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# x>y x^2 - 1> y^2 -4y +x - 1 Which of the following

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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, 10:30
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.
Director
Joined: 05 Jan 2008
Posts: 707
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Re: inequalities [#permalink]  31 Jan 2008, 03:54
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
_________________

Persistence+Patience+Persistence+Patience=G...O...A...L

Director
Joined: 01 Jan 2008
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Re: inequalities [#permalink]  31 Jan 2008, 06:25
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

Manager
Joined: 02 Jan 2008
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Re: inequalities [#permalink]  31 Jan 2008, 09:39
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
Intern
Joined: 03 Mar 2007
Posts: 31
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Re: inequalities [#permalink]  31 Jan 2008, 13:50
1
KUDOS
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
CEO
Joined: 29 Mar 2007
Posts: 2585
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Re: inequalities [#permalink]  31 Jan 2008, 15:46
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
Manager
Joined: 02 Jan 2008
Posts: 159
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Re: inequalities [#permalink]  31 Jan 2008, 19:01
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.
CEO
Joined: 29 Mar 2007
Posts: 2585
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Kudos [?]: 243 [0], given: 0

Re: inequalities [#permalink]  31 Jan 2008, 19:07
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
VP
Joined: 22 Nov 2007
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Re: inequalities [#permalink]  18 Mar 2008, 05:19
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
CEO
Joined: 29 Mar 2007
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Re: inequalities [#permalink]  18 Mar 2008, 07:37
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
Manager
Joined: 11 Apr 2008
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Schools: Kellogg(A), Wharton(W), Columbia(D)
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Re: inequalities [#permalink]  27 May 2008, 20:46
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.
CEO
Joined: 29 Mar 2007
Posts: 2585
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Kudos [?]: 243 [0], given: 0

Re: inequalities [#permalink]  28 May 2008, 07:53
Wow. looking back at these older problems. Def answer is B.

Just try values. y cannot be 0.
Re: inequalities   [#permalink] 28 May 2008, 07:53
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