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# M02-08

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:17
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Difficulty:

45% (medium)

Question Stats:

69% (01:26) correct 31% (02:02) wrong based on 167 sessions

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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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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:

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

$$5y \gt 0$$;

$$y \gt 0$$.

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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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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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09 Oct 2014, 05:39
Oh! Sweet thank you very much!
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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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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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25 Mar 2015, 03:36
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.
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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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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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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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Clipper Ledgard
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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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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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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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28 Aug 2016, 06:17
gmatravi wrote:
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?

How y > 8 will be true if y > 0? What if y = 1, is 1 greater than 8?
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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.

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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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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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# M02-08

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