GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video
close
It is currently 08 May 2021, 10:47
Register
GMAT Club Tests
Decision Tracker
My Rewards
New posts
New comers' posts

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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . +
13 posts Go to First Unread Post
Print view
NEW TOPIC
POST REPLY
Downloads
My Bookmarks
Reviews
SORT BY:
Date
Kudos
Tags:
Find Similar Topics 

Show Tags
Hide Tags

User avatar
V
BrentGMATPrepNow
GMAT Club Legend
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 5438
Location: Canada
Top Member of the Month
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  08 May 2017, 06:06
45
Bookmarks
Expert Reply
Top Contributor
00:00
A
B
C
D
E

Difficulty:

95% (hard)

Question Stats:

36% (02:34) correct 64% (02:17) wrong based on 347 sessions

Hide Show timer Statistics

If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions
ShowHide Answer
Official Answer
Official Answer and Stats are available only to registered users. Register/Login.

_________________
Image

A focused approach to GMAT mastery
Signature Read More
Most Helpful Expert Reply
User avatar
V
BrentGMATPrepNow
GMAT Club Legend
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 5438
Location: Canada
Top Member of the Month
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  10 May 2017, 13:07
14
Kudos
9
Bookmarks
Expert Reply
Top Contributor
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²


We have several differences of squares hiding in the expression 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²

1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² = 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100)
= (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
= (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
= (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

IMPORTANT: within the sum, 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100, we have all of the ODD integers from 1 to 99 inclusive, and we have all of the EVEN integers from 2 to 100 inclusive.

So, we can say that 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 = K + J

So, we're replace 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 with K + J.
We get: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100) = (-1)(K + J)
= -K - J

Answer:
Show Spoiler
C


Cheers,
Brent
_________________
Image

A focused approach to GMAT mastery
Signature Read More
Most Helpful Community Reply
User avatar
V
rohit8865
Director
Director
Joined: 05 Mar 2015
Posts: 951
Reviews Badge
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  08 May 2017, 06:53
9
Kudos
1
Bookmarks
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions



simply put j= 2+4
k= 1+3

then 1^2- 2^2 + 3^2 - 4^2 = 1-4+9-16= -10


also j= 2+4= 6 && k = 1+3 =4
j^2= 36 && k^2 =16

just plug in values to option to get -10 as our answer

only option C does

Ans C
General Discussion
User avatar
V
chetan2u
Math Expert
Joined: 02 Aug 2009
Posts: 9813
Reviews Badge Top Member of the Month
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  08 May 2017, 07:25
7
Kudos
Expert Reply
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions



Hi,

\(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\) .....
Here take all inpairs \(1^2-2^2\), \(3^2-4^2\), and so on till \(99^2-100^2\)..
1^2-2^2=(1-2)(1+2)=-1(1+2)=-1-2...
3^2-4^2=(3-4)(3+4)=-1(3+4)=-3-4..
So the equation becomes -1-2-3-4-......-99-100=-(1+2+3+4+...+99+100)= -[(2+4+6....+98+100)+(1+3+5+...+97+99)]=-[(j)+(k)]=-j-k
C
avatar
P
sobby
Director
Director
Joined: 14 Nov 2014
Posts: 543
Location: India
Schools: Rotman '20 (S), Kelley '21 (S)
GMAT 1: 700 Q50 V34
GPA: 3.76
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  08 May 2017, 09:52
2
Kudos
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions


We can break the problem into a^2 - b ^2 = 1^2 - 2^2 = (1+2) (1-2) = -3
similarly other pair will give = -7 ,next pair will give = -11
final pair will give -199
Now the question stem is reduced to below seq:
-3 -7-11.....-199

-3 = -(1+2)
-7 = -(3+4)
-11 = -(5+6)

-(1+3+5....)-(2+4+6...)
-(j)-(k)
-(j+k)..Ans
User avatar
V
BillyZ
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1299
Location: Malaysia
Concentration: General Management, Strategy
Schools: Sloan '23 (D), Tuck '23 (D), Darden '23 (WL), Ross '23 (D), Kellogg '23 (D), Wharton '23 (D), Booth '23 (D)
GMAT 1: 750 Q51 V40 (Online)
GPA: 3.53
GMAT ToolKit User Reviews Badge
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  11 May 2017, 21:52
GMATPrepNow wrote:
If \(J = 2 + 4 + 6 + 8 + . . . 98 + 100\), and \(K = 1 + 3 + 5 + 7 + . . . + 97 + 99\), then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions


From Matt (Veritas Prep)

We could also cheat with a pattern:

\(n² - (n + 1)² => n² - (n² + 2n + 1) => -(2n + 1) => -(n + n + 1)\) for any value of n.

Since we've got 1² - 2² + 3² - 4² ..., we've really got -(1 + 2) -(3 + 4) .... -(99 + 100), or -1 -2 -3 -4 .... - 99 - 100, or -(1 + 2 + 3 + ... + 100), or -(K + J), or -K - J.
_________________
"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Rules for posting in verbal forum | Please DO NOT post short answer in your post!

Advanced Search : https://gmatclub.com/forum/advanced-search/
Signature Read More
User avatar
D
shashankism
Director
Director
Joined: 13 Mar 2017
Affiliations: IIT Dhanbad
Posts: 710
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE:Engineering (Energy and Utilities)
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  11 May 2017, 22:28
1
Kudos
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions


There can be many solutions possible. But we will go with the basic solution, though it might be lengthy one to understand the concepts..
We can use tricks to solve problem as suggested by other members in exam.. Learning tricks is also very important.

J = 2+4+6+8+....+98+100
K = 1+3+5+7+....++97+99

\(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)
= (1-2)(1+2) + (3-4) (3+4)+ (5-6)(5+6) +.......+ (97-98)(97+98) +(99-100)(99+100)
= -1[(1+2)+(3+4)(5+6) +......+ (97+98)+(99+100)]
= -1[ (1+3+5+...+97+99) + (2+4+6+...+98+100)]
= -1(K+J)
= -K-J

Show Spoiler
Anwer : C

_________________

Learn Quant at my blog: WebMaggu.com

Signature Read More
avatar
B
DasAshish1
Intern
Intern
Joined: 28 Mar 2013
Posts: 14
GMAT 1: 680 Q49 V32
GMAT ToolKit User Reviews Badge
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  11 May 2017, 22:39
1
Kudos
Answer is C .
We can pair any 2 consecutive term and apply (a+b) (a-b) in stead of a^2 - b ^2.


Sent from my Moto G (4) using GMAT Club Forum mobile app
User avatar
B
MayurAgrawal
Intern
Intern
Joined: 05 Mar 2018
Posts: 3
Location: India
Concentration: Entrepreneurship, Strategy
GPA: 2.7
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  30 Apr 2018, 20:01
GMATPrepNow wrote:
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²


We have several differences of squares hiding in the expression 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²

1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² = 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100)
= [color=red](-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
= (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
= (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

IMPORTANT: within the sum, 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100, we have all of the ODD integers from 1 to 99 inclusive, and we have all of the EVEN integers from 2 to 100 inclusive.

So, we can say that 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 = K + J

So, we're replace 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 with K + J.
We get: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100) = (-1)(K + J)
= -K - J

Answer:
Show Spoiler
C


Cheers,
Brent


Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write
(1-2)(1+2) = (-1)(1 + 2)
(3-4)(3+4) =(-1)(3+4)... and so on...?
User avatar
V
BrentGMATPrepNow
GMAT Club Legend
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 5438
Location: Canada
Top Member of the Month
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  01 May 2018, 07:16
2
Kudos
Expert Reply
Top Contributor
MayurAgrawal wrote:
GMATPrepNow wrote:
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²


We have several differences of squares hiding in the expression 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²

1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100² = 1² - 2² + 3² - 4² + 5² - 6² + . . . . . + 97² - 98² + 99² - 100²
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100)
= (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
= (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
= (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

IMPORTANT: within the sum, 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100, we have all of the ODD integers from 1 to 99 inclusive, and we have all of the EVEN integers from 2 to 100 inclusive.

So, we can say that 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 = K + J

So, we're replace 1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100 with K + J.
We get: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100) = (-1)(K + J)
= -K - J

Answer:
Show Spoiler
C


Cheers,
Brent


Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write
(1-2)(1+2) = (-1)(1 + 2)
(3-4)(3+4) =(-1)(3+4)... and so on...?


Sure thing.
Let's take (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100) and break it into its individual parts:
(1 - 2)(1 + 2) = (-1)(1 + 2) because 1 - 2 = -1
(3 - 4)(3 + 4) = (-1)(3 + 4) because 3 - 4 = -1
(5 - 6)(5 + 6) = (-1)(5 + 6) because 5 - 6 = -1
.
.
.

(97 - 98)(97 + 98) = (-1)(97 + 98) because 97 - 98 = -1
(99 - 100)(99 + 100) = (-1)(99 + 100) because 99 - 100 = -1

So, we get: (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
From here, we can factor out the -1 to get: (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
Which is the same as: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

Does that help?

Cheers,
Brent
_________________
Image

A focused approach to GMAT mastery
Signature Read More
User avatar
B
MayurAgrawal
Intern
Intern
Joined: 05 Mar 2018
Posts: 3
Location: India
Concentration: Entrepreneurship, Strategy
GPA: 2.7
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  01 May 2018, 17:00
Quote:
Quote:
Hi Brent, I did't understand the step 3. Can you please elaborate, how did you write
(1-2)(1+2) = (-1)(1 + 2)
(3-4)(3+4) =(-1)(3+4)... and so on...?


Sure thing.
Let's take (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + . . . . . + (97 - 98)(97 + 98) + (99 - 100)(99 + 100) and break it into its individual parts:
(1 - 2)(1 + 2) = (-1)(1 + 2) because 1 - 2 = -1
(3 - 4)(3 + 4) = (-1)(3 + 4) because 3 - 4 = -1
(5 - 6)(5 + 6) = (-1)(5 + 6) because 5 - 6 = -1
.
.
.

(97 - 98)(97 + 98) = (-1)(97 + 98) because 97 - 98 = -1
(99 - 100)(99 + 100) = (-1)(99 + 100) because 99 - 100 = -1

So, we get: (-1)(1 + 2) + (-1)(3 + 4) + (-1)(5 + 6) + . . . . . + (-1)(97 + 98) + (-1)(99 + 100)
From here, we can factor out the -1 to get: (-1)[(1 + 2) + (3 + 4) + (5 + 6) + . . . . . + (97 + 98) + (99 + 100)]
Which is the same as: (-1)(1 + 2 + 3 + 4 + . . . . . 97 + 98 + 99 + 100)

Does that help?

Cheers,
Brent


Silly Me. I totally missed subtraction. Thank you very much for clearing doubt. :)
User avatar
P
Pritishd
UNC Kenan Flagler Moderator
Joined: 18 Jul 2015
Posts: 240
GMAT 1: 530 Q43 V20
WE:Analyst (Consumer Products)
GMAT ToolKit User
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  24 Dec 2019, 06:37
GMATPrepNow wrote:
If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + 97 + 99, then \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + . . . . . + 97^2 - 98^2 + 99^2 - 100^2 =\)

A) J² - K²
B) -50(J² - K²)
C) -K - J
D) K² - J²
E) (-J - K)²

*kudos for all correct solutions


Brent's solution in this thread above is amazing and is the way to go. I am sharing here the approach for working backwards from the answer choices.

1. Take a sample from the two series (highlighted above)
\(J = 2 + 4 + 6 + 8 = 20\)
\(K = 1 + 3 + 5 + 7 = 16\)

Now, \(1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 - 8^2 = -36\)

2. Plug \(J = 20\), \(K = 16\) in the choices to check which one yields \(-36\)
As you will realize that only C i.e. \(-K - J\) works
\(-K - J = -20-16 = -36\)

Ans. C
_________________
Cheers. Wishing Luck to Every GMAT Aspirant | Press +1 if this post helped you!
Signature Read More
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 18967
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink] New post  16 Jan 2021, 02:10
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
Signature Read More
GMAT Club Bot
gmatclubot
Re: If J = 2 + 4 + 6 + 8 + . . . 98 + 100, and K = 1 + 3 + 5 + 7 + . . . + [#permalink]
16 Jan 2021, 02:10
NEW TOPIC
POST REPLY
Downloads
My Bookmarks
Reviews
Moderators:
chetan2u
Math Expert
9813 posts
Bunuel
Math Expert
74521 posts
yashikaaggarwal
Senior Moderator - Masters Forum
2457 posts
generis
Senior SC Moderator
4719 posts

cron

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

Main navigation

GMAT Resources

Partners

MBA Resources

www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions