Find all School-related info fast with the new School-Specific MBA Forum

It is currently 24 Oct 2014, 17:05

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Meg and Bob are among the 5 participants in a cycling race.

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Director
Director
avatar
Joined: 22 Aug 2007
Posts: 573
Followers: 1

Kudos [?]: 13 [0], given: 0

Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 22 Oct 2007, 19:14
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

66% (02:27) correct 34% (01:06) wrong based on 72 sessions
Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?

A. 24
B. 30
C. 60
D. 90
E. 120

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Jul 2014, 08:28, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Kaplan Promo CodeKnewton GMAT Discount CodesManhattan GMAT Discount Codes
VP
VP
avatar
Joined: 10 Jun 2007
Posts: 1467
Followers: 6

Kudos [?]: 108 [0], given: 0

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 22 Oct 2007, 19:22
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

please post your solution too, thank you.



JXXXX => Bob can be in any X...Total 4! = 24
XJXXX => Bob can be in the last 3 Xs...Total 3! = 6
XXJXX => 2! = 2
XXXJX => 1! = 1

24+6+2+1 = 33
VP
VP
avatar
Joined: 28 Mar 2006
Posts: 1388
Followers: 2

Kudos [?]: 19 [0], given: 0

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 22 Oct 2007, 19:31
bkk145 wrote:
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

please post your solution too, thank you.



JXXXX => Bob can be in any X...Total 4! = 24
XJXXX => Bob can be in the last 3 Xs...Total 3! = 6
XXJXX => 2! = 2
XXXJX => 1! = 1

24+6+2+1 = 33


Yep I missed factorials for arranagements
VP
VP
avatar
Joined: 10 Jun 2007
Posts: 1467
Followers: 6

Kudos [?]: 108 [0], given: 0

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 22 Oct 2007, 20:34
Let me redeem myself:

JXXXX => Bob can be in any X...Total 4! = 24

XJXXX => Bob can be in the last 3 Xs...Total 3! = 6. However, since the first person can change too and it cannot be Bob, it can only be 3 other people. So Total = 6*3 = 18

XXJBX => 3! = 6 total
XXJXB => 3! = 6 total
So this yields total of 12

XXXJB => 3! = 6


Ans = 24+18+12+6 = 60
2 KUDOS received
SVP
SVP
User avatar
Joined: 24 Sep 2005
Posts: 1898
Followers: 10

Kudos [?]: 110 [2] , given: 0

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 16 Nov 2007, 22:24
2
This post received
KUDOS
Refer to one method promoted by Johnr, i figured out a quick solution to this problem:

with 5 participants, there're 5! ways to arrange them.

Because Bob and Jen have identical roles in these 5 participants, the chance that each of them finishes before the another is 50%. Thus, the number of ways in which Jen finishes before Bob is 5! * 50% = 60 ways
Director
Director
avatar
Joined: 09 Aug 2006
Posts: 767
Followers: 1

Kudos [?]: 52 [0], given: 0

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 16 Nov 2007, 22:31
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

please post your solution too, thank you.


J has to finish before B (given)

J finishes 1st then B can finish in 4 ways and other three can finish in 3! ways to give us 4*3! ways = 24
J finishes 2nd then B can finish in 3 ways and other three can finish in 3! ways to give us 3*3! ways = 18
J finishes 3rd then B can finish in 2 ways and other three can finish in 3! ways to give us 2*3! ways = 12
J finishes 4th then B can finish in 1 way (last) and other three can finish in 3! ways to give us 1*3! ways = 6

Total = 24 + 18 + 12+ 6 = 60
Director
Director
avatar
Joined: 01 May 2007
Posts: 795
Followers: 1

Kudos [?]: 79 [0], given: 0

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 17 Nov 2007, 16:07
What's the assessment of level of difficulty of this prob? Range one would see this problem in terms of GMAT scoring?
Director
Director
User avatar
Joined: 26 Feb 2006
Posts: 907
Followers: 4

Kudos [?]: 44 [0], given: 0

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 17 Nov 2007, 16:42
IrinaOK wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

please post your solution too, thank you.


for me, the question is ambigious in the above red part, basically "infront of Bob". does it mean "JB" or "JXXXB"?

if "infront of Bob" means only "JB", then there are 24 (4x3!) ways.
if "infront of Bob" means only "JXXXB", then there are 60 [3!(1+2+3+4)]ways.
2 KUDOS received
Manager
Manager
User avatar
Joined: 27 May 2009
Posts: 222
Followers: 5

Kudos [?]: 35 [2] , given: 2

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 07 Sep 2009, 11:34
2
This post received
KUDOS
There is another way to get this solved. This method can be generalized to solve such kind of problems.

The steps are:
1. Take a specific case that satisfies the given condition.
2. Find the possible arrangements for that case.
3. Find the number of such cases.

Let me elaborate with the solution.

1. We can choose any two positions out of the 5 positions. Say i choose a specific case where Jen is 1st & Bob is 4th.
J _ _ B _
2. The three blank spaces can be filled in 3! ways.
3. Now 2 positions can be selected from 5 positions ( as done in step 1 ) in 5C2 ways

So, total ways: 3! * 5C2 = 60
_________________

I do not suffer from insanity. I enjoy every minute of it.

Manager
Manager
avatar
Joined: 27 Oct 2008
Posts: 186
Followers: 1

Kudos [?]: 80 [0], given: 3

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 28 Sep 2009, 10:25
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?


Soln: total number of ways is 5! = 120 ways

Of these , in exactly half of the arrangements, will Jen finish ahead of Bob. Hence
= 120/2
= 60 ways
Senior Manager
Senior Manager
avatar
Joined: 03 Nov 2005
Posts: 399
Location: Chicago, IL
Followers: 3

Kudos [?]: 25 [0], given: 17

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 01 Oct 2009, 21:03
We can count Bob & Jen as one contestant.

So the total number of permutations is 4!, which is 24.

Did not read the question right.
_________________

Hard work is the main determinant of success

1 KUDOS received
VP
VP
avatar
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1364
Followers: 12

Kudos [?]: 144 [1] , given: 10

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 02 May 2011, 22:30
1
This post received
KUDOS
out of 5! = 120 ways.half of the ways are when B is ahead of J.
hence 60
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

1 KUDOS received
Manager
Manager
avatar
Joined: 09 Aug 2010
Posts: 107
Followers: 1

Kudos [?]: 20 [1] , given: 7

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 03 May 2011, 00:20
1
This post received
KUDOS
There are always 2 ways to arrange Jen and Bob.
First possibility: J(...)B (with or without people in between)
Second possibility: B(...)J (with or without people in between)

Solution: 5!/2 = 60
CEO
CEO
User avatar
Joined: 09 Sep 2013
Posts: 2858
Followers: 207

Kudos [?]: 43 [0], given: 0

Premium Member
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 26 Jul 2014, 05:55
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

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23409
Followers: 3613

Kudos [?]: 28921 [1] , given: 2871

Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 26 Jul 2014, 08:29
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
IrinaOK wrote:
Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?

A. 24
B. 30
C. 60
D. 90
E. 120


Total # of ways the race can be finished is 5!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 5!/2=60.

Answer: C.

Similar questions to practice:
mother-mary-comes-to-me-86407.html (or: mary-and-joe-126407.html);
six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
in-how-many-different-ways-can-the-letters-a-a-b-91460.html
goldenrod-and-no-hope-are-in-a-horse-race-with-6-contestants-82214.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 26 Jul 2014, 08:29
    Similar topics Author Replies Last post
Similar
Topics:
28 Experts publish their posts in the topic Meg and Bob are among the 5 participants in a cycling race. marcodonzelli 13 07 Jan 2008, 05:05
Q17: Meg and Bob are among the 5 participants in a cycling jet1445 2 03 Jun 2007, 19:40
2 Experts publish their posts in the topic Meg and Bob are among the 5 participants in a cycling race. girikorat 8 23 Oct 2006, 09:43
Meg and Bob are among the 5 participants in a cycling race. TOUGH GUY 5 10 Dec 2005, 15:22
Meg and Bob are among the 5 participants in a cycling race. Vithal 7 05 Jun 2005, 13:11
Display posts from previous: Sort by

Meg and Bob are among the 5 participants in a cycling race.

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.