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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: 2734

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Location: New York City
x and y are consecutive positive integers and x>y. X^2 - [#permalink]

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06 Nov 2007, 22:49
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 represents all the possible values of Y?

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

Kudos [?]: 1045 [0], given: 4

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1045 [0], given: 4

Location: New York City

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06 Nov 2007, 23:01
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?????

Kudos [?]: 1045 [0], given: 4

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 166 [0], given: 0

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07 Nov 2007, 00:15
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

Kudos [?]: 166 [0], given: 0

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 842 [0], given: 19

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07 Nov 2007, 01:34
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

Kudos [?]: 842 [0], given: 19

Current Student
Joined: 28 Dec 2004
Posts: 3351

Kudos [?]: 319 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

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07 Nov 2007, 08:11
I agree y>0

try any value of y..i.e 1/2 /1/4..5 it holds

Kudos [?]: 319 [0], given: 2

SVP
Joined: 28 Dec 2005
Posts: 1545

Kudos [?]: 178 [0], given: 2

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02 Dec 2007, 16:04
i get until x(x-1) > y(y-4) .... how do you get to the next step ?

Kudos [?]: 178 [0], given: 2

Senior Manager
Joined: 06 Aug 2007
Posts: 360

Kudos [?]: 35 [0], given: 0

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02 Dec 2007, 18:31
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.

Kudos [?]: 35 [0], given: 0

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 166 [0], given: 0

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03 Dec 2007, 00:00
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.

Kudos [?]: 166 [0], given: 0

Manager
Joined: 03 Sep 2006
Posts: 233

Kudos [?]: 25 [0], given: 0

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03 Dec 2007, 00:36
Ans B (the same method as GMAT TIGER)

Kudos [?]: 25 [0], given: 0

SVP
Joined: 28 Dec 2005
Posts: 1545

Kudos [?]: 178 [0], given: 2

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15 Dec 2007, 12:16

Ended up with B. woo hoo

Kudos [?]: 178 [0], given: 2

15 Dec 2007, 12:16
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