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:

Statement (1) does not give any useful information that i can see

(n+1)!-(n-1)! is greater than (n+1)!/(n-1)! for any positive number except 1 (as (n-1)! would be 0...i think!...thus making (n+1)!/(n-1)! undefined).
If n is divisible by 2...the lowest possible value of n is two...
this is sufficient to say (n+1)!-(n-1)! > (n+1)!/(n-1)!

Statement (1) does not give any useful information that i can see

(n+1)!-(n-1)! is greater than (n+1)!/(n-1)! for any positive number except 1 (as (n-1)! would be 0...i think!...thus making (n+1)!/(n-1)! undefined). If n is divisible by 2...the lowest possible value of n is two... this is sufficient to say (n+1)!-(n-1)! > (n+1)!/(n-1)!

thus, my answer is B

the whole equation reduces to (n-1)! * (n(n+1) -1) ... this needs to be compared with (n+1)!/(n-1)! = n(n+1)

I) Square root of n is a positive integer
n can be 1 or 2 or any number greater than 2

for n =1 0! =1 (not zero) - LHS will be less than RHS
for n = 2, LHS < RHS
for n - 3, LHS = 2*11 ; RHS = 12 => LHS > RHS
and for any number greater than 2, LHS > RHS

We need (2) to determine that n is greater than 2

(2) only this is not sufficient either as n can be equal to 2

I will go with C too
The equation reduces to (n-1)![n(n+1) -1] > n(n+1)
From statement 1: n can be 1, 4, 9,...
when n is 1: (n-1)![n(n+1) -1] is not > than n(n+1)
when n is 4: (n-1)![n(n+1) -1] is > than n(n+1)
when n is 9: (n-1)![n(n+1) -1] is > than n(n+1)
so statement 1 is not sufficient

From statement 2: n can be any even number such as 2, 4, 6, 8....
when n is 2: (n-1)![n(n+1) -1] is not > than n(n+1)
when n is 4: (n-1)![n(n+1) -1] is > than n(n+1)
so statement 2 is not sufficient

But both put together: n can be 4, 16, .....
As shown above the condition is satisfied for any number greater than 2
Hence C

Statement 1, it repets nothing but the original condition that n is a positive integer. It behaves differently for 1 and any higher positive integer. Hence Insufficient. Hence rule out option D as well.

Statement 2, says n is even. substraction operation on two positive integer ( Large positive integer- small positive integer) is always greater than division operation on of the same two numbers ( large positive number/small positive number). Hence B is sufficient.

substraction operation on two positive integer ( Large positive integer- small positive integer) is always greater than division operation on of the same two numbers ( large positive number/small positive number).

3 - 2 is not greater than 3 /2

Clarification: I meant in question's context (with factorial ). we will never have a situation 3-2 and 3/2 according to statement 2.
Please let me know why not B is correct as per my logic.

When n<=2, (n-1)!<=1
(n-1)!*[n(n+1)-1]-n(n+1)<=n(n+1)-1-n(n+1)=-1<0

When n>=3, (n-1)!>=2
(n-1)!*[n(n+1)-1]-n(n+1)>=2n(n+1)-1-n(n+1)=n(n+1)-1>0

(1) n is square of an integer, n could be 1 or 4 or greater, insufficient
(2) n is even, n could be 2 or 4 or greater, insufficient
Combined, n could only be 4 or greater, sufficient. (It's always >0 when n>=3)

(C)
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.