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# m02 #21

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16 May 2011, 19:08
Yep C,
This topic really shouldnt have so many posts.
This is a very simple question.
You have no information on y, hence y could have been zero and that can only be averted with the presence of statment 1.
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18 May 2012, 06:55
[i[/img]
dzyubam wrote:
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?

Ya
Since it's given question stem that x is +ve, we need not consider the -ve value of x
so the vale of x we will take the value of x as 5
and the the value of the expression will come y to the power 2 multiplied with 121
Now Y has the roll, suppose it's less then 1 say .1, the value of the entire expression will be 1.21 which is less than 75 , so the value of y has to be more than 1 or less than -1 to yield the the value of the expression >75
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20 May 2013, 05:01
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jimmiejaz wrote:
If $$x$$ is a positive integer, is $$y^2(x^3-x+1)>75?$$

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

[Reveal] Spoiler: OA
C

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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????

Jimmie Jaz is correct; the answer is E. True, Y is greater than 1 but if Y were a very small fraction (i.e. 1001/1000, 1.001) then raised to the 242 power the product would still be less than 75.

1.001^242= 1.27
1.27<75
Statement's 1 and 2 are Insufficient.
E
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20 May 2013, 05:03
To be more specific if y is >1.018 than the answer would be C.

1.018^242 = 74.98
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05 May 2014, 06:51
Given
Statement 1 y>1 and x is a positive integer.
So x could be any value from 1 onward.
depending on the value of x, the vale of equation changes. so we do no get a concrete value.

Statement 2
(X-1)^2=16
from this equation we got two values of x
i.e 5 and -3
discard the -ve value of x, because it is given in the queston
y^2(5^3-5+1)= y^2(121).
still we don't know the exact value
if y is 0, then we get 0<75
of if y is fraction we get different values for different values of y
so statement 2 is also not sufficient
Combine statement 1 and 2 we get the vale of equation 121 or more
i.e >75
so C is the right answer

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Last edited by daboo343 on 06 May 2014, 02:27, edited 2 times in total.
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05 May 2014, 11:18
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I believe the answer should be E.
Since,y >1.
It is not given that y is integer.3

from Statement 1,we know y>1.:-
If we consider y=1.1 then the y^2(x^3-x+1)>75 is true.
If we consider y=2,the y^2(x^3-x+1)>75 is false.

Thus both the statements doesn't help to come to a definite answer.
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06 May 2014, 01:53
ankushbassi wrote:
I believe the answer should be E.
Since,y >1.
It is not given that y is integer.3

from Statement 1,we know y>1.:-
If we consider y=1.1 then the y^2(x^3-x+1)>75 is true.
If we consider y=2,the y^2(x^3-x+1)>75 is false.

Thus both the statements doesn't help to come to a definite answer.

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

(1) $$y>1$$. No info about x. Not sufficient.

(2) $$(x-1)^2=16$$. Take the square root: $$|x-1|=4$$ --> $$x=-3$$ or $$x=5$$. Discard the first root because we are told that $$x$$ is a positive integer, so we have that $$x=5$$. Thus the question becomes: is $$121y^2>75$$? If y=0, then the answer is NO but if y=1, then the answer is YES. Not sufficient.

(1)+(2) From (2) the question became: is $$121y^2>75$$? (1) says that $$y>1$$. For any y greater than 1, $$121y^2$$ will be greater than 75. Sufficient.

Hope it's clear.
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06 May 2014, 02:08
(1) y>1 --> not sufficient (bkz if x = y = 1 --> y^2(x^3-x+1) =1)
(2) (x-1)^2=16
--> |x-1| = 4
--> x positive -> x = 5
-> y^2(x^3-x+1) = 121 y^2

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16 May 2014, 06:52
If x is a positive integer, is $$y^2(x^3-x+1)>75$$?

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

Statement 1
The inequality cannot be uniquely evaluated with only information: $$x >0$$ and $$y > 1$$.....Not sufficient......A B C D E

Statement 2

$$(x-1)^2=16$$ Or,$$|x-1|=4$$ Or, $$x=5$$ [ The other solution of $$x = -3$$ is discarded as $$x > 0$$]

Substituting $$x=5$$ and $$(x-1)^2 =16$$ into the expression $$x^3-x+1$$ ,

$$x^3-x+1 = (x+1)(x-1)^2= 5*16=80$$

As no information on y is available, the inequality $$y^2(x^3-x+1)>75$$ can not be uniquely evaluated......Not sufficient...AB C D E

Using both statement (1) and (2), it is possible to evaluate the expression $$y^2(x^3-x+1)$$ or $$(y^2) * 80$$ to be always greater than $$75$$ even for the minimum value of $$y$$......Both statements together is sufficient...............AB C D E

Re: m02 #21   [#permalink] 16 May 2014, 06:52

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# m02 #21

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