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

 It is currently 26 Sep 2018, 07:51

### 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

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

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49544
Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

20 Dec 2017, 02:17
00:00

Difficulty:

75% (hard)

Question Stats:

51% (02:23) correct 49% (01:53) wrong based on 98 sessions

### HideShow timer Statistics

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

_________________
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1218
Location: India
GPA: 3.82
Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

23 Dec 2017, 03:11
3
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
Math Expert
Joined: 02 Aug 2009
Posts: 6816
Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

23 Dec 2017, 04:39
1
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
_________________

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

PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1218
Location: India
GPA: 3.82
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

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 . yup I completely ignored that condition
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12461
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

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

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

# 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/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** 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 Joined: 16 Oct 2010 Posts: 8301 Location: Pune, India Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] ### Show Tags 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 > GMAT self-study has never been more personalized or more fun. Try ORION Free! Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2835 Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] ### Show Tags 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 _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions PS Forum Moderator Joined: 25 Feb 2013 Posts: 1218 Location: India GPA: 3.82 Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] ### Show Tags 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 Joined: 12 Sep 2015 Posts: 2908 Location: Canada Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] ### Show Tags 22 Feb 2018, 10:55 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 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 _________________ Brent Hanneson – GMATPrepNow.com Sign up for our free Question of the Day emails CEO Joined: 12 Sep 2015 Posts: 2908 Location: Canada Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] ### Show Tags 22 Feb 2018, 11:03 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 _________________ Brent Hanneson – GMATPrepNow.com Sign up for our free Question of the Day emails EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12461 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 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 # 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/

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

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo &nbs [#permalink] 22 Feb 2018, 13:23
Display posts from previous: Sort by

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

## Events & Promotions

 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®.