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m02#8

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m02#8 [#permalink] New post 03 Nov 2008, 15:24
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If x and y are consecutive integers (x>y) and x^2-1>y^2-4y+x-1, then which of the following must be true?

A. y\geq{0}
B. y>0
C. y>1
D. y>7
E. y>8

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Last edited by Bunuel on 05 Nov 2013, 06:04, edited 1 time in total.
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Re: m02#8 [#permalink] New post 03 Nov 2008, 21:09
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Although the inequality looks complex, if we replace x with (y+1) in the inequality then we get y > 0.
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Re: m02#8 [#permalink] New post 04 Nov 2008, 08:12
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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
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Re: m02#8 [#permalink] New post 04 Nov 2010, 05:55
x^2-1=y^2-4y+x-1 => x^2-x=y^2-4y => x^2-x+\frac{1}{4}-\frac{1}{4}=y^2-4y+4-4 => (x-\frac{1}{2})^2-\frac{1}{4}=(y-2)^2-4 => (x-\frac{1}{2})^2>(y-2)^2-\frac{15}{4}

since y is the smaller positive integer than x, the above is always true. so y can be any number larger than zero (option E is a subset of option D; option D is a subset of option C; option C is a subset of option B).

an alternative way to solve this is as follows:

since y is smaller than x, but larger than zero, option A can be deleted.

then since the question asks for all possible numbers of y, it could be a smaller one rather than a larger value. we pick from option C as a start. if C fails, we move to B. if C works ok, we move to D.
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Re: m02#8 [#permalink] New post 04 Nov 2010, 10:53
bigfernhead wrote:
If x and y are consecutive positive integers, and:
\left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}

which of the following represents all the possible values of y ?

(A) y \ge 0
(B) y \gt 0
(C) y \gt 1
(D) y \gt 7
(E) y \gt 8

[Reveal] Spoiler: OA
B

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I picked numbers to solve this equation.

The first thing I did was change the 2nd equation to look like this so that all of the variables are isolated on each side.

(x^2) - x - 1 > (y^2) - 4y - 1

Then I picked the numbers in the answer boxes to represent Y. Since it is asking for all of the possible solutions I started off with the smallest value for Y.

If Y = 0; then x has to equal 1 to satisfy the first equation x>y and is consecutive. This did not work because after plugging the numbers into the 2nd equation I got -1>0 which is not true. Cross out A.

Then I tried Y = 1; x = 2. This worked so all of the other answers will work too. I did try Y = 7; x = 8 just to be 100% sure though.
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Re: m02#8 [#permalink] New post 04 Nov 2010, 16:05
sonnco wrote:
I picked numbers to solve this equation.

The first thing I did was change the 2nd equation to look like this so that all of the variables are isolated on each side.

(x^2) - x - 1 > (y^2) - 4y - 1

Then I picked the numbers in the answer boxes to represent Y. Since it is asking for all of the possible solutions I started off with the smallest value for Y.

If Y = 0; then x has to equal 1 to satisfy the first equation x>y and is consecutive. This did not work because after plugging the numbers into the 2nd equation I got -1>0 which is not true. Cross out A.

Then I tried Y = 1; x = 2. This worked so all of the other answers will work too. I did try Y = 7; x = 8 just to be 100% sure though.
it's a quicker way, but i think you don't even have to try y=0 because y is said to be a positive interger, which means that y is at least +1.
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Re: m02#8 [#permalink] New post 21 Nov 2010, 16:42
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Has to be B. Here is the most simple reason why.

X and Y are consecutive integers. Therefore if Y = 0 then X has to equal 1. So if Y = 0 how could the second statement be true? O is not greater than 1. Therefore Y must be > 0. B.
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Re: m02#8 [#permalink] New post 12 Dec 2010, 19:03
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As x > y and x,y are consecutive

x = y+1

ans - y > 0
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Re: m02#8 [#permalink] New post 04 Feb 2011, 03:04
Can somebody explain why im not getting results?
below is my solution :

\left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}

initially i am adding 2 inequalities

x2 -1+x-x+1>y2-4y+y
x2>y2-3y
x=y+1
y2+2y+1>y2-3y

is it wrong to add inequalities above?
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Re: m02#8 [#permalink] New post 04 Apr 2011, 11:53
tinki wrote:
tinki wrote:
Can somebody explain why im not getting results?
below is my solution :

\left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}

initially i am adding 2 inequalities

x2 -1+x-x+1>y2-4y+y
x2>y2-3y
x=y+1
y2+2y+1>y2-3y

is it wrong to add inequalities above?



Anybody? why i cent get final answer with above? thx for feedback


Interesting. I am not too sure about my explanation, but just to add my thinking.

x is a greater value than y; i.e. 1 more;
if you add a greater value to the LHS and a smaller value to RHS, the inequality will still be correct but the range of the inequality changes.

e.g.
x>10
5>2
x+5>10+2
x>7

Thus, there is nothing wrong with the inequality but as we can see, the range changed.
We started with; x>10
Now, x>7.
Similar thing is happening in your example.

Well, the interesting thing to note is that; whether y>(-1/5) or y>0; the least value for y is going to be 1. So far; y>=1, we have the entire range.
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Re: m02#8 [#permalink] New post 13 Nov 2011, 16:43
B. Simplify to x (X - 1) > y (y - 4) and plug in numbers for Y from answers with the restriction that X>Y and x and y are consecutive positive integers.
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Re: m02#8 [#permalink] New post 18 Nov 2011, 02:37
if we replace x by Y+1 in inequality equation then we get 5y>0, hence Y>0.
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Re: m02#8 [#permalink] New post 03 Jun 2012, 05:14
The OA states that "since x and y are consecutive positive integers (...)" but this is not stated by the question, it is actually our expected conclusion. In my view the OA should remove "positive" to avoid confusion
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Re: m02#8 [#permalink] New post 03 Jun 2012, 05:44
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J82 wrote:
The OA states that "since x and y are consecutive positive integers (...)" but this is not stated by the question, it is actually our expected conclusion. In my view the OA should remove "positive" to avoid confusion


Actually it is: "If x and y are consecutive positive integers, and ..."
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Re: m02#8 [#permalink] New post 16 Jun 2012, 09:21
Substituting x=y+1 into the second equations renders 5y>0, which gives y>0

Therefore B.
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Re: m02#8 [#permalink] New post 08 Nov 2012, 06:34
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bigfernhead wrote:
If x and y are consecutive positive integers, and:
\left{ \begin{eqnarray*} x &\gt& y\\ x^2 - 1 &\gt& y^2 - 4y + x - 1\\ \end{eqnarray*}

which of the following represents all the possible values of y ?

(A) y \ge 0
(B) y \gt 0
(C) y \gt 1
(D) y \gt 7
(E) y \gt 8

[Reveal] Spoiler: OA
B

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BELOW IS REVISED VERSION OF THIS QUESTION:
If x and y are consecutive integers (x>y) and x^2-1>y^2-4y+x-1, then which of the following must be true?

A. y\geq{0}
B. y>0
C. y>1
D. y>7
E. y>8

Since x and y are consecutive positive integers and x>y then x=y+1. Substitute x with y+1 in the given equation: (y+1)^2-1>y^2-4y+(y+1)-1 --> 5y>0 --> y>0.

Answer: B.
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Re: m02#8 [#permalink] New post 16 Dec 2013, 19:48
i feel like they forgot to mention that they are positive integers. Wait actually that wouldn't matter because only dividing and multiplying with negative changes the inequality.
Re: m02#8   [#permalink] 16 Dec 2013, 19:48
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