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

 It is currently 15 Dec 2018, 10:33

### 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

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # 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 Joined: 02 Sep 2009 Posts: 51218 Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] ### Show Tags 20 Dec 2017, 01:17 12 00:00 Difficulty: 75% (hard) Question Stats: 59% (01:27) correct 41% (01:28) wrong based on 149 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 _________________ ##### Most Helpful Community Reply PS Forum Moderator Joined: 25 Feb 2013 Posts: 1217 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, 02:11 1 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 Math Expert Joined: 02 Aug 2009 Posts: 7107 Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] ### Show Tags 23 Dec 2017, 03:39 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: 1217 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, 03:49 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: 13087 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, 12:21 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 # 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, 12:45.
Last edited by EMPOWERgmatRichC on 22 Feb 2018, 12:21, edited 1 time in total.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8678
Location: Pune, India
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

20 Feb 2018, 03:09
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

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

22 Feb 2018, 08:15
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.

_________________

Jeffery Miller

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

PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1217
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, 08:23
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

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: 11 Sep 2015
Posts: 3238
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

22 Feb 2018, 09:55
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

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

CEO
Joined: 11 Sep 2015
Posts: 3238
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

22 Feb 2018, 10:03
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

Cheers,
Brent
_________________

Test confidently with gmatprepnow.com

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13087
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, 12:23
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!*****

Intern
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

20 Oct 2018, 20:17
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
Joined: 17 May 2016
Posts: 10
Concentration: Economics, Operations
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

22 Oct 2018, 01:39
Hi karishma,
I am not able to understand why have we divide 120/ 6 ?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8678
Location: Pune, India
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

23 Oct 2018, 01:12
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

Intern
Joined: 17 May 2016
Posts: 10
Concentration: Economics, Operations
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

### Show Tags

23 Oct 2018, 01:26
karishma ,
Thanks a lot for clearing this. I am was missing a very simple point.
Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo &nbs [#permalink] 23 Oct 2018, 01:26
Display posts from previous: Sort by