GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Oct 2018, 08:00

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

((n-1)!+n!+(n+1)!)/n!

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
Joined: 06 Nov 2013
Posts: 3
((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post Updated on: 18 Feb 2016, 15:26
2
12
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

62% (02:13) correct 38% (02:19) wrong based on 176 sessions

HideShow timer Statistics

\(\frac{(n-1)!+n!+(n+1)!}{n!}\)

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

Originally posted by jahanafsana on 08 Mar 2015, 18:25.
Last edited by ENGRTOMBA2018 on 18 Feb 2016, 15:26, edited 1 time in total.
Reformatted the question and added the OA.
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6973
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 08 Mar 2015, 18:41
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!


Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18


hi jahanafsana,
it depends how you have reduced the equation...
1) first is you open each term .... find common terms... get an equation and then substitute values..
2) second , which may be slightly less time consuming.....
\(\frac{((n-1)!+n!+(n+1)!)}{n!}\)
=\(\frac{(n-1)!}{n!}\)+\(\frac{n!}{n!}\)+\(\frac{(n+1)!}{n!}\)
=\(\frac{1}{n}\)+1+n+1.... now you can easily see n+2 should be greater than or equal to 16.... 14 is the value...
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 09 Mar 2015, 23:50
2
2
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!


Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18



\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Director
Director
User avatar
S
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 523
Location: India
GMAT 1: 780 Q51 V46
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 11 Mar 2015, 23:58
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.
_________________

For more info on GMAT and MBA, follow us on @AskCrackVerbal

Intern
Intern
avatar
Joined: 06 Nov 2013
Posts: 3
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 12 Mar 2015, 09:44
CrackVerbalGMAT wrote:
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.



Thank you for your reply. It helped.
Intern
Intern
avatar
Joined: 06 Nov 2013
Posts: 3
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 12 Mar 2015, 09:46
CrackVerbalGMAT wrote:
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.


Thanks a lot for your detail reply. It is clear now.
Manager
Manager
avatar
Joined: 03 Jan 2015
Posts: 80
GMAT ToolKit User
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 18 Feb 2016, 13:51
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!


Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18



\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)


Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)

Could you show how you simplified this part in steps? I seem to get a different answer.
Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2633
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 18 Feb 2016, 15:20
saiesta wrote:
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!


Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18



\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)


Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)

Could you show how you simplified this part in steps? I seem to get a different answer.


You can adopt 2 methods to verify: either substitute n=3 and cross check. For algebraic method look below:

Realize that (n+1)! = (n+1)*n*(n-1)(n-2)... = (n+1)*n!

Similarly, n!=n*(n-1)! ---> (n-1)!= n!/n

Thus, \(\frac{(n - 1)! + n! + (n + 1)!}{n!} =\frac{(n!/n)+n!+(n+1)n!}{n!}\) = (cancelling n! from both the numerator and denominator you get) \(\frac{1}{n}+1+(n+1)\)

Hope this helps.
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6973
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 18 Feb 2016, 20:46
saiesta wrote:
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!


Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18



\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)


Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)

Could you show how you simplified this part in steps? I seem to get a different answer.


\(\frac{((n-1)!+n!+(n+1)!)}{n!}\)
=\(\frac{(n-1)!}{n!}\)+\(\frac{n!}{n!}\)+\(\frac{(n+1)!}{n!}\)
=\(\frac{1}{n}\)+1+n+1....
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8399
Location: Pune, India
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 18 Feb 2016, 20:59
2
saiesta wrote:
VeritasPrepKarishma wrote:
jahanafsana wrote:
(n-1)!+n!+(n+1)!
n!


Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18



\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)


Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)

Could you show how you simplified this part in steps? I seem to get a different answer.


Note that
\((n-1)! = 1*2*3*...*(n-1)\)
\(n! = 1*2*3*...*(n-1)*n\)

So n! just has an extra n in it.

When you divide (n-1)! by n!, you are just left with an n in the denominator.

Similarly,
\(n! = 1*2*3*...*(n-1)*n\)
\((n+1)! = 1*2*3*...*(n-1)*n*(n+1)\)
So (n+1)! only has an extra (n+1).

When you divide (n+1)! by n!, you are left with (n+1) in the denominator.

\(\frac{(n - 1)! + n! + (n + 1)!}{n!}\)

\(= \frac{(n - 1)!}{n!} + \frac{n!}{n!} + \frac{(n+1)!}{n!}\)

\(= \frac{1}{n} + 1 + (n+1)\)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Manager
Manager
avatar
Joined: 03 Jan 2015
Posts: 80
GMAT ToolKit User
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 19 Feb 2016, 13:01
VeritasPrepKarishma Thank you. Your explanation has been quite useful.
Director
Director
User avatar
P
Joined: 13 Mar 2017
Posts: 622
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 11 May 2017, 02:49
jahanafsana wrote:
\(\frac{(n-1)!+n!+(n+1)!}{n!}\)

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.


\(\frac{(n-1)!+n!+(n+1)!}{n!}\)
=1/n+1+n+1
=n+2+(1/n)
16<n+2+1/n<17
->14<n+1/n<15
n=14
Answer A.

_________________

CAT 2017 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu


Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)



What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 8505
Premium Member
Re: ((n-1)!+n!+(n+1)!)/n!  [#permalink]

Show Tags

New post 31 Jul 2018, 17:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Bot
Re: ((n-1)!+n!+(n+1)!)/n! &nbs [#permalink] 31 Jul 2018, 17:31
Display posts from previous: Sort by

((n-1)!+n!+(n+1)!)/n!

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.