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VP
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x>y x^2 - 1> y^2 -4y +x - 1 Which of the following [#permalink]
 27 Jan 2008, 11:30
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
Let us first simplify the inequality. Cancel out the (-1) from both sides and move the x over to the left hand side:
x^2- x>y^2 - 4y
from then I can't catch any passage of the explaantion:
x(x-1)>y(y-4)
(y+1)(y+1-1)>y(y-4)
1>4
then it is said that Then it is said in the explanation that when we divided both parts of the equation by , we assumed that it is greater than 0. Since at the end we are left without a variable, this means that the equation works for any value where y>0
what's the meaning of this? explanations appreciated.
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Director
Joined: 05 Jan 2008
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My approach to such questions would be get to the solving the equation, then you have x(x-1)>y(y-4) now using the values from the answer choices, if y>=0(let us take y=0) then x=1,y=0 1(0)>0(-4) ->False so 1 cannot be true i f y>0 then y=1 and x=2 2(2-1)>1(1-4)=>2>-3 =>True if y>1, then y=2 and x=3 3(2)>2(2-4) =>6>-4 => True if y>7, then y=8 and x=9 9(9-1)>8(8-4)=>72>32 =>True if y>8 then y=9 and x=10 10(9)>9(5)=>90>45 => True Thus OA would be B as for all the values of Y above 0 the equation holds good. Hope this helps. Regards, Prasanna
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Director
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marcodonzelli wrote: 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
As you mentioned the second inequality can be simplified to x(x-1) > y(y-4) 1 and 2 don't work because because (x,y) = (1/2,1/100) doesn't satisfy if y > 1, then since x > y and x(x-1) > y*(x-1) > y*(y-1)-3*y = y*(y-4) - true The answer is C.
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Manager
Joined: 02 Jan 2008
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Marco, I think you are missing a part of the question.
x and y are consecutive positive integers and x>y
--
So, we have: x=n+1, y=n
x^2-1>y^2-4y+x-1
(x+1)(x-1) - (x-1) >y^2-4y
(x-1)(x+1-1) >y(y-4)
(x-1)(x) >y(y-4)
sub now for x and y
(n+1-1)(n+1) >n(n-4)
n(n+1) > n(n-4)
n > -4n
5n > 0 => n>0 ; y>0
B
alt. you can sub n and n+1 directly in inequality
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Intern
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The fact that the equation can be broken down to x^2 - x > y^2 - 4y means that any positive number (any y>0) makes the equation on the right negative
so the answer must be B
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CEO
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srp wrote: Marco, I think you are missing a part of the question.
x and y are consecutive positive integers and x>y
--
So, we have: x=n+1, y=n
x^2-1>y^2-4y+x-1
(x+1)(x-1) - (x-1) >y^2-4y
(x-1)(x+1-1) >y(y-4)
(x-1)(x) >y(y-4)
sub now for x and y
(n+1-1)(n+1) >n(n-4)
n(n+1) > n(n-4)
n > -4n
5n > 0 => n>0 ; y>0
B
alt. you can sub n and n+1 directly in inequality how do you know x and y are consecutive + integers? If they aren't then I agree w/ Marco
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Manager
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GMATBLACKBELT wrote: srp wrote: Marco, I think you are missing a part of the question.
x and y are consecutive positive integers and x>y
--
So, we have: x=n+1, y=n
x^2-1>y^2-4y+x-1
(x+1)(x-1) - (x-1) >y^2-4y
(x-1)(x+1-1) >y(y-4)
(x-1)(x) >y(y-4)
sub now for x and y
(n+1-1)(n+1) >n(n-4)
n(n+1) > n(n-4)
n > -4n
5n > 0 => n>0 ; y>0
B
alt. you can sub n and n+1 directly in inequality how do you know x and y are consecutive + integers? If they aren't then I agree w/ Marco I know this because I faced this question in "GmatClub challenge-M02" day before y'day and this is how I solved.
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CEO
Joined: 29 Mar 2007
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srp wrote: GMATBLACKBELT wrote: srp wrote: Marco, I think you are missing a part of the question.
x and y are consecutive positive integers and x>y
--
So, we have: x=n+1, y=n
x^2-1>y^2-4y+x-1
(x+1)(x-1) - (x-1) >y^2-4y
(x-1)(x+1-1) >y(y-4)
(x-1)(x) >y(y-4)
sub now for x and y
(n+1-1)(n+1) >n(n-4)
n(n+1) > n(n-4)
n > -4n
5n > 0 => n>0 ; y>0
B
alt. you can sub n and n+1 directly in inequality how do you know x and y are consecutive + integers? If they aren't then I agree w/ Marco I know this because I faced this question in "GmatClub challenge-M02" day before y'day and this is how I solved. Oh 
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VP
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CaspAreaGuy wrote: junkbondswap wrote: The fact that the equation can be broken down to x^2 - x > y^2 - 4y means that any positive number (any y>0) makes the equation on the right negative
so the answer must be B Very good! Thanks yes, I didn't consider that x and y were consecutive positive integers
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CEO
Joined: 29 Mar 2007
Posts: 2618
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marcodonzelli wrote: 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
Let us first simplify the inequality. Cancel out the (-1) from both sides and move the x over to the left hand side:
x^2- x>y^2 - 4y
from then I can't catch any passage of the explaantion:
x(x-1)>y(y-4)
(y+1)(y+1-1)>y(y-4)
1>4
then it is said that Then it is said in the explanation that when we divided both parts of the equation by , we assumed that it is greater than 0. Since at the end we are left without a variable, this means that the equation works for any value where y>0
what's the meaning of this? explanations appreciated. You can try values. x>y. Thus lets say that x=.1 and y=0. This cannot work b/c the left side would be negative. Assume we have: x(x-1)>y(y-4) We need to make it so the left side has a the minimum value, while keeping it bigger than the right side. Make x=1. Thus the left side is 0. This is the min value we can create here for the left side. Since y cannot equal 0, the next best thing is y>0. thus B
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Manager
Joined: 11 Apr 2008
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marcodonzelli wrote: 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
Let us first simplify the inequality. Cancel out the (-1) from both sides and move the x over to the left hand side:
x^2- x>y^2 - 4y
from then I can't catch any passage of the explaantion:
x(x-1)>y(y-4)
(y+1)(y+1-1)>y(y-4)
1>4
then it is said that Then it is said in the explanation that when we divided both parts of the equation by , we assumed that it is greater than 0. Since at the end we are left without a variable, this means that the equation works for any value where y>3/2
what's the meaning of this? explanations appreciated. In m02 the explanation reads y>3/2 instead of y>0. I think its only a typo. Can someone pls confirm.
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CEO
Joined: 29 Mar 2007
Posts: 2618
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Wow. looking back at these older problems. Def answer is B.
Just try values. y cannot be 0.
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