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# m02 #21

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14 Nov 2008, 12:33
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If $$x$$ is a positive integer, is $$y^2(x^3-x+1)>75?$$

(1) $$y>1$$
(2) $$(x-1)^2=16$$

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[Reveal] Spoiler:
Hi,

I think the answer/ explanation of this question is wrong.

Here is the official explanation:
---------------------------------------------
We need to know two things here

The value of x
Whether y is greater than 1
From S1, we only know that y>1

From S2, we only have the value of x. we also know that x is positive. So, we have to only consider the positive root.
Combining the two statements, we have the required information.
------------------------------------------------
However, i think the answer is E, since we dont have any information on whether y is an integer or not. It can be possible that y is a fraction.
Am i missing something guys????

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14 Nov 2008, 13:27
When you consider the statements together, you must use Statement 1 which tells you that 1 is greater than 1. It could be a fraction, but it will be a fraction greater than 1, thus making the statements together, still sufficient.
jimmiejaz wrote:
Hi,

I think the answer/ explanation of this question is wrong.

If x is a positive integer, is y^2(x^3-x+1)>75?
1. y>1
2. (x-1)^2=16

Here is the official explanation:
---------------------------------------------
We need to know two things here

The value of y
Whether y is greater than 1
From S1, we only know that y>1

From S2, we only have the value of x. we also know that x is positive. So, we have to only consider the positive root.
Combining the two statements, we have the required information.
------------------------------------------------
However, i think the answer is E, since we dont have any information on whether y is an integer or not. It can be possible that y is a fraction.
Am i missing something guys????

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27 Nov 2008, 07:28
Hi. From S2 we know that x=5. If x=5 we know that (x^3-x+1)=125-5+1=121. As long as y>1, we are sure that y^2(x^3-x+1)>75.
Hope this makes sense.

jimmiejaz wrote:
Hi,

I think the answer/ explanation of this question is wrong.

If x is a positive integer, is y^2(x^3-x+1)>75?
1. y>1
2. (x-1)^2=16

Here is the official explanation:
---------------------------------------------
We need to know two things here

The value of x
Whether y is greater than 1
From S1, we only know that y>1

From S2, we only have the value of x. we also know that x is positive. So, we have to only consider the positive root.
Combining the two statements, we have the required information.
------------------------------------------------
However, i think the answer is E, since we dont have any information on whether y is an integer or not. It can be possible that y is a fraction.
Am i missing something guys????

_________________

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27 Nov 2008, 08:14
jimmiejaz wrote:
Hi,

I think the answer/ explanation of this question is wrong.

If x is a positive integer, is y^2(x^3-x+1)>75?
1. y>1
2. (x-1)^2=16

Here is the official explanation:
---------------------------------------------
We need to know two things here

The value of x
Whether y is greater than 1
From S1, we only know that y>1

From S2, we only have the value of x. we also know that x is positive. So, we have to only consider the positive root.
Combining the two statements, we have the required information.
------------------------------------------------
However, i think the answer is E, since we dont have any information on whether y is an integer or not. It can be possible that y is a fraction.
Am i missing something guys????

Both statement togather are suff. we need two things i.e. values of x and y.

Statement 1 tells us the range of y.
Statement 2 tells us the value of x.
so togather the value of y^2 (x^3 - x + 1) is at least > 121.

suff. C.
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10 Jan 2009, 19:40
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In statement 2 why don't we account for:
x-1 = -4 ==> x = -3
and x-1 = 4 ==> x = 5
This would yield E as the answer. Am I missing something?

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12 Jan 2009, 03:12
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We know from the question stem that $$x$$ is a positive integer.

x-ALI-x wrote:
In statement 2 why don't we account for:
x-1 = -4 ==> x = -3
and x-1 = 4 ==> x = 5
This would yield E as the answer. Am I missing something?

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28 Feb 2009, 21:57
Just a useful reminder - you can write mathematical expressions in the posts by enclosing them in the [m] math [/m] tags.

If you don't want to write those out, just select the line or part of the text that you want to be converted into math, and hit the "m" button in the menu (the one directly below the big "B") and the software will automatically insert the tags.

You can also include a reference to the test question by hitting the "t" button.
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30 Apr 2010, 05:17
by expanding $$(x-1)^2=16$$ using foil, I created a quadratic equation $$x^2-2x-15=0$$ This produced the expression $$(x-5)*(x+3)=0$$ .... Plugging the solutions x=5 and x=-3 into the question stem I produced a "maybe" or "yes and no" answer to the question stem and rendered an answer of E! Please help!

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11 May 2010, 01:48
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You forget that $$x$$ is a positive integer, as stated in the question stem.
jbark55 wrote:
by expanding $$(x-1)^2=16$$ using foil, I created a quadratic equation $$x^2-2x-15=0$$ This produced the expression $$(x-5)*(x+3)=0$$ .... Plugging the solutions x=5 and x=-3 into the question stem I produced a "maybe" or "yes and no" answer to the question stem and rendered an answer of E! Please help!

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11 May 2010, 07:20
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This probem is similar to one given in the OG.

If x is a positive integer, is y^2(x^3-x+1)>75?
1. y>1
2. (x-1)^2=16

1. Insufficient. Y can be any number >1 and no X value is given.
2. Insufficient however statement 2 does give good information

(x-1)^2=16 can be simplified to read:

take the sqrt of both sides. x-1 = 4

x = 5... however we dont know what x is.

so we take statement 1 and combine with statement 2 and we can solve the equation.

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11 May 2010, 07:44
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$$(x-1)^2 = 16$$
shall be simplified as $$x-1=+-4$$
or $$x=+5$$, $$x=-3$$
if we consider $$(x-1)^2=4^2$$ and consider $$x-1 = 4$$
We miss the second root ie $$x =-3$$<quadratic equation must have 2 roots>.
In this question we are spared even if we consider $$x-1=4$$ ,
However if question had not specified x as positive integer, answer would have been E.
$$y^2(x^3-x+1)$$ +ve in case $$x=+5$$ and -ve in case $$x=-3$$.

Good question...
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11 May 2010, 09:51
S1 is not sufficient as no info about x
S2 is not sufficient as no info about y

S2implies x=5, so as long as S1 is true (i.e. y>1) the inequality in the stem is holds good.

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11 May 2010, 13:04
we already know statement (1) or (2) proved insufficient
when considered individually.

When Combined, we get y^2(121)
since y>1, the expression becomes (1 + k)(121)....k is some
positive value being part of y. That is: 121 + 121k > 75
correct response is C
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11 May 2010, 14:52
If x is a positive integer, is y^2(x^3-x+1)>75

1. y>1
2. (x-1)^2=16

St 1 alone is not suff(B & D elimitated)

St2:
=>x=5,-3 (Since x is +ve , x=5)
St 2 alone is not suff(A & D eliminated)

Now, Combining both
y>1 & x=5

y^2(x^3-x+1)>75
=> OA is C.

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11 May 2010, 22:43
First I took statement 2 , We get x = 5 since x =-3 is ruled out. Applying x = 5 in the eqn we get y (121) > 75.For this expr to be true I want atleast y > 1 which is in statment 1. So each statement alone is insuff whereas both the statements combined can yield a result..

I will go with option C

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20 May 2010, 04:55
So immediately we are told more about x than y. We know that x is a positive integer. This being so, the whole blob of Xcubed whatever can be nothing less than 1. X cannot be less than 1 and so xcubed minus x plus 1 can be nothing less than 1. Now we need to know more about y.

Statement 1. tells us that y is greater than 1. This coupled with what we know about x is insufficient to tell us if the whole thing is greater than 75. Statement 1 did not tell us anything more than what we knew about x (that it is equal to or greater than 1), and not enough about y. Not sufficient.

Statement 2. (x-1)squared =16. This still tells us nothing about y and thus will likely be insufficient by itself.

But lets look further. FOILing out the (X-1)(x-1) = x^2 -2x+1=16 --->X^2 -2x-15=0. Now factoring this we get (x+3)(x-5)=0. So x could be either -3 or 5. We know that x cannot be negative from the initial statement. So x must be 5. Then plugging 5 back into the x cubed thing we get 125-5+1=121. So (y^2)121. Y is greater than 1 so the product must be greater than 75. If we put statements 1 and 2 together we get C. 1 and 2 are sufficient together.

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20 May 2010, 04:58
Something important that occurs to me about this question. To go through that whole process may take to long to be SURE of what the answer is. If you can do this math quickly and under two minutes okay. At some point it may be better to get this ? wrong than to waste a huge amt of time on it.

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13 May 2011, 03:34
(1) is insufficient

y = 2 x = 1

4 * 1 < 75

y = 77 x = 1

(77)^2 * 1 > 75

(2)

x = 4 + 1 = 5

But no information about y, hence insufficient

(1) + (2)

y = 2, x = 5

4 * (125 - 5 + 1) = 484 > 75
So y will always increase, hence value will be always > 75.

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13 May 2011, 04:45
xALIx wrote:
In statement 2 why don't we account for:
x-1 = -4 ==> x = -3
and x-1 = 4 ==> x = 5
This would yield E as the answer. Am I missing something?

The question clearly says that x is a positive integer than x-1 can only be equal to 4 in that case.
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13 May 2011, 05:15
1. Not sufficient
as we don't know anything about x.

2. Not sufficient
As we don't know anything about y.

Together
Sufficient as y>1,x=5

Given expression>75

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