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

It is currently 19 Oct 2018, 21:20

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

A rectangular table seats 4 people on each of two sides, with every pe

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

Hide Tags

Manager
Manager
User avatar
Joined: 11 Feb 2008
Posts: 76
A rectangular table seats 4 people on each of two sides, with every  [#permalink]

Show Tags

New post 24 Mar 2010, 02:59
1
12
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

54% (01:48) correct 46% (02:19) wrong based on 147 sessions

HideShow timer Statistics

A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7

_________________

-------------------------------------------------------------------------------------------
Amar
http://amarnaik.wordpress.com

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
A rectangular table seats 4 people on each of two sides, with every  [#permalink]

Show Tags

New post 24 Mar 2010, 13:41
3
anaik100 wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7


Person A can take any place, probability that person B will take the opposite seat is 1/7 (as there are 7 seats left).

Answer: C.
_________________

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

Manager
Manager
User avatar
G
Joined: 22 Jun 2016
Posts: 246
Reviews Badge
A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 24 Jul 2016, 10:04
2
A B C D
------------
| TABLE |
------------
E F G H

A to H are 8 people.

Prob to select any 1 person = 1

Prob to select the person opposite to the chosen person = 1/7

For ex. If we select A as the person than prob of choosing E is 1/7.

Hence, answer will be C.

_________________

P.S. Don't forget to give Kudos :)

Intern
Intern
avatar
Joined: 31 Aug 2014
Posts: 4
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 28 Jul 2016, 11:27
2
We are given :

+ + + +
TABLE
+ + + +

Probability = # Favorable Outcomes / # Total Outcomes

Total Outcomes = 8 * 7 (Since a spot on the table for 2 people can be selected in 8 * 7 ways) = 56

Favorable Outcomes = 8 * 1 (One person can pick a seat in 8 ways, but to sit opposite to that person, there is only 1 option, thus 8*1) = 8

Probability = 8/56 = 1/7, thus the Answer is C.

Thanks for the Kudos! :)
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8397
Location: Pune, India
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 28 Jul 2016, 22:38
3
Bunuel wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7


The question asks for the probability of any two people (say A and B) facing each other.

Method 1:
A can take any seat. After that B should take the seat facing A which is 1 of the 7 remaining seats. So probability of A and B facing each other = 1/7

Method 2:
In how many ways can 8 people can sit on a rectangular table with 4 chairs on 2 opposite sides? The first person can sit in 4 ways (there are 4 distinct seats with respect to the table. Think circular arrangement). Now there are 7 distinct seats and 7 people so they can sit in 7! ways.
Total number of ways = 4*7!

In how many ways can 8 people sit such that A and B face each other? A goes and sits in 4 ways (as before). B sits directly opposite him in 1 way only. Now we have 6 distinct seats and 6 people. They can sit in 6! ways.

Probability = 4*6!/4*7! = 1/7
_________________

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!

Manager
Manager
User avatar
B
Status: Profile 1
Joined: 20 Sep 2015
Posts: 66
GMAT 1: 690 Q48 V37
GPA: 3.2
WE: Information Technology (Investment Banking)
GMAT ToolKit User
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 31 Jul 2017, 02:28
1
1
anaik100 wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7



Consider those 2 persson A and B so -

Fav condition:
A can sit in any of seat around the table no of ways - 8
B has to take seat opposite to A no of ways - 1
other member can take seat other 6 seats no of ways - 6!

All condition:
all possible combination without restriction no of ways - 8!

\(prob = fav/all => 8*1*6!/8!\)
\(prob = 1/7\)
Director
Director
User avatar
P
Joined: 13 Mar 2017
Posts: 622
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 09 Aug 2017, 08:08
VeritasPrepKarishma wrote:
Bunuel wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7


The question asks for the probability of any two people (say A and B) facing each other.

Method 1:
A can take any seat. After that B should take the seat facing A which is 1 of the 7 remaining seats. So probability of A and B facing each other = 1/7

Method 2:
In how many ways can 8 people can sit on a rectangular table with 4 chairs on 2 opposite sides? The first person can sit in 4 ways (there are 4 distinct seats with respect to the table. Think circular arrangement). Now there are 7 distinct seats and 7 people so they can sit in 7! ways.
Total number of ways = 4*7!

In how many ways can 8 people sit such that A and B face each other? A goes and sits in 4 ways (as before). B sits directly opposite him in 1 way only. Now we have 6 distinct seats and 6 people. They can sit in 6! ways.

Probability = 4*6!/4*7! = 1/7


To solve this question we can do it by 2 methods....
Let the 2 persons facing each other be A & B
Method 1 : Place A at a position... Now there are 7 spaces for B to seat but in only one case it is opposite to B. So probability of B facing A = 1/7.

Method 2 : Counting total no. of arrangements
So, Total no. of ways of seating all the 8 people = 8!
(VeritasPrepKarishma I disagree with you here for total no. of ways. Lets A is seated at corner seat then left and right seating of A is very different in nature. Hence it may not be considered as circular arrangement.)

Total no. of desired ways in which A & B faces each other = 4 (pair of opposite seats) * 2 (arranging A and B on the pair of opposite seat) * 6! (arranging other 6 people on the 6 seats) = 4*2*6!

So required probability = 4*2*6!/8! = 1/7

Answer C
_________________

CAT 2017 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu


Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)



What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".

Manager
Manager
avatar
B
Joined: 08 Sep 2016
Posts: 118
A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 28 Sep 2018, 17:00
Hello. Can someone tell me if my approach is valid?

Total possible ways are 8C2 = 28

Favorable outcome (choosing 1 person from 1 side) = 4c1= 4

4C1/ 8C2 = 4/28 = 1/7
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8397
Location: Pune, India
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 28 Sep 2018, 19:59
shashankism wrote:
VeritasPrepKarishma wrote:
Bunuel wrote:
A rectangular table seats 4 people on each of two sides, with every person directly facing another person across the table. If eight people choose their seats at random, what is probability that any two of them directly face other?

(A) 1/56
(B) 1/8
(C) 1/7
(D) 15/56
(E) 4/7


The question asks for the probability of any two people (say A and B) facing each other.

Method 1:
A can take any seat. After that B should take the seat facing A which is 1 of the 7 remaining seats. So probability of A and B facing each other = 1/7

Method 2:
In how many ways can 8 people can sit on a rectangular table with 4 chairs on 2 opposite sides? The first person can sit in 4 ways (there are 4 distinct seats with respect to the table. Think circular arrangement). Now there are 7 distinct seats and 7 people so they can sit in 7! ways.
Total number of ways = 4*7!

In how many ways can 8 people sit such that A and B face each other? A goes and sits in 4 ways (as before). B sits directly opposite him in 1 way only. Now we have 6 distinct seats and 6 people. They can sit in 6! ways.

Probability = 4*6!/4*7! = 1/7


To solve this question we can do it by 2 methods....
Let the 2 persons facing each other be A & B
Method 1 : Place A at a position... Now there are 7 spaces for B to seat but in only one case it is opposite to B. So probability of B facing A = 1/7.

Method 2 : Counting total no. of arrangements
So, Total no. of ways of seating all the 8 people = 8!
(VeritasPrepKarishma I disagree with you here for total no. of ways. Lets A is seated at corner seat then left and right seating of A is very different in nature. Hence it may not be considered as circular arrangement.)

Total no. of desired ways in which A & B faces each other = 4 (pair of opposite seats) * 2 (arranging A and B on the pair of opposite seat) * 6! (arranging other 6 people on the 6 seats) = 4*2*6!

So required probability = 4*2*6!/8! = 1/7

Answer C



Look at this:

Attachment:
Fig1.jpeg
Fig1.jpeg [ 22.95 KiB | Viewed 322 times ]

If the first person comes and sits at A, it is exactly the same as sitting at D (table corner on his left and a person on his right, the shorter side of the rectangle on the left). Similarly, positions B and C are exactly the same.
Using same logic for the shorter sides of the rectangle too, we will get 4 unique places to sit for the first person, not 8 (with respect to the table)
_________________

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!

CR Forum Moderator
avatar
G
Joined: 25 Apr 2018
Posts: 239
Premium Member CAT Tests
Re: A rectangular table seats 4 people on each of two sides, with every pe  [#permalink]

Show Tags

New post 28 Sep 2018, 22:14
hdavies wrote:
Hello. Can someone tell me if my approach is valid?

Total possible ways are 8C2 = 28

Favorable outcome (choosing 1 person from 1 side) = 4c1= 4

4C1/ 8C2 = 4/28 = 1/7


Bunuel chetan2u VeritasKarishma

Can you please comment on the above approach whether it is right?
_________________

Please give kudos if you found my posts helpful!

GMAT Club Bot
Re: A rectangular table seats 4 people on each of two sides, with every pe &nbs [#permalink] 28 Sep 2018, 22:14
Display posts from previous: Sort by

A rectangular table seats 4 people on each of two sides, with every pe

  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®.