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M02-08

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New post 16 Sep 2014, 00:17
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If \(x\) and \(y\) are consecutive integers (\(x \gt y\)) and \(x^2-1 \gt y^2-4y+x-1\), then which of the following must be true?


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

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New post 16 Sep 2014, 00:17
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Official Solution:


If \(x\) and \(y\) are consecutive integers (\(x \gt y\)) and \(x^2-1 \gt y^2-4y+x-1\), then which of the following must be true?


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


Since \(x\) and \(y\) are consecutive integers and \(x \gt y\) then \(x=y+1\). Substitute \(x\) with \(y+1\) in the given equation:

\((y+1)^2-1 \gt y^2-4y+(y+1)-1\);

\(5y \gt 0\);

\(y \gt 0\).


Answer: B
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Re: M02-08  [#permalink]

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New post 09 Oct 2014, 05:21
Bunuel wrote:

Since x and y are consecutive positive integers and x \gt y then x=y+1.



In the question stem is only written consecutive integers. How can we tell that x and y are positive?
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New post 09 Oct 2014, 05:26
martinnotceoyet wrote:
Bunuel wrote:

Since x and y are consecutive positive integers and x \gt y then x=y+1.



In the question stem is only written consecutive integers. How can we tell that x and y are positive?


The word "positive" there was wrong. Edited. Check again.
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New post 09 Oct 2014, 05:39
Oh! Sweet thank you very much!
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Re: M02-08  [#permalink]

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New post 24 Mar 2015, 03:30
Hi,
Lets make this a little complicated. Assume that, in the given problem, its is just mentioned that " x and y are positive integers". Then the approach will be a little different.

LHS : x^2 - 1 = (x+1)(x-1)
Since x is positive, we will only consider (x-1)........Eq1
RHS : y^2 -4y +x-1........Eq2

Equating both: (x-1)>y^2 -4y +x-1
We get y(y-4)> 0
Therefore y>0 as the best case and must be true.
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Re: M02-08  [#permalink]

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New post 24 Mar 2015, 16:31
I agree that b must be true. However, i be is true, then A must also be true.
If y > 0, then it is also true that Y >= 0.
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New post 25 Mar 2015, 03:36
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New post 25 Mar 2015, 12:59
Bunuel wrote:
cledgard wrote:
I agree that b must be true. However, i be is true, then A must also be true.
If y > 0, then it is also true that Y >= 0.


That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.



This is a problem of semantics. I agree that Y cannot be 0.
What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.
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New post 26 Mar 2015, 03:52
cledgard wrote:
Bunuel wrote:
cledgard wrote:
I agree that b must be true. However, i be is true, then A must also be true.
If y > 0, then it is also true that Y >= 0.


That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.



This is a problem of semantics. I agree that Y cannot be 0.
What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.


No, that's not correct.

The question asks: which of the following must be true?

A says \(y\geq{0}\), so it allows y to be 0. But y cannot be 0 because in this case \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true. Thus A is not always true.
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Re: M02-08  [#permalink]

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New post 26 Mar 2015, 07:39
I agree that b must be true. However, if b is true, then A must also be true.
If y > 0, then it is also true that Y >= 0.

Quote:
That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.



This is a problem of semantics. I agree that Y cannot be 0.
What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.[/quote]

Quote:
No, that's not correct.

The question asks: which of the following must be true?

A says \(y\geq{0}\), so it allows y to be 0. But y cannot be 0 because in this case \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true. Thus A is not always true.



Of course 0 is not true. You must not test O (zero) because it is not a valid answer. What I said is that for any valid integer of Y , y>0 is true, but also Y>=0 is true. This is what the the question asks for.
Let' see it as a sets problem. B is the set of all positive integers, and A is the set of all non-negative integers. This way A has the same elements of B, as well as the 0; therefore, any element that is in B is also in A.
B says that y>0, and that is true. A says that Y is equal or more than 0. If Y is more than 0 -excluding 0- it means that Y>= 0 (more or equal, one or the other). We can even say that Y must be more than -10, and it will be true. Y does not have to be 0 in order to be equal or more than 0.

Study the question carefully.
The question did not ask for all possible values of Y;

The question is: Is A true for all values of Y? The answer is YES
The question is not: : Is Y true for all values of A? The answer is NOT
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New post 26 Mar 2015, 08:43
cledgard wrote:
I agree that b must be true. However, if b is true, then A must also be true.
If y > 0, then it is also true that Y >= 0.

Quote:
That's not true. If y=0 and x=y+1=1, then \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true.



This is a problem of semantics. I agree that Y cannot be 0.
What I say is that if Y is more than 0, then it must also be true that Y is more than -1 , or -2, or it is equal or more than 0.

If you give me any valid number for Y , answers Y>0 and y>= 0 are always true.


Quote:
No, that's not correct.

The question asks: which of the following must be true?

A says \(y\geq{0}\), so it allows y to be 0. But y cannot be 0 because in this case \(x^2-1 \gt y^2-4y+x-1\) does NOT hold true. Thus A is not always true.



Of course 0 is not true. You must not test O (zero) because it is not a valid answer. What I said is that for any valid integer of Y , y>0 is true, but also Y>=0 is true. This is what the the question asks for.
Let' see it as a sets problem. B is the set of all positive integers, and A is the set of all non-negative integers. This way A has the same elements of B, as well as the 0; therefore, any element that is in B is also in A.
B says that y>0, and that is true. A says that Y is equal or more than 0. If Y is more than 0 -excluding 0- it means that Y>= 0 (more or equal, one or the other). We can even say that Y must be more than -10, and it will be true. Y does not have to be 0 in order to be equal or more than 0.

Study the question carefully.
The question did not ask for all possible values of Y;

The question is: Is A true for all values of Y? The answer is YES
The question is not: : Is Y true for all values of A? The answer is NOT[/quote]

Edited option A to avoid ambiguity. Thank you.
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New post 10 Jan 2016, 19:36
I went about it with a mix of algebra and number plugging:
Since X= Y+1

The inequality turns out to be:

Y^2+2Y > Y^2-4y+y
2y>-3y

You might notice right here that Y will increase by 2 on the left and decrease by -3 on the right with any number greater than 0 keeping the statement true. I didn't realize it right away so I just tried a few numbers: 0,1, and finally -1 (made it not true)

Also, you could further reduce the inequality like Bunuel did just to double prove it:
2y > -3y
5y>0
y>0
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New post 28 Aug 2016, 06:13
I got the answer as y>0 but it is a must be true question.

So i chose y>8, which wil also be true.

Why not E?
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New post 29 Aug 2016, 07:26
Bunuel wrote:
If \(x\) and \(y\) are consecutive integers (\(x \gt y\)) and \(x^2-1 \gt y^2-4y+x-1\), then which of the following must be true?


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



Rearranging the inequality: x(x-1) > y (y-4) & x>y

Another way is to plug in values depending on answer choices

A. \(y \le 0\) ........Put y= 0 and x =1...........then 0>0 eliminate A

B. y>0 ........................Put y=1 and x = 2...........then 2>-3 .....Keep

To cover choice C,D &E ..put y=9 & x=10..... then 90> 45....Keep

Then for any y>0, the inequality hold true.

Answer B
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New post 14 Nov 2017, 21:05
4 of the answer choices are:

2. y>0
3. y>1
4. y>7
5. y > 8


now since it is a "must be true" question, doesn't 3,4,5 reduce to 2? That is, if y > 8, it MUST be true that y > 0.
What am I missing here?
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New post 14 Nov 2017, 21:29
sevenplusplus wrote:
4 of the answer choices are:

2. y>0
3. y>1
4. y>7
5. y > 8


now since it is a "must be true" question, doesn't 3,4,5 reduce to 2? That is, if y > 8, it MUST be true that y > 0.
What am I missing here?


Yes, if y > 8, then y > 0 too. But the correct answer is y > 0 (y must be greater than 0) and this does not mean that y must necessarily be greater than 8.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M02-08 &nbs [#permalink] 14 Nov 2017, 21:29
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