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Problem Solving for 780+ Aspirants [#permalink]
21 Mar 2009, 07:25

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

67% (02:19) correct
33% (00:00) wrong based on 5 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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. LetPsbe the probability that two marbles drawn at random from the box are the same color, and letPdbe the probability that they are diﬀerent colors. Find|Ps − Pd|.

Re: Problem Solving for 780+ Aspirants [#permalink]
27 Mar 2009, 00:17

1

This post received KUDOS

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 =\(\frac{1000}{2001}\)

Re: Problem Solving for 780+ Aspirants [#permalink]
18 Apr 2009, 15:32

1

This post received KUDOS

nightwing79 wrote:

GMAT TIGER wrote:

nightwing79 wrote:

so I had exactly the same working as you did except

"x = 5x6x6x6 = 6x180= 1080"

how did you get that? hmmmm

Just get \(y\)'s value in integer. To do that, \(x\) has to be \(5*6*6*6\). Since \(y = 4sqrt(5x^3 / 6)\) has one \(5\), one \(6\) in denominator, and three \(x\), make the value of \((5x^3 / 6)\) a squre of a square. To do that, \(x\) requires \(5^3\) to make \(5^4\) and \(6^9\) to make \(6^8\) after cancealing a 6. Hence 4th sqrt power of \((5x^3 / 6)\) is a minimum integer vale.

Good stuff!

However - why does it have to be 6^9 why not 6^5 to make 6^4 hence square root of would still be an integer value.

The key is root to the power of 4 - have to remember that and then looking to understand what a min integer value would look like - good learning.

look forward to your correspondence.

che dg

How do you get that? If you can, then thats the answer. But again, how?:roll: _________________

Walker...i think we should multiply it by 2 for same color and different color individually because we might get 2 blacks or 2 reds. So prob will be 2(1000/2001) and 2(1001/2001)

Re: Problem Solving for 780+ Aspirants [#permalink]
23 Mar 2009, 02:28

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

Re: Problem Solving for 780+ Aspirants [#permalink]
23 Mar 2009, 02:40

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.... _________________

Re: Problem Solving for 780+ Aspirants [#permalink]
11 Apr 2009, 03:17

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

Re: Problem Solving for 780+ Aspirants [#permalink]
11 Apr 2009, 19:03

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

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...