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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.

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

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 _________________

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 Give Kudos, If you like my post

Last edited by daboo343 on 06 May 2014, 02:27, edited 2 times in total.

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.

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...............ABCDE