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m02 #21

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Re: m02 #21 [#permalink] New post 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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Re: m02 #21 [#permalink] New post 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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Re: m02 #21 [#permalink] New post 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.
The correct answer is C.
------------------------------------------------
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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Re: m02 #21 [#permalink] New post 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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Re: m02 #21 [#permalink] New post 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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Re: m02 #21 [#permalink] New post 05 May 2014, 11:18
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Bunuel,Can you please help.
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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Re: m02 #21 [#permalink] New post 06 May 2014, 01:53
Expert's post
ankushbassi wrote:
Bunuel,Can you please help.
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.

Answer: C.

Hope it's clear.
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Re: m02 #21 [#permalink] New post 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


Answer: C
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Re: m02 #21 [#permalink] New post 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

Answer (C).
Re: m02 #21   [#permalink] 16 May 2014, 06:52
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