+1) > y(y+4)
(y+1) (y+1-1)> y(y-4)
y+1 > y-4
1 > 4

can someone elaborate on why y+1 > y-4 becomes 1 > 4?????

x^2 - 1 > y^2 - 4y + x - 1
x^2 - x > y^2 - 4y
x (x-1) > y (y-4)
(y+1) (y+1-1)> y(y-4)
y(y+1) > y(y-4)
y(y+1) - y(y-4) > 0
y (y+1-y+4) > 0
y (y+1-y+4) > 0
y (5) > 0
y > 0
Current Student
Joined: 28 Dec 2004
Posts: 3387
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 14

Kudos [?]: 186 [0], given: 2

I agree y>0

try any value of y..i.e 1/2 /1/4..5 it holds
SVP
Joined: 28 Dec 2005
Posts: 1579
Followers: 2

Kudos [?]: 91 [0], given: 2

i get until x(x-1) > y(y-4) .... how do you get to the next step ?
Senior Manager
Joined: 06 Aug 2007
Posts: 368
Followers: 1

Kudos [?]: 23 [0], given: 0

pmenon wrote:
i get until x(x-1) > y(y-4) .... how do you get to the next step ?

pmenon..since x>y and they are consecutive so x = y + 1.
SVP
Joined: 01 May 2006
Posts: 1798
Followers: 8

Kudos [?]: 102 [0], given: 0

Re: inequalities [#permalink]  02 Dec 2007, 23:00
bmwhype2 wrote:
Fig wrote:
bmwhype2 wrote:
bmwhype2 wrote:
x and y are consecutive positive integers and x>y

X^2 - 1 > Y^2 - 4y + x - 1

Which represents all the possible values of Y?

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

simplifying the equation we get
X(x+1) > y(y+4)
(y+1) (y+1-1)> y(y-4)
y+1 > y-4
1 > 4

can someone elaborate on why y+1 > y-4 becomes 1 > 4?????

We have 1 > -4 not 1 > 4.... That means it's always true for y > 0

why?

Because:
o The domain of definition for y is "all positive integers"... So y > 0
o 1 > -4 remains valid all time... So on the domain of definition for y.

Manager
Joined: 03 Sep 2006
Posts: 233
Followers: 1

Kudos [?]: 7 [0], given: 0

Ans B (the same method as GMAT TIGER)
SVP
Joined: 28 Dec 2005
Posts: 1579
Followers: 2

Kudos [?]: 91 [0], given: 2

Ended up with B. woo hoo
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