|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[0], given: 0
|
Problem Solving for 780+ Aspirants [#permalink]
 21 Mar 2009, 08:25
Question Stats:
66% (00:00) correct
33% (00:00) wrong based on 0 sessions
Folks, with over 6 yrs of experience as a GMAT Mentor and Content Developer I hope I can make a difference in your GMAT preparation through this thread.As I manage similar threads in other forums you many find these questions on other websites as well of course you will find the same avatar in all those places I request the users to adhere to the following rules to avoid confusion in the thread
1. Do not post your questions here. Please post them in the relevant thread on this forum
2. Every question posted here will have an unique number and please use this in your PMs.
3. Instead of simply giving your answer try to support your answer with some explanation
Let's start...and here is the first question. There are 1001 red marbles and 1001 black marbles in a box. Let Ps be the probability that two marbles drawn at random from the box are the same color, and let Pd be the probability that they are different colors. Find |Ps − Pd|.(A) 0(B) \frac{1}{2002}(C) \frac{1}{2001}(D) \frac{2}{2001}(E) \frac{1}{1000}
_________________
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)
|
|
|
|
|
|
|
|
|
Intern
Joined: 09 Mar 2009
Posts: 3
Location: Netherlands
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
23 Mar 2009, 03:28
This will be my first post here, sorry for my bad spelling (iam not a native english speaker). First to support the chanse to have two red(r) marbles or two black(b) marbles (Ps). P( (b*b) + (r*r) ) first marble P1(b) = 1001/2002 second marble p2(b) = 1000/2001 (1 less black marble and 1 less overall marbels) The same calculation for the red marbles. ((1001/2002)*(1000/2001)) = ((1/2)*(1000/2001)) = 1000/4002 (b*b) so: P( (1000/4002) + (1000/4002) ) = 2000/4002 (This makes Ps) Now for Pd: The following combinations (r*b) and (b*r) P( (r*b) + (b*r) ) P( b*r ) = (1001/2002) * (1001/2001) = ((1/2)*(1001/2002))= 1001/4002 Again the same goes for r*b P( (1001/4002) + (1001/4002) ) = 2002/4002 (this makes Pd) Now to answer the question: |Ps - Pd| (2000/4002) - (2002/4002) = (2/4002) = (1/2001) So i think it would be answer C. Than again, its my first post so comments plz 
|
|
|
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
23 Mar 2009, 03:40
Sjorsl wrote: This will be my first post here, sorry for my bad spelling (iam not a native english speaker). First to support the chanse to have two red(r) marbles or two black(b) marbles (Ps). P( (b*b) + (r*r) ) first marble P1(b) = 1001/2002 second marble p2(b) = 1000/2001 (1 less black marble and 1 less overall marbels) The same calculation for the red marbles. ((1001/2002)*(1000/2001)) = ((1/2)*(1000/2001)) = 1000/4002 (b*b) so: P( (1000/4002) + (1000/4002) ) = 2000/4002 (This makes Ps) Now for Pd: The following combinations (r*b) and (b*r) P( (r*b) + (b*r) ) P( b*r ) = (1001/2002) * (1001/2001) = ((1/2)*(1001/2002))= 1001/4002 Again the same goes for r*b P( (1001/4002) + (1001/4002) ) = 2002/4002 (this makes Pd) Now to answer the question: |Ps - Pd| (2000/4002) - (2002/4002) = (2/4002) = (1/2001) So i think it would be answer C. Than again, its my first post so comments plz Welcome dude...A very good attempt...Why don't you cut down the calculation part? Think of it....
_________________
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)
|
|
|
|
|
|
SVP
Joined: 28 Dec 2005
Posts: 1612
Followers: 1
Kudos [?]:
53
[0], given: 2
|
Re: Problem Solving for 780+ Aspirants [#permalink]
23 Mar 2009, 15:49
im getting 0 as my answer. is that right ? ill explain my logic if it is.
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3591
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 230
Kudos [?]:
1298
[1] , given: 346
|
Re: Problem Solving for 780+ Aspirants [#permalink]
23 Mar 2009, 16:01
1
This post received
KUDOS
Because of symmetry (1001 red = 1001 black) it doesn't matter what marble we take first. For second marble: \frac{1000}{2001} - the same color \frac{1001}{2001} - the different color |Ps - Pd| = |\frac{1000}{2001} - \frac{1001}{2001}| = \frac{1}{2001}
_________________
iPhone/iPod/iPad: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
Android: GMAT ToolKit (NEW!). POLL: What tool do you need next?
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Director
Joined: 25 Oct 2006
Posts: 657
Followers: 7
Kudos [?]:
104
[0], given: 6
|
Re: Problem Solving for 780+ Aspirants [#permalink]
23 Mar 2009, 22:54
Agree with walker.. |2002/2002*1000/2001 - 2002/2002*1001/2001| = 1/2001
_________________
If You're Not Living On The Edge, You're Taking Up Too Much Space
|
|
|
|
|
|
Intern
Joined: 09 Mar 2009
Posts: 3
Location: Netherlands
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
24 Mar 2009, 02:20
walker wrote: Because of symmetry (1001 red = 1001 black) it doesn't matter what marble we take first. For second marble:
\frac{1000}{2001} - the same color
\frac{1001}{2001} - the different color
|Ps - Pd| = |\frac{1000}{2001} - \frac{1001}{2001}| = \frac{1}{2001}  once you see it its so obvious!
|
|
|
|
|
|
Intern
Joined: 25 Mar 2009
Posts: 2
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
25 Mar 2009, 06:36
Sjorsl I concur
Quite simple actually!
|
|
|
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
27 Mar 2009, 01:17
Since the number of black marbles = number of red marbles, we need not consider the first ball in the calculation. So, the probability of getting different balls = probability of picking up the second ball from the lot other than from which the first ball is picked= \frac{1001}{2001} Similarly the probability of getting same colored ball=probability of picking up the second ball from the same lot from which the first ball is picked = \frac{1000}{2001}Hence |{P_s - P_d}|= \frac{1}{2001}
_________________
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)
|
|
|
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
27 Mar 2009, 22:32
|
|
|
|
|
|
Director
Joined: 01 Aug 2008
Posts: 775
Followers: 3
Kudos [?]:
32
[0], given: 99
|
Re: Problem Solving for 780+ Aspirants [#permalink]
28 Mar 2009, 00:07
cicerone wrote: Folks, here comes the next one..This is from Geometry....  IMO A. the distance from small circle center to the origin point of XY plane is = root2 r. 16 = r + r Root 2) r = 16(root 2 -1)
|
|
|
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
01 Apr 2009, 03:49
|
|
|
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[1] , given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
01 Apr 2009, 04:16
1
This post received
KUDOS
Here is the next one... N, the set of natural numbers is portioned into subsets
S1={1}
S2={2,3}
S3={4,5,6}
S4={7,8,9,10} and so on.
The sum of the numbers in subset S50 is(A) 61,250
(B) 65,525
(C) 42,455
(D) 62,525
(E) None of these
_________________
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)
|
|
|
|
|
|
Current Student
Joined: 13 Jan 2009
Posts: 375
Location: India
Followers: 16
Kudos [?]:
42
[1] , given: 1
|
Re: Problem Solving for 780+ Aspirants [#permalink]
01 Apr 2009, 11:31
1
This post received
KUDOS
With the same logic
S50 = 50*49/2 + 1 = 1226
SUM = 1226+1227....(1226+49)
= 1226*50 + (1+2...49)
= 61250 + 49*50/2
D
|
|
|
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
11 Apr 2009, 04:17
Folks, sorry for the delay.(I was on a vacation)... Here is the next one..... There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there?
A. 4
B. 6
C. 8
D. 10
E. Cannot be determined
_________________
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2530
Followers: 41
Kudos [?]:
357
[0], given: 19
|
Re: Problem Solving for 780+ Aspirants [#permalink]
11 Apr 2009, 20:03
cicerone wrote: Folks, sorry for the delay.(I was on a vacation)...
Here is the next one.....
There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there?
A. 4
B. 6
C. 8
D. 10
E. Cannot be determined b + g = c + 1 b/2 + g = c - 3 b = 8
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2530
Followers: 41
Kudos [?]:
357
[0], given: 19
|
Re: Problem Solving for 780+ Aspirants [#permalink]
11 Apr 2009, 20:12
cicerone wrote: There are 1001 red marbles and 1001 black marbles in a box. Let Ps be the probability that two marbles drawn at random from the box are the same color, and let Pd be the probability that they are different colors. Find |Ps − Pd|.
(A) 0
(B) \frac{1}{2002}
(C) \frac{1}{2001}
(D) \frac{2}{2001}
(E) \frac{1}{1000} I found the question ambigious. It would be out of ambiguity if it had made clear about how the balls are drawn. The question assumed that the repetition is not allowed but that was not clearly stipulated in the question.
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Intern
Joined: 11 Apr 2009
Posts: 4
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
12 Apr 2009, 07:16
=> (2 * (1001 C 2 )/ 2002 C 2) - (1001 C 1 * 1001 C 1 ) / 2002 C 2
=> (2 * 1000 /2001) - ((1001 * 1001) /((2002 * 2001)/2))
=> 1000/2001 - 1001/2001
=> 1/2001
Ans - C
|
|
|
|
|
|
Intern
Joined: 11 Apr 2009
Posts: 4
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
12 Apr 2009, 07:36
b+g-1-3 = b/2 + g b- b/2 = 4 b = 8 GMAT TIGER wrote: cicerone wrote: Folks, sorry for the delay.(I was on a vacation)...
Here is the next one.....
There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there?
A. 4
B. 6
C. 8
D. 10
E. Cannot be determined b + g = c + 1 b/2 + g = c - 3 b = 8 |
|
|
|
|
|
Senior Manager
Joined: 28 Aug 2006
Posts: 302
Followers: 7
Kudos [?]:
39
[0], given: 0
|
Re: Problem Solving for 780+ Aspirants [#permalink]
16 Apr 2009, 04:13
sureshbala wrote: Folks, sorry for the delay (I am on a vacation).
Anyway, here is the next one.....
There are some boys and some girls. If each child sits on one chair, one boy will have no chair to sit. If two boys sit on one chair and one girl occupies one chair, then three chairs are unoccupied. How many boys are there?
A. 4
B. 6
C. 8
D. 10
E. Cannot be determined Folks, this can be answered very quickly if you can conclude the following two results. Result 1: Since the number of chairs occupied by the girls in both the cases is same there is no need to consider the number of girls. Result 2: In the second case we are able to seat 2 boys per chair and no boy is left unseated, hence the number of boys must be even. Let the number of boys be 2x.Seats occupied in the first case = 2x-1Seats occupied in the second case = xGiven that 2x-1-x = 3So x = 4Hence the number of boys 2x = 8.
_________________
Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)
|
|
|
|
|
|
|
Re: Problem Solving for 780+ Aspirants
[#permalink]
16 Apr 2009, 04:13
|
|
|
|
|
|
|
|
|
|
|
close
