Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 _________________

You have to have the darkness for the dawn to come.

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