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

It is currently 15 Oct 2018, 18:36

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

In racing over a given distance d at uniform speed, A can be

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

Hide Tags

Board of Directors
User avatar
P
Joined: 01 Sep 2010
Posts: 3305
In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post Updated on: 09 Jul 2012, 04:26
14
32
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

39% (02:53) correct 61% (02:28) wrong based on 813 sessions

HideShow timer Statistics

In racing over a given distance d at uniform speed, A can beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal

A. not determined by the give information
B. 58
C. 100
D. 116
E. 120

_________________

COLLECTION OF QUESTIONS AND RESOURCES
Quant: 1. ALL GMATPrep questions Quant/Verbal 2. Bunuel Signature Collection - The Next Generation 3. Bunuel Signature Collection ALL-IN-ONE WITH SOLUTIONS 4. Veritas Prep Blog PDF Version 5. MGMAT Study Hall Thursdays with Ron Quant Videos
Verbal:1. Verbal question bank and directories by Carcass 2. MGMAT Study Hall Thursdays with Ron Verbal Videos 3. Critical Reasoning_Oldy but goldy question banks 4. Sentence Correction_Oldy but goldy question banks 5. Reading-comprehension_Oldy but goldy question banks


Originally posted by carcass on 09 Jul 2012, 03:42.
Last edited by Bunuel on 09 Jul 2012, 04:26, edited 2 times in total.
Edited the question.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49858
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 09 Jul 2012, 04:24
29
13
carcass wrote:
In racing over a given distance d at uniform speed, A can be beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal

A. not determined by the give information
B. 58
C. 100
D. 116
E. 120


Given:
A can beat B by 20 yards,
A can beat C by 28 yards,
B can beat C by 10 yards,

So, when A is on the finish line B is 20 yards back and C is 28 yards back.

Hence, 20 yards before the finish line B is 8 yards ahead of C, and since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards.

Answer: C.

Hope it's clear.
_________________

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

Most Helpful Community Reply
Manager
Manager
User avatar
Affiliations: Project Management Professional (PMP)
Joined: 30 Jun 2011
Posts: 152
Location: New Delhi, India
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 09 Jul 2012, 07:14
10
2
Bunuel wrote:
carcass wrote:
In racing over a given distance d at uniform speed, A can be beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal

A. not determined by the give information
B. 58
C. 100
D. 116
E. 120


Given:
A can beat B by 20 yards,
A can beat C by 28 yards,
B can beat C by 10 yards,

So, when A is on the finish line B is 20 yards back and C is 28 yards back.

Hence, 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards.

Answer: C.

Hope it's clear.

Hi
Can we also resolve by comparing time taken,
i.e. d/a=(d-20)/b
d/b=(d-10)/c
d/a=(d-28)/c

we can resolve d=100 from these eqn..
_________________

Best
Vaibhav

If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks

General Discussion
Intern
Intern
avatar
Joined: 10 Apr 2012
Posts: 1
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 20 Jul 2012, 20:39
I want to ask bunnel how did he get this
" 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards.."
can you pls explain a bit in detail.
thanks
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49858
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 22 Jul 2012, 03:19
1
hardwoker1010 wrote:
I want to ask bunnel how did he get this
" 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards.."
can you pls explain a bit in detail.
thanks


Consider the case when A, B and C are racing each other. When A is on the finish line B must be 20 yards from the finish line and C must be 28 yards from the finish line. Hence when B is 20 yards from the finish line he's 28-20=8 yards ahead of C.

Hope it's clear.
_________________

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

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49858
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 27 Jun 2013, 02:25
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Distance/Rate Problems: distance-speed-time-word-problems-made-easy-87481.html

All DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
All PS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=64

_________________

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

Director
Director
User avatar
B
Joined: 17 Dec 2012
Posts: 629
Location: India
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 27 Jun 2013, 17:50
16
4
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100
_________________

Srinivasan Vaidyaraman
Sravna Holistic Solutions
http://www.sravnatestprep.com

Holistic and Systematic Approach

Intern
Intern
avatar
Joined: 27 May 2013
Posts: 1
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 04 Jul 2013, 10:44
2
2
Think it into a other way.
1) When A completes, B is 20 Behind and C is 28 Behind. C is 8 behind B.
2) When B completes, C is 10 Behind.

What it refers is that in last 20 yards B has gained 2 yards, which refers that B is gaining 1 yard over C in every 10 yard it covers, so in last 80 yards it has gained 8 yards, so total is 80+20 = 100 :D
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 429
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 03 Aug 2013, 12:16
5
1
In racing over a given distance d at uniform speed, A can beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal

We are given nothing about time and nothing about rate. Solving using a traditional formula (i.e. distance = rate*time) might be difficult. Is that correct?

A must be 28 yards ahead of C. We know that A beats both B and C.

When A hits the finish line, it is 20 yards ahead of B and 28 yards ahead of C. This means at this time B is 8 yards ahead of C but it beats C by 10 yards. In 20 yards, it has to gain a two further yards between itself and C. Because A, B, C travel at a constant rate, B gains on C by 2 yards every 20 yards. This means that when B and C started out they were even and after 20 yards B was two ahead of C, after 40 yards B was 4 ahead of C, etc. B would have to travel 100 yards for it to gain 10 yards on C.

Answer: C. 100
Intern
Intern
avatar
Joined: 13 Aug 2013
Posts: 21
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 16 Aug 2013, 03:13
3
1
Another method to solve such questions : Let us consider the case when all three are racing each other. A is ahead of B by 20 yards at the finish and ahead of c by 28 yards, which means B is ahead of C by 8 yards at that instant. Now B defeats C by 10 yards, means the duration during which B covered 20, C covered 8+10=18. Hence, the ratio of their speeds is 10:9. That means in a race of 10yards B defeats c by 1 yard , so to defeat him by 10 yards the race should be of 10x10 =100 yards.
Director
Director
avatar
Joined: 29 Nov 2012
Posts: 775
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 26 Sep 2013, 01:49
Bunuel wrote:

Given:
A can beat B by 20 yards,
A can beat C by 28 yards,
B can beat C by 10 yards,

So, when A is on the finish line B is 20 yards back and C is 28 yards back.



Can you explain this part with a diagram

<--------C---------------B-----------------A ( finish line)

So the distance between A and B is 20 and A and C is 28 and not b and c is this correct?
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49858
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 26 Sep 2013, 01:56
fozzzy wrote:
Bunuel wrote:

Given:
A can beat B by 20 yards,
A can beat C by 28 yards,
B can beat C by 10 yards,

So, when A is on the finish line B is 20 yards back and C is 28 yards back.



Can you explain this part with a diagram

<--------C-------(8)--------B-------(20)----------A ( finish line)

So the distance between A and B is 20 and A and C is 28 and not b and c is this correct?


Yes, that's correct: when A is on the finish line B is 20 yards back and C is 28 yards back.
_________________

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

Intern
Intern
avatar
Joined: 14 Oct 2013
Posts: 11
Location: India
Concentration: General Management
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 10 Jan 2014, 07:04
SravnaTestPrep wrote:
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100



Hi Sravna,

I dont follow how the ratio of the speeds are constant? what about the time factor while equating? as per the discussions in the various for this questions here it is re-iterated time and again that B gains 2 yards in the last 20 yards? if that is the case how can the speed ratios be constant?

please help. confused :? :roll:
Intern
Intern
avatar
Joined: 15 Apr 2014
Posts: 10
Concentration: Strategy, Marketing
GMAT 1: 700 Q49 V36
WE: Marketing (Advertising and PR)
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 13 May 2014, 09:28
narangvaibhav wrote:
Bunuel wrote:
carcass wrote:
In racing over a given distance d at uniform speed, A can be beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal

A. not determined by the give information
B. 58
C. 100
D. 116
E. 120


Given:
A can beat B by 20 yards,
A can beat C by 28 yards,
B can beat C by 10 yards,

So, when A is on the finish line B is 20 yards back and C is 28 yards back.

Hence, 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards.

Answer: C.

Hope it's clear.

Hi
Can we also resolve by comparing time taken,
i.e. d/a=(d-20)/b
d/b=(d-10)/c
d/a=(d-28)/c

we can resolve d=100 from these eqn..



You can'tresolve this..... 3equation and four variables....need one more equation here
_________________

May everyone succeed in their endeavor. God Bless!!!

Intern
Intern
avatar
Joined: 17 May 2014
Posts: 40
In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 18 May 2014, 04:39
1
rawjetraw wrote:
SravnaTestPrep wrote:
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100



Hi Sravna,

I dont follow how the ratio of the speeds are constant? what about the time factor while equating? as per the discussions in the various for this questions here it is re-iterated time and again that B gains 2 yards in the last 20 yards? if that is the case how can the speed ratios be constant?

please help. confused :? :roll:



Lemme try and explain.

In a given times A runs d, B runs d-20, and C runs d-28. Since the time is same their speed a, b, c are in the ratio of the distance the travelled
Thus
a:b:c = d:d-20:d-28
Also, we know that b:c= d: d-10

Thus, equating b:c in both the cases, we get

d-20/d-28 = d/d-10

Solving this we get d= 100.

A more intuitive approach has been given by Bunuel in this tread. The approach works as the speeds are constant and are in direct proportion with the distance if time traveled is same.

Hope it helps!!!
Intern
Intern
avatar
Joined: 17 May 2014
Posts: 40
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 18 May 2014, 04:46
enders wrote:
narangvaibhav wrote:
Bunuel wrote:
In racing over a given distance d at uniform speed, A can be beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal

A. not determined by the give information
B. 58
C. 100
D. 116
E. 120


Given:
A can beat B by 20 yards,
A can beat C by 28 yards,
B can beat C by 10 yards,

So, when A is on the finish line B is 20 yards back and C is 28 yards back.

Hence, 20 yards before the finish line C is 8 yards ahead of B, and since the final difference between B and C is 10 yards, then B gains 2 yards every 20 yard. To gain the final difference of 10=2*5 yards B should run total of 20*5=100 yards.

Answer: C.

Hope it's clear.
Hi
Can we also resolve by comparing time taken,
i.e. d/a=(d-20)/b
d/b=(d-10)/c
d/a=(d-28)/c

we can resolve d=100 from these eqn..



You can'tresolve this..... 3equation and four variables....need one more equation here


You can resolve this very easily. Also, we don't need to find all the four variables. We can't find the speed of a, b and c individually as we don't know about the distance

The first one can be written as a/b = d/d-20
second one is b/c = d/d-10

Multiplying these two yeilds a/c = d^2/(d-10)(d-20)

Third one also says a/c = d/d-28

We equate these 2 and get the result as 100.
A more intuitive way has been already discusses. You should use that way.
Director
Director
User avatar
B
Joined: 17 Dec 2012
Posts: 629
Location: India
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 17 Jun 2014, 02:07
rawjetraw wrote:
SravnaTestPrep wrote:
1. When A touches d, B is at d-20 and C is at d-28
2. When B completes d, C completes d-10
3. Equate d-20 / d-28 = d / d-10 because the ratio of the speed of B to the speed of C is a constant.
4. d= 100



Hi Sravna,

I dont follow how the ratio of the speeds are constant? what about the time factor while equating? as per the discussions in the various for this questions here it is re-iterated time and again that B gains 2 yards in the last 20 yards? if that is the case how can the speed ratios be constant?

please help. confused :? :roll:


Hi,

Very sorry for such a late reply.

Let us say that B travels a distance x and in the same time C travels distance y. The ratio of the distance = x/y. This is also the ratio of their speed since time is the same.

In the actual problem we have this ratio as d-20/d-28. This gives the ratio of their speed.

Similarly when B reaches d, c reaches d-10. The time taken by both to reach the respective distance is the same. This d/d-10 also gives the ratio of the speed of B and C. Thus we can equate the two and find d.
_________________

Srinivasan Vaidyaraman
Sravna Holistic Solutions
http://www.sravnatestprep.com

Holistic and Systematic Approach

Manager
Manager
User avatar
Joined: 10 Jun 2015
Posts: 118
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 16 Jun 2015, 04:38
carcass wrote:
In racing over a given distance d at uniform speed, A can beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equal

A. not determined by the give information
B. 58
C. 100
D. 116
E. 120

(C) is the answer
I applied substitution method.
I considered the distance as 100 yards and A takes 1 hour to cover the distance. So his speed is 100 yards per hour. B's speed is 80 yards per hour.
Therefore B will take 75 minutes to cover 100 yards. In 75 minutes C covers 90 yards. Hence, in 60 minutes C covers 72 yards. That is his speed.
VP
VP
avatar
P
Joined: 07 Dec 2014
Posts: 1102
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 06 Nov 2015, 18:23
Let d=distance of race.
Between A's finish and B's finish, B runs 20 yards while C runs 18 yards.
B's rate to C's rate=10/9.
When B finishes, he is 10 yards ahead of C.
10/9=d/(d-10)
d=100 yards
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 8400
Premium Member
Re: In racing over a given distance d at uniform speed, A can be  [#permalink]

Show Tags

New post 04 Apr 2018, 05:35
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: In racing over a given distance d at uniform speed, A can be &nbs [#permalink] 04 Apr 2018, 05:35
Display posts from previous: Sort by

In racing over a given distance d at uniform speed, A can be

  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®.