|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 26 Apr 2010
Posts: 111
Concentration: Strategy, Entrepreneurship
|
Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
 Updated on: 11 Jul 2013, 01:35
19
This post was
BOOKMARKED
D
E
Question Stats:
47% (01:37) correct 53% (01:45) wrong based on 536 sessions
HideShow timer Statistics
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member? (1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30. (2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
I appreciate the kudos if you find this post helpful! +1
Originally posted by martie11 on 20 Sep 2011, 21:12.
Last edited by Bunuel on 11 Jul 2013, 01:35, edited 5 times in total.
Edited the question.
|
|
|
|
|
|
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1914
|
Re: 700+ DS...Company X is forming a two-person committee... [#permalink]
Show Tags
21 Sep 2011, 00:57
1
This post received
KUDOS
martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
i) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
ii) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.
Using both statements: Case I: M Only:2, B only:15, M&B:10 2*15=30 Satisfies Both M&B=10 Satisfies So, \(Committees=C^{10}_1*C^{15}_1+C^{2}_1*C^{10}_1+C^{10}_1*C^{9}_1=150+10+90=250\) Case II: M Only:30, B only:1, M&B:10 1*30=30 Satisfies Both M&B=10 Satisfies So, \(Committees=C^{10}_1*C^{30}_1+C^{1}_1*C^{10}_1+C^{10}_1*C^{9}_1=300+10+90=400\) Not Sufficient. We could have taken other examples as well: {6,5} as 6*5=30 {3,10} as 3*10=30 Ans: "E"
_________________
~fluke
GMAT Club Premium Membership - big benefits and savings
|
|
|
|
|
|
Manager
Joined: 18 Oct 2010
Posts: 80
|
Re: 700+ DS...Company X is forming a two-person committee... [#permalink]
Show Tags
21 Sep 2011, 01:32
to answer this question, the big question is how many managers are there in the company X and the form: P(m)+P(b)-P(m&b)-None
statement 1 lets you know none is 30 surely insufficient
statement 2 lets you know both P(m&b)= 10 insufficient
both cant give you the answer of the question how many manahers there are in the company X
so E
|
|
|
|
|
|
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1154
|
Re: 700+ DS...Company X is forming a two-person committee... [#permalink]
Show Tags
21 Sep 2011, 23:33
managers =m , board members = b and common member = mb now required combination = (m+b)C2 - (m-mb)C1*(b-mb)C1 a (m-mb)*(b-mb)= 30 hence not sufficient. b mbC2 = 10 meaning - mb*(mb-1)/2 = 10 thus mb = 5 not sufficient. a+b (m-5)*(b-5) = 30 meaning m,b can be 20,7 or 15,8. hence not sufficient. thus E it is.
_________________
Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!
|
|
|
|
|
|
Manager
Status: Target MBA
Joined: 20 Jul 2010
Posts: 170
Location: Singapore
|
Re: 700+ DS...Company X is forming a two-person committee... [#permalink]
Show Tags
29 Sep 2011, 09:43
martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
i) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
ii) The number of committees that can be formed such that each committee member is both a manager and a board member is 10.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
Thank you. E should be the OA since we do not know the number of managers and board members
_________________
Thanks and Regards,
GM.
|
|
|
|
|
|
Manager
Joined: 20 Jun 2012
Posts: 93
Location: United States
Concentration: Finance, Operations
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
17 Oct 2013, 03:31
martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10. I know everyone got it right but there is more to this question .. this is a good question .. here: no. of managers - M Board members - B it can be represented in a venn Diagram ... a+c = M(a - only managers, c - managers as well as board member) b+c = B)b - only board members, c - manafers as well as board members) statement1: aC1.bC2 = 30(selected one manager and one board member) a.b = 60 .. now there are many combinations for a.b such as 1*60, 2*30, 3*20 ...... hence a+b can be 61,32,23 ..... not sufficient statement2: cC2=10 (selected 2 persons which are both manager and board member) c(c-1) = 20 .. only possible answer for c is 5 .. ... If the question would've asked no. of members which are both manager and board member we could've answered it with B.
_________________
Forget Kudos ... be an altruist
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44654
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
21 Mar 2014, 02:40
2
This post received
KUDOS
Expert's post
5
This post was
BOOKMARKED
vishalrastogi wrote: martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10. Bunuel - Pl help, not able to understand the solution ! Thanks. The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}. The number of managers ONLY = x; The number of board members ONLY = y; The number of both managers and board members = z. Total = x+y+z. Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member: \(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\) (1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient. (2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient. (1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient: x=1, y=30, z=5; x=2, y=15, z=5; x=3, y=10, z=5; x=5, y=6, z=5; x=6, y=5, z=5; x=10, y=3, z=5; x=15, y=2, z=5; x=30, y=1, z=5. Answer: E. Hope it's clear.
_________________
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
|
|
|
|
|
|
Senior Manager
Joined: 28 Apr 2014
Posts: 260
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
08 May 2014, 05:52
1
This post was
BOOKMARKED
Bunuel wrote: vishalrastogi wrote: martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10. Bunuel - Pl help, not able to understand the solution ! Thanks. The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}. The number of managers ONLY = x; The number of board members ONLY = y; The number of both managers and board members = z. Total = x+y+z. Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member: \(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\) (1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient. (2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient. (1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient: x=1, y=30, z=5; x=2, y=15, z=5; x=3, y=10, z=5; x=5, y=6, z=5; x=6, y=5, z=5; x=10, y=3, z=5; x=15, y=2, z=5; x=30, y=1, z=5. Answer: E. Hope it's clear. Bunuel a couple of doubts over here. When you say The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.Why is z different ? Won't it be made from x,y ? Also not clear on why the probability aspects came into your solution ? Isn't this a question of combinations ?
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44654
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
08 May 2014, 08:12
himanshujovi wrote: Bunuel wrote: vishalrastogi wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.
The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.
Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.
(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.
Answer: E.
Hope it's clear. Bunuel a couple of doubts over here. When you say The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.Why is z different ? Won't it be made from x,y ? Also not clear on why the probability aspects came into your solution ? Isn't this a question of combinations ? Managers and board members form overlapping sets. Attachment: 
Untitled.png [ 4.91 KiB | Viewed 4324 times ]
The number of managers ONLY = x; The number of board members ONLY = y; The number of both managers and board members = z. Total = x+y+z.
_________________
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
|
|
|
|
|
|
Intern
Joined: 22 Jun 2013
Posts: 38
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
10 May 2014, 01:53
Bunuel wrote: vishalrastogi wrote: martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10. Bunuel - Pl help, not able to understand the solution ! Thanks. The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}. The number of managers ONLY = x; The number of board members ONLY = y; The number of both managers and board members = z. Total = x+y+z. Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member: \(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\) (1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient. (2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient. (1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient: x=1, y=30, z=5; x=2, y=15, z=5; x=3, y=10, z=5; x=5, y=6, z=5; x=6, y=5, z=5; x=10, y=3, z=5; x=15, y=2, z=5; x=30, y=1, z=5. Answer: E. Hope it's clear. Hello Bunuel Question Asks us to find the Number of Committees, You have given you solutn by finding the probability. Although doest makes much difference ! I tried a slight different method. Plz tell me where i am going wrong ? Sorry for wasting your time  The number of managers ONLY = x; The number of board members ONLY = y; The number of both managers and board members = z. Total = x+y+z. Let t= x+y. t shows the total number of members other than the one who are both Managers & Board Members. We have to find out: How many committees can be formed such that at least one committee member is both a manager and a board member. So it can be := \(C^1_z . C^1_t + C^2_z\) where \(C^1_z . C^1_t\) = one member from the members who are both Director & board member, Other from the remaining i.e. t \(C^2_z\) = both memebers from the members who are both Director & board member. (1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 t represents the total of neither so we get : \(C^2_t = 30\) --> \(\frac{(t-1)t}{2}=30\) Not sufficient. (2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient. (Same as yours) 1+2) We had to find : \(C^1_z . C^1_t + C^2_z\) From eq.2 We know Z from eq.1 It seems we can find out t - although its not giving integer values. I dont knw why. This way i was getting C as the answer Please help Thanks a lot Again !! 
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44654
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
10 May 2014, 06:21
niyantg wrote: Bunuel wrote: vishalrastogi wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.
The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.
Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.
(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.
Answer: E.
Hope it's clear. Hello Bunuel Question Asks us to find the Number of Committees, You have given you solutn by finding the probability. Although doest makes much difference ! I tried a slight different method. Plz tell me where i am going wrong ? Sorry for wasting your time The number of managers ONLY = x; The number of board members ONLY = y; The number of both managers and board members = z. Total = x+y+z. Let t= x+y. t shows the total number of members other than the one who are both Managers & Board Members. We have to find out: How many committees can be formed such that at least one committee member is both a manager and a board member. So it can be := \(C^1_z . C^1_t + C^2_z\) where \(C^1_z . C^1_t\) = one member from the members who are both Director & board member, Other from the remaining i.e. t \(C^2_z\) = both memebers from the members who are both Director & board member. (1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 t represents the total of neither so we get : \(C^2_t = 30\) --> \(\frac{(t-1)t}{2}=30\) Not sufficient. (2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient. (Same as yours) 1+2) We had to find : \(C^1_z . C^1_t + C^2_z\) From eq.2 We know Z from eq.1 It seems we can find out t - although its not giving integer values. I dont knw why. This way i was getting C as the answer Please help Thanks a lot Again !! \(C^2_t = 30\) in your solution includes the cases when both members are managers as well as the cases when both members are board members, while we are told that one member must be from the managers and the other member from the board members. So, the first statement should be translated as \(x*y=30\). Does this make sense?
_________________
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
|
|
|
|
|
|
Intern
Joined: 22 Jun 2013
Posts: 38
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
11 May 2014, 22:56
Quote: \(C^2_t = 30\) in your solution includes the cases when both members are managers as well as the cases when both members are board members, while we are told that one member must be from the managers and the other member from the board members. So, the first statement should be translated as \(x*y=30\).
Does this make sense? Ohhhh ! I fotgot to take into consideration the condition given in the Question. Thanksss A lot Bunuel You are GREAT !
|
|
|
|
|
|
Senior Manager
Joined: 28 Apr 2014
Posts: 260
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
12 May 2014, 04:00
Bunuel wrote: vishalrastogi wrote: martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. Howmany committees can be formed such that at least one committee member is both a manager and a board member?
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10. Bunuel - Pl help, not able to understand the solution ! Thanks. The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}. The number of managers ONLY = x; The number of board members ONLY = y; The number of both managers and board members = z. Total = x+y+z. Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member: \(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\) (1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient. (2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient. (1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient: x=1, y=30, z=5; x=2, y=15, z=5; x=3, y=10, z=5; x=5, y=6, z=5; x=6, y=5, z=5; x=10, y=3, z=5; x=15, y=2, z=5; x=30, y=1, z=5. Answer: E. Hope it's clear. Sorry for being silly but are we asked probability or total number of combinations ?
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44654
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
12 May 2014, 04:13
himanshujovi wrote: Bunuel wrote: vishalrastogi wrote:
Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
The pool from which we are selecting the two-person committee consists of managers and/or board members: {total} = {managers only} + {board members only} + {both}.
The number of managers ONLY = x;
The number of board members ONLY = y;
The number of both managers and board members = z.
Total = x+y+z.
Now, we are asked to find the probability that the two-person committee has at least one committee member who is both a manager and a board member (at least one from z), which is 1 minus the probability that the two-person committee has no member who is both a manager and a board member:
\(P = 1 - \frac{x+y}{x+y+z}*\frac{x+y-1}{x+y+z-1}\)
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30 --> \(x*y=30\). Not sufficient.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10 --> \(C^2_z=10\) --> \(\frac{(z-1)z}{2}=10\) --> \(z=5\). Not sufficient.
(1)+(2) \(x*y=30\) and \(z=5\). Still not sufficient:
x=1, y=30, z=5;
x=2, y=15, z=5;
x=3, y=10, z=5;
x=5, y=6, z=5;
x=6, y=5, z=5;
x=10, y=3, z=5;
x=15, y=2, z=5;
x=30, y=1, z=5.
Answer: E.
Hope it's clear.
Sorry for being silly but are we asked probability or total number of combinations ? Does it change the solution in any way?
_________________
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
Joined: 30 Mar 2013
Posts: 113
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
16 Oct 2014, 13:29
diebeatsthegmat wrote: to answer this question, the big question is how many managers are there in the company X and the form: P(m)+P(b)-P(m&b)-None
statement 1 lets you know none is 30 surely insufficient
statement 2 lets you know both P(m&b)= 10 insufficient
both cant give you the answer of the question how many manahers there are in the company X
so E slight typo i think. Neither is added back in sets, isn't it? Technically, even in probability it should be added back. Can someone correct me please?
|
|
|
|
|
|
Intern
Joined: 22 Feb 2017
Posts: 2
Concentration: Strategy, Technology
|
Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
04 Mar 2017, 09:48
stunn3r wrote: martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10. I know everyone got it right but there is more to this question .. this is a good question .. here: no. of managers - M Board members - B it can be represented in a venn Diagram ... a+c = M(a - only managers, c - managers as well as board member) b+c = B)b - only board members, c - manafers as well as board members) statement1: aC1.bC2 = 30(selected one manager and one board member) a.b = 60 .. now there are many combinations for a.b such as 1*60, 2*30, 3*20 ...... hence a+b can be 61,32,23 ..... not sufficient statement2: cC2=10 (selected 2 persons which are both manager and board member) c(c-1) = 20 .. only possible answer for c is 5 .. ... If the question would've asked no. of members which are both manager and board member we could've answered it with B. can you please explain how is cC2=10 translated into c(c-1) = 20 ?
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44654
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
05 Mar 2017, 03:27
HassanGmat17 wrote: stunn3r wrote: martie11 wrote: Company X is forming a two-person committee by selecting one member from the managers of Company X and the other member from the board members of Company X. How many committees can be formed such that at least one committee member is both a manager and a board member?
(1) The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
(2) The number of committees that can be formed such that each committee member is both a manager and a board member is 10. I know everyone got it right but there is more to this question .. this is a good question .. here: no. of managers - M Board members - B it can be represented in a venn Diagram ... a+c = M(a - only managers, c - managers as well as board member) b+c = B)b - only board members, c - manafers as well as board members) statement1: aC1.bC2 = 30(selected one manager and one board member) a.b = 60 .. now there are many combinations for a.b such as 1*60, 2*30, 3*20 ...... hence a+b can be 61,32,23 ..... not sufficient statement2: cC2=10 (selected 2 persons which are both manager and board member) c(c-1) = 20 .. only possible answer for c is 5 .. ... If the question would've asked no. of members which are both manager and board member we could've answered it with B. can you please explain how is cC2=10 translated into c(c-1) = 20 ? \(C^2_c=\frac{c!}{(c-2)!*2!}=\frac{(c-2)!*(c-1)*c}{(c-2)!*2!}=\frac{(c-1)*c}{2!}\) \(\frac{(c-1)*c}{2!}=10\) --> \((c-1)*c=20\). Hope it's clear.
_________________
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
|
|
|
|
|
|
Senior Manager
Joined: 02 Apr 2014
Posts: 471
|
Re: Company X is forming a two-person committee by selecting one [#permalink]
Show Tags
24 Nov 2017, 12:23
Let a - number of persons who are only managers
b - number of persons who are both manager and board member
c - number of persons who are only board member
Question: N(atleast one committee member is both a manager and a board member) = N(both are both manager and board members) + N(one member is both manager and board memeber AND other member is only manager) +
N(one member is both manager and board memeber AND other member is only board member)
= \(bC2 + bC1 * aC1 + bC1 * cC1 = bC2 + b * a + b * c = bC2 + b * (a+c)\) = ?
Statement 1:
The number of committees that can be formed such that neither committee member is both a manager and a board member is 30.
= number of persons only manager * number of persons only board member = 30
= \(aC1 * cC1 = 30\) => \(ac = 30\), we don't know about b -> Insuff
Statement 2:
The number of committees that can be formed such that each committee member is both a manager and a board member is 10.
= bC2 = 10, we can find b = 5, we don't know about a, c InSuff
Statement 1 + 2:
\(bC2 + b * (a+c) = 10 + 5 (a + c)\)
\(ac = 30, => a = 15, c = 2, OR a = 6, c = 5\)
accordingly we get different values of \(10 + 5 (a + c)\)
Still not sufficient
Answer (E)
|
|
|
|
|
|
|
Re: Company X is forming a two-person committee by selecting one
[#permalink]
24 Nov 2017, 12:23
|
|
|
|
|
|
|
|
|
|
|
close
