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

It is currently 21 Nov 2018, 12:20

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
Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
  • All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

     November 22, 2018

     November 22, 2018

     10:00 PM PST

     11:00 PM PST

    Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
  • Key Strategies to Master GMAT SC

     November 24, 2018

     November 24, 2018

     07:00 AM PST

     09:00 AM PST

    Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

In a group of 11 members, 5 have 4 siblings each and 6 have one

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

Hide Tags

Manager
Manager
User avatar
G
Joined: 07 Jun 2017
Posts: 174
Location: India
Concentration: Technology, General Management
GMAT 1: 660 Q46 V38
GPA: 3.6
WE: Information Technology (Computer Software)
GMAT ToolKit User Reviews Badge
In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 30 Oct 2017, 21:17
5
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

48% (02:00) correct 52% (02:15) wrong based on 97 sessions

HideShow timer Statistics

In a group of 11 members, 5 have 4 siblings each and 6 have one sibling each. If two members are picked randomly, what is the probability that they are siblings?

A. \(\frac{16}{55}\)

B. \(\frac{13}{55}\)

C. \(\frac{10}{55}\)

D. \(\frac{6}{11}\)

E. \(\frac{17}{110}\)

_________________

Regards,
Naveen
email: nkmungila@gmail.com
Please press kudos if you like this post

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50730
In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 30 Oct 2017, 22:33
4
nkmungila wrote:
In a group of 11 members, 5 have 4 siblings each and 6 have one sibling each. If two members are picked randomly, what is the probability that they are siblings?

A. \(\frac{16}{55}\)

B. \(\frac{13}{55}\)

C. \(\frac{10}{55}\)

D. \(\frac{6}{11}\)

E. \(\frac{17}{110}\)


5 have 4 siblings each implies that there is one group of 5 siblings {1 - 2 - 3 - 4 - 5} (notice that each of these 5 have 4 siblings);
6 have one sibling each implies that there are three groups of 2 siblings {6 - 7}, {8 - 9}, {10 - 11}.

The total number of ways to choose two people out of 11 is 11C2 = 11!/(2!9!) = 55.

In order to get two siblings we can choose any two from {1 - 2 - 3 - 4 - 5}, so 5C2 = 10 cases or any of the three sibling pairs: {6 - 7}, {8 - 9} or {10 - 11}, so 3 cases.

P = (10 + 3)/55 = 13/55.

Answer: B.
_________________

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

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50730
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 30 Oct 2017, 22:34
Bunuel wrote:
nkmungila wrote:
In a group of 11 members, 5 have 4 siblings each and 6 have one sibling each. If two members are picked randomly, what is the probability that they are siblings?

A. \(\frac{16}{55}\)

B. \(\frac{13}{55}\)

C. \(\frac{10}{55}\)

D. \(\frac{6}{11}\)

E. \(\frac{17}{110}\)


5 have 4 siblings each implies that there is one group of 5 siblings {1 - 2 - 3 - 4 - 5};
6 have one sibling each implies that there are three groups of 2 siblings {6 - 7}, {8 - 9}, {10 - 11}.

The total number of ways to choose two people out of 11 is 11C2 = 11!/(2!9!) = 55.

In order to get two siblings we can choose any two from {1 - 2 - 3 - 4 - 5}, so 5C2 = 10 cases or any of the three sibling pairs: {6 - 7}, {8 - 9} or {10 - 11}, so 3 cases.

P = (10 + 3)/55 = 13/55.

Answer: B.


Similar question: https://gmatclub.com/forum/in-a-room-fi ... 87550.html
_________________

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

Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 7046
In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 30 Oct 2017, 22:48
nkmungila wrote:
In a group of 11 members, 5 have 4 siblings each and 6 have one sibling each. If two members are picked randomly, what is the probability that they are siblings?

A. \(\frac{16}{55}\)

B. \(\frac{13}{55}\)

C. \(\frac{10}{55}\)

D. \(\frac{6}{11}\)

E. \(\frac{17}{110}\)



Hi....

5 have 4 siblings each means these 5 are siblings
6have 1 sibling each means there are 3 pair of siblings.

So two ways to solve..

1) finding prob of picking siblings..
As also explained above by bunuel...
Picking 2 out of 5 siblings =5C2=10
Picking 2 of the two siblings=2C2=1, three such pairs so 3 ways..
Total 10+3=13..
Ways to pick up 2 out of 11=11C2=11*10/2=55

Prob =13/55

2) prob of not picking siblings
Pick up 1 of these 5siblings and SECOND can be any of remaining 6, so 5*6=30
Pick up 1 of the two siblings and SECOND can be any of 11-2 or 9, so 2*9..
Three such pairs=2*9*3=54

Total 30+54=84..
Way to pick up 2 =11*10
So prob of not picking=84/11*10=84/110=42/55
So prob of picking=1-42/55=13/55

B
_________________

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

Target Test Prep Representative
User avatar
P
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4170
Location: United States (CA)
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 01 Nov 2017, 16:11
nkmungila wrote:
In a group of 11 members, 5 have 4 siblings each and 6 have one sibling each. If two members are picked randomly, what is the probability that they are siblings?

A. \(\frac{16}{55}\)

B. \(\frac{13}{55}\)

C. \(\frac{10}{55}\)

D. \(\frac{6}{11}\)

E. \(\frac{17}{110}\)


Of the 5 people who have 4 siblings each, they must be each others’ siblings. Of the 6 people who have 1 sibling each, there must be 3 pairs of people who are siblings of each another.

When 2 people are picked randomly, the probability that they are 2 people from the 5-people siblings is:

5/11 x 4/10 = 20/110

The probability that they are 2 people from any of the 3 pairs of siblings is:

(2/11 x 1/10) x 3 = 6/110

Thus, the overall probability is 20/110 + 6/110 = 26/110 = 13/55.

Answer: B
_________________

Scott Woodbury-Stewart
Founder and CEO

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

SVP
SVP
avatar
P
Joined: 12 Dec 2016
Posts: 1672
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
GMAT ToolKit User Premium Member
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 07 Jan 2018, 00:23
I am really bad at probability. I still get confused with A. I even get paranoid, and comes up with something else.
This question is extremely important, but I got a wrong answer.
examPAL Representative
User avatar
G
Joined: 07 Dec 2017
Posts: 813
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 07 Jan 2018, 00:38
chesstitans wrote:
I am really bad at probability. I still get confused with A. I even get paranoid, and comes up with something else.
This question is extremely important, but I got a wrong answer.


Can you be a bit more specific about what you find difficult?

In this question, the 'trick' is to realize that there are 4 groups - 1 containing 5 people and 3 containing two people.
Once you've realized that, the solution method is fairly standard - just choose people from your groups and divide by the total possibilities.

So the hard part of the question is selecting the right 'choice model'.
Which model should you use? It's best to learn this via experience - by solving lots of questions in the field.
_________________

Image
Sign up for 7-day free trial
Image

I am a CR Expert - Ask Me ANYTHING about CR
I am a DS Expert - Ask Me ANYTHING about DS

The best DEAL of the year has finally arrived! Save up to $320 on your GMAT course.

Expires on Nov 24th.
Image

SVP
SVP
avatar
P
Joined: 12 Dec 2016
Posts: 1672
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
GMAT ToolKit User Premium Member
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 07 Jan 2018, 20:10
DavidTutorexamPAL wrote:
chesstitans wrote:
I am really bad at probability. I still get confused with A. I even get paranoid, and comes up with something else.
This question is extremely important, but I got a wrong answer.


Can you be a bit more specific about what you find difficult?

In this question, the 'trick' is to realize that there are 4 groups - 1 containing 5 people and 3 containing two people.
Once you've realized that, the solution method is fairly standard - just choose people from your groups and divide by the total possibilities.

So the hard part of the question is selecting the right 'choice model'.
Which model should you use? It's best to learn this via experience - by solving lots of questions in the field.


the problem is not which math model I use, but the logical sense that I try to comprehend from applying the models. I feel frustrated.
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8580
Location: Pune, India
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one  [#permalink]

Show Tags

New post 08 Jan 2018, 03:35
chesstitans wrote:
DavidTutorexamPAL wrote:
chesstitans wrote:
I am really bad at probability. I still get confused with A. I even get paranoid, and comes up with something else.
This question is extremely important, but I got a wrong answer.


Can you be a bit more specific about what you find difficult?

In this question, the 'trick' is to realize that there are 4 groups - 1 containing 5 people and 3 containing two people.
Once you've realized that, the solution method is fairly standard - just choose people from your groups and divide by the total possibilities.

So the hard part of the question is selecting the right 'choice model'.
Which model should you use? It's best to learn this via experience - by solving lots of questions in the field.


the problem is not which math model I use, but the logical sense that I try to comprehend from applying the models. I feel frustrated.


It is important to realise what each statement means:

5 have 4 siblings each - There are a total of 11 people. Say A has 4 siblings so it means we have a group of 5 brother-sister. Each of those 5 brother-sister has exactly 4 siblings. So we have 5 people with 4 siblings each.
6 have one sibling each - Say B has a sibling C. So C also has sibling B. So we have 2 people who have a sibling each. Similarly, we will have 2 more pairs of 2 siblings each. In all we have 3 pairs of 2 siblings each.

So here are sibling groups:
A-P-Q-R-S
B-C
D-E
F-G

Probability of selecting a sibling pair:
We could select one of 5 with a probability of 5/11.
We could then select a sibling for her with a probability of 4/10
Probability = (5/11)*(4/10) = 20/110

OR

We could select one of the other 6 with a probability of 6/11
Her sibling is 1 of the leftover 10 so we can select him with a probability of 1/10
Probability = (6/11)*(1/10) = 6/110

Total Probability = 26/110 = 13/55

Answer (B)
_________________

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!

GMAT Club Bot
Re: In a group of 11 members, 5 have 4 siblings each and 6 have one &nbs [#permalink] 08 Jan 2018, 03:35
Display posts from previous: Sort by

In a group of 11 members, 5 have 4 siblings each and 6 have one

  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®.