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

It is currently 21 May 2019, 20:56

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

Runners V, W, X, Y, and Z are competing in the Bayville local triathlo

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

Hide Tags

 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 55230
Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 20 Dec 2017, 02:17
14
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

55% (02:16) correct 45% (01:57) wrong based on 141 sessions

HideShow timer Statistics

Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7684
Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 23 Dec 2017, 04:39
2
1
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120



Hi..

There are Two restrictions..
1) \(V>W>Z\)
2) \(X>Y\)

Let's just take first restriction...
1) \(V>W>Z\)
Total ways for all \(5 = 5!=120\)
V,W,Z can be arranged in 3! or 6 ways AND only one of it will be V>W>Z
So \(\frac{120}{6}=20\) ways..

2) now half of 20 will have x before y and half y before x..
So \(\frac{20}{2}=10\)
B...

niks18, you have missed out on first restriction and so you are getting D as answer above
_________________
Most Helpful Community Reply
Retired Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1214
Location: India
GPA: 3.82
GMAT ToolKit User Reviews Badge
Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 23 Dec 2017, 03:11
4
4
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120


No of ways 5 runners can run the race \(= 5!=120\)

Now order of finishing the race for V, W & Z is V...W...Z. VWZ among themselves can have \(3!=6\) positions but only \(1\) position is allowed.

So number of ways runners can run a race \(= \frac{120}{6}=20\)

Half the time X beats Y (and half the time Y beats X)

so no of ways X finishes before Y \(=\frac{20}{2} = 10\)

Option B
General Discussion
Retired Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1214
Location: India
GPA: 3.82
GMAT ToolKit User Reviews Badge
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 23 Dec 2017, 04:49
chetan2u wrote:
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120



Hi..

There are Two restrictions..
1) V>W>Z
2) X>Y

Let's just take first restriction...
1) V>W>Z
Total ways for all 5 = 5!=120
V,W,Z can be arranged in 3! or 6 ways AND only one of it will be V>W>Z
So 120/6=20 ways..
2) now half of 20 will have x before y and half y before x..
So 20/2=10
B...

niks18, you have missed out on first restriction and so you are getting D as answer above


Thanks chetan2u for highlighting :thumbup: . yup I completely ignored that condition :(
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14188
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post Updated on: 22 Feb 2018, 13:21
Hi All,

Assuming that there are no "ties" - and IF there were no 'restrictions', then the 5 runners would have 5! = 120 different outcomes. However, there ARE some restrictions that we have to deal with:

1) V finishes before W and W finishes before Z. By placing these 3 runners in these three 'relative positions', we have to eliminate all of the possibilities in which this outcome does not occur. That can be done by dividing the total by 3! = 6. Thus 120 options are reduced to 120/6 = 20 options
2) Runner X will finish ahead of Runner Y HALF the time, so the number of options is further reduced by half: 20/2 = 10

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

Originally posted by EMPOWERgmatRichC on 19 Feb 2018, 13:45.
Last edited by EMPOWERgmatRichC on 22 Feb 2018, 13:21, edited 1 time in total.
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 9230
Location: Pune, India
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 20 Feb 2018, 04:09
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120


Check this post:

https://www.veritasprep.com/blog/2011/1 ... s-part-ii/

We extend the same concept to this question. VWZ can be arranged in 3! ways but only 1 way is acceptable. So we divide 120 by 3! to get 20.
X finishes before Y in half of these cases so answer is 10.
_________________
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 >
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2823
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 22 Feb 2018, 09:15
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120



Without any restrictions, we have 5! = 120 arrangements. There are 3! = 6 arrangements for V, W and Z but only one counts because we want “V > W > Z,” so we only have 1/6 x 120 = 20 arrangements possible when “V > W > Z.” Of these 20 arrangements, half of them are “X > Y” and the other half are “Y > X.” Thus there are 20/2 = 10 arrangements where “X > Y” i.e., X finishes before Y.

Answer: B
_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Retired Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1214
Location: India
GPA: 3.82
GMAT ToolKit User Reviews Badge
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 22 Feb 2018, 09:23
1
EMPOWERgmatRichC wrote:
Hi All,

Assuming that there are no "ties", the 5 runners would have 5! = 120 different outcomes. Since Runner X will finish ahead of Runner Y half the time, the number of options is 120/2 = 60

Final Answer:

GMAT assassins aren't born, they're made,
Rich


Hi EMPOWERgmatRichC

Your solution ignores one essential condition that VWZ also has to maintain a particular order, hence you are getting answer as option D.
CEO
CEO
User avatar
V
Joined: 12 Sep 2015
Posts: 3722
Location: Canada
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 22 Feb 2018, 10:55
Top Contributor
1
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120


A different approach...

Take the task of arranging the 5 runners and break it into stages.

GIVEN: V finishes before W and W finishes before Z
So, the order is: Z - W - V
Our goal is to now place the remaining 2 runners (runner X and runner Y)

NOTE: I'm going to IGNORE the restriction that says X must finish before Y
You'll see why shortly.

Stage 1: place runner X into the existing order.
Notice that if we already have the arrangement Z - W - V, then we can place spaces in the areas where runner X might go.
We have: _Z_W_V_
Since there are 4 spaces where we can place runner X, we can complete stage 1 in 4 ways

Stage 2: place runner Y into the existing order.
At this point, we have placed runners Z, W, V and X
Let's pretend for a moment, that the arrangement is ZXWV
From here, we can place spaces in the areas where runner Y might go.
We have: _Z_X_W_V_
Since there are 5 spaces where we can place runner Y, we can complete stage 2 in 5 ways

By the Fundamental Counting Principle (FCP), we can the 2 stages (and thus arrange all 5 runners) in (4)(5) ways (= 20 ways)

IMPORTANT: If we IGNORE the restriction that says X must finish before Y, then there are 20 possible arrangements.
However, in HALF of those 20 arrangements, X is ahead of Y, and in the other HALF of those 20 arrangements, Y is ahead of X.

So, the number of arrangements in which X is ahead of Y = 20/2 = 10

Answer: B

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEOS



_________________
Test confidently with gmatprepnow.com
Image
CEO
CEO
User avatar
V
Joined: 12 Sep 2015
Posts: 3722
Location: Canada
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 22 Feb 2018, 11:03
1
Top Contributor
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120


Another option is to systematically list and count the possible outcomes
To begin, Z, W and V must have the following order: Z-W-V

Now insert an X and a Y so that X is ahead of Y.
We get:
1) YZWVX
2) ZYWVX
3) ZWYVX
4) ZWVYX
5) YZWXV
6) ZYWXV
7) ZWYXV
8) YZXWV
9) ZYXWV
10) YXZWV


Answer: B

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
Image
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14188
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 22 Feb 2018, 13:23
Hi niks18,

Nice catch! Looking over my notes, it appears that I didn't write anything down about V, W and Z. There's a big 'takeaway' there about making sure that proper note-taking is done consistently!

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
Intern
Intern
avatar
B
Joined: 08 Aug 2018
Posts: 40
Location: India
GMAT 1: 720 Q49 V40
GPA: 4
WE: Engineering (Energy and Utilities)
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 20 Oct 2018, 21:17
chetan2u wrote:
Bunuel wrote:
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??

A. 5
B. 10
C. 30
D. 60
E. 120



Hi..

There are Two restrictions..
1) \(V>W>Z\)
2) \(X>Y\)

Let's just take first restriction...
1) \(V>W>Z\)
Total ways for all \(5 = 5!=120\)
V,W,Z can be arranged in 3! or 6 ways AND only one of it will be V>W>Z
So \(\frac{120}{6}=20\) ways..

2) now half of 20 will have x before y and half y before x..
So \(\frac{20}{2}=10\)
B...

niks18, you have missed out on first restriction and so you are getting D as answer above


For my understanding, which principle has been used to calculate Restriction 1 (5!/3!)? Mississipi Formula? Can we express it in terms of combinations?
Intern
Intern
avatar
B
Joined: 17 May 2016
Posts: 16
Location: Canada
Concentration: Economics, Operations
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 22 Oct 2018, 02:39
Hi karishma,
I am not able to understand why have we divide 120/ 6 ?
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 9230
Location: Pune, India
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 23 Oct 2018, 02:12
1
Piggu18 wrote:
Hi karishma,
I am not able to understand why have we divide 120/ 6 ?


5 runners - V, W, X, Y, Z

They can complete the race in 5! ways = 120 ways.

VWXYZ
WXVYZ
ZXVWY
...
etc

If we focus on only 3 of these V, W and Z, there are different arrangements possible.
VWZ
WVZ
ZWV
etc...

These 3 can be arranged in 3! ways = 3*2*1 = 6 ways
But of these 6, only 1 way is acceptable to us: VWZ

So out of every 6 ways in the 120 ways, only 1 way is acceptable to us. So we divide 120 by 6 to get 20 acceptable ways.
_________________
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 >
Intern
Intern
avatar
B
Joined: 17 May 2016
Posts: 16
Location: Canada
Concentration: Economics, Operations
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 23 Oct 2018, 02:26
karishma ,
Thanks a lot for clearing this. I am was missing a very simple point.
Intern
Intern
avatar
B
Joined: 02 Jan 2017
Posts: 46
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 24 Apr 2019, 05:33
EMPOWERgmatRichC wrote:
Hi All,

Assuming that there are no "ties" - and IF there were no 'restrictions', then the 5 runners would have 5! = 120 different outcomes. However, there ARE some restrictions that we have to deal with:

1) V finishes before W and W finishes before Z. By placing these 3 runners in these three 'relative positions', we have to eliminate all of the possibilities in which this outcome does not occur. That can be done by dividing the total by 3! = 6. Thus 120 options are reduced to 120/6 = 20 options
2) Runner X will finish ahead of Runner Y HALF the time, so the number of options is further reduced by half: 20/2 = 10

Final Answer:

GMAT assassins aren't born, they're made,
Rich


Why are we dividing it by 6!. I cannot understand . Shouldnt it be subtraction
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14188
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

Show Tags

New post 24 Apr 2019, 12:56
Hi AlN,

You could certainly calculate ALL of the ways that "don't fit" what we're looking for - and then subtract that number from 120 - but that would be a LOT of work.

Here's another way to consider the given information:

Let's focus on just the 3 runners: V, W and Z. If there were no 'restrictions' on how they finished (and there were no "ties"), then there would be 6 possible outcomes:

VWZ
VZW
WVZ
WZV
ZWV
ZVW

This prompt DOES give us the restriction that V has to finish before W and W has to finish before Z. Thus, of those 6 possibilities, 5 of them have to be removed. Here, we can do that by dividing 6 by 3!, since 6/3! = 1... the 1 option that "fits" what we were told. Including a 4th or 5th runner does not change that math.

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
GMAT Club Bot
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo   [#permalink] 24 Apr 2019, 12:56
Display posts from previous: Sort by

Runners V, W, X, Y, and Z are competing in the Bayville local triathlo

  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®.