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

It is currently 20 Jul 2018, 17:29

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

A group of 8 friends want to play doubles tennis. How many

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

Hide Tags

2 KUDOS received
Manager
Manager
avatar
Joined: 11 Jul 2010
Posts: 210
GMAT ToolKit User
A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 14 Nov 2010, 04:18
2
17
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

42% (00:39) correct 58% (01:57) wrong based on 156 sessions

HideShow timer Statistics

A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

Hypo:

Instead, if the problem were to say: A group of 9 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

Can the answer be arrived at as follows?

select 8 out of 9 people, first: 9C8 = 9

Then select the sets of 2 people

8!/2^4*4! = 105

answer = 9*105 ---> is this correct? can anyone who knows tell me? Many thanks.
Most Helpful Expert Reply
Expert Post
3 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47157
A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 15 Nov 2010, 01:57
3
7
AkritiMehta wrote:
Hi how do we get the answer for the Ist question as 105. Pls explain


A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups


There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

General Discussion
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 775
Location: London
GMAT ToolKit User Reviews Badge
Re: Hypo Combinations Question  [#permalink]

Show Tags

New post 14 Nov 2010, 09:50
gmat1011 wrote:
Disclaimer: This is a hypothetical set up for purposes of understanding the concept. I don't have a definitive answer, so don't sue me :-D

Known Problem: discussed on this forum a few times in the past
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

Hypo:

Instead, if the problem were to say: A group of 9 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

Can the answer be arrived at as follows?

select 8 out of 9 people, first: 9C8 = 9

Then select the sets of 2 people

8!/2^4*4! = 105

answer = 9*105 ---> is this correct? can anyone who knows tell me? Many thanks.


So the answer to Q1 is 105

And then what you are saying is you choose the one person who sits out (9 ways) TIMES the number of ways to form the teams

This approach is correct
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

GMAT Club Premium Membership - big benefits and savings

Intern
Intern
avatar
Joined: 21 Jun 2010
Posts: 4
Re: Hypo Combinations Question  [#permalink]

Show Tags

New post 14 Nov 2010, 22:08
Hi how do we get the answer for the Ist question as 105. Pls explain
Manager
Manager
avatar
Joined: 11 Jul 2010
Posts: 210
GMAT ToolKit User
Re: Hypo Combinations Question  [#permalink]

Show Tags

New post 15 Nov 2010, 06:45
Thanks Shrouded for confirming that.

Thanks Bunuel.
Manager
Manager
User avatar
Joined: 13 Jul 2010
Posts: 149
Re: Hypo Combinations Question  [#permalink]

Show Tags

New post 16 Nov 2010, 19:12
Bunuel wrote:
AkritiMehta wrote:
Hi how do we get the answer for the Ist question as 105. Pls explain


A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

\(\frac{C^2_8*C^2_6*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups


There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.


Bunuel - the formula you cite above is this a derivation of the standard combinations formula? I have never seen the exponent in the denominator? Thanks.
Senior Manager
Senior Manager
User avatar
Joined: 18 Jan 2010
Posts: 254
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 27 May 2016, 23:20
Bunuel wrote:
AkritiMehta wrote:
Hi how do we get the answer for the Ist question as 105. Pls explain


A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

\(\frac{C^2_8*C^2_6*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups


There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.



I do not agree that we need to divide it by 4!

8C2 * 6C2*4C2 is a regular combination formula

We choose 2 persons out of 8. 1st team is formed. There is no arrangement, just selection. (8C2)

We then choose 2 persons out of remaining 6. 2nd team is formed. There is no arrangement, just selection. (6C2)

We then choose 2 persons out of remaining 4. 3rd team is formed. There is no arrangement, just selection. (4C2)

We then choose 2 persons out of remaining 2. 4th team is formed. There is no arrangement, just selection. (2C2)

Product: 8C2 * 6C2*4C2

This works out to 8! / (2!.2!.2!.2!) = 7!/2 = 2520
Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6258
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 28 May 2016, 00:10
1
1
adiagr wrote:
Bunuel wrote:
AkritiMehta wrote:
Hi how do we get the answer for the Ist question as 105. Pls explain


A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

\(\frac{C^2_8*C^2_6*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+groups


There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.



I do not agree that we need to divide it by 4!

8C2 * 6C2*4C2 is a regular combination formula

We choose 2 persons out of 8. 1st team is formed. There is no arrangement, just selection. (8C2)

We then choose 2 persons out of remaining 6. 2nd team is formed. There is no arrangement, just selection. (6C2)

We then choose 2 persons out of remaining 4. 3rd team is formed. There is no arrangement, just selection. (4C2)

We then choose 2 persons out of remaining 2. 4th team is formed. There is no arrangement, just selection. (2C2)

Product: 8C2 * 6C2*4C2

This works out to 8! / (2!.2!.2!.2!) = 7!/2 = 2520


Hi,
the solution requires that the answer be divided by 4!...
reason is to eliminate the repetitions that arise out of the standard formula...
Explanation of the same in a simpler scenario..

say 4 people, ABCD in 2 groups...
standard formula \(= 4C2*2C2 = 6*1 = 6\)....
write these cases-
1) AB and CD
2) AC and BD
3) AD and BC

so only three ! where are other three?
say in 4c2 we chose BC , so AD in second group..
so BC and AD... BUT this is SAME as 3 above - AD and BC..
so when we choose AD, we are automatically choosing the other group BC..
so our answer \(= \frac{6}{2!}= 3\)

Hope it helps you
_________________

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

Senior Manager
Senior Manager
User avatar
Joined: 18 Jan 2010
Posts: 254
A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 28 May 2016, 00:35
Thanks for explanation. I would say that there is an obvious ambiguity in the language of the question. If you see that (in your example) the number of groups that you formed are still 6. That is what the question has asked. Infact after selection you are taking arrangements into account and arrangements work out to be similar and you apply correction for the same. I am fine with it.

My limited point is that the question does not adequately clarifies that it is an arrangement question in the way it is phrased.

Because of this ambiguity, I believe that this question is unlikely to be a GMAT OG / GMAT prep / Real GMAT problem.
Expert Post
1 KUDOS received
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2679
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 13 Jun 2017, 12:13
1
1
gmat1011 wrote:
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105


First, we need to select the groups:

Since we have 8 people, the first team can be formed in 8C2 = (8 x 7)/2! = 28 ways. Since there are now 6 people left, the second team can be selected in 6C2 = (6 x 5)/2! = 15 ways. Since there are 4 people left, the third team can be selected in 4C2 = (4 x 3)/2! = 6 ways. Since there are 2 people left, the final team can be selected in 2C2 = 1 way.

Thus, the total number of ways to select the 4 teams is 28 x 15 x 6 x 1 = 2520, if the order of selecting these teams matters. However, the order of selection doesn’t matter, so we have to divide by 4! = 24. Thus, the total number of ways to select the teams when order doesn’t matter is 2520/24 = 105.

Answer: E
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Director
Director
User avatar
P
Joined: 13 Mar 2017
Posts: 610
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 21 Aug 2017, 00:18
gmat1011 wrote:
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

Hypo:

Instead, if the problem were to say: A group of 9 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

Can the answer be arrived at as follows?

select 8 out of 9 people, first: 9C8 = 9

Then select the sets of 2 people

8!/2^4*4! = 105

answer = 9*105 ---> is this correct? can anyone who knows tell me? Many thanks.



We need to select 4 teams of group of 2 peoples..
Total 8 people are there.

So, No. of ways of selecting 1st group = 8C2 = 28
No. of ways of selecting 2nd group = 6C2 = 15
No. of ways of selecting 3rd group = 4C2 = 6
No. of ways of selecting 4th group = 2C2 = 1

So, if we have to find the number of ways keeping in view that 1st,2nd, 3rd and 4th groups are specific like representing their countries..
But here this is not the case. Groups are equivalent and we don't need to consider the arrangements of group .

So, Final number of ways in which the group can be divided into 4 teams of 2 people = 28*15*6*1/4! = (28 *15*6) /(4*3*2) = 105

Answer E
_________________

CAT 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".

1 KUDOS received
Manager
Manager
avatar
G
Joined: 30 Dec 2016
Posts: 158
Premium Member Reviews Badge
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 22 Aug 2017, 03:47
1
Bunuel wrote:
AkritiMehta wrote:
Hi how do we get the answer for the Ist question as 105. Pls explain


A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups


There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.


Hi Bunuel
I guess there is a slight typo error . All the combinations should be C28∗C26∗C24∗C224!=105C82∗C62∗C42∗C224!=105
C^2_4* is missing i guess.

kudos if my post helped .
_________________

Regards
SandySilva


____________
Hit kudos if my post helped (:

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47157
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

Show Tags

New post 22 Aug 2017, 03:49
sandysilva wrote:
Bunuel wrote:
AkritiMehta wrote:
Hi how do we get the answer for the Ist question as 105. Pls explain


A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups


There is also direct formula for this:

1. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).

2. The number of ways in which \(mn\) different items can be divided equally into \(m\) groups, each containing \(n\) objects and the order of the groups is important is \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.


Hi Bunuel
I guess there is a slight typo error . All the combinations should be C28∗C26∗C24∗C224!=105C82∗C62∗C42∗C224!=105
C^2_4* is missing i guess.

kudos if my post helped .

________________
Edited. Thank you.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Re: A group of 8 friends want to play doubles tennis. How many &nbs [#permalink] 22 Aug 2017, 03:49
Display posts from previous: Sort by

A group of 8 friends want to play doubles tennis. How many

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

Events & Promotions

PREV
NEXT


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®.