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# Problem Solving for 780+ Aspirants

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Problem Solving for 780+ Aspirants [#permalink]

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21 Mar 2009, 07:25
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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

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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|.

$$(A) 0$$

$$(B) \frac{1}{2002}$$

$$(C) \frac{1}{2001}$$

$$(D) \frac{2}{2001}$$

$$(E) \frac{1}{1000}$$
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Re: Problem Solving for 780+ Aspirants [#permalink]

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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)

|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
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Re: Problem Solving for 780+ Aspirants [#permalink]

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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)

|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....
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Re: Problem Solving for 780+ Aspirants [#permalink]

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23 Mar 2009, 14:49
im getting 0 as my answer. is that right ? ill explain my logic if it is.
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Re: Problem Solving for 780+ Aspirants [#permalink]

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23 Mar 2009, 15:01
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Expert's post
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

$$|Ps - Pd| = |\frac{1000}{2001} - \frac{1001}{2001}| = \frac{1}{2001}$$
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Re: Problem Solving for 780+ Aspirants [#permalink]

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23 Mar 2009, 21:54
Agree with walker..

|2002/2002*1000/2001 - 2002/2002*1001/2001| = 1/2001
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Re: Problem Solving for 780+ Aspirants [#permalink]

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24 Mar 2009, 01:20
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!
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Re: Problem Solving for 780+ Aspirants [#permalink]

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25 Mar 2009, 05:36
Sjorsl I concur

Quite simple actually!
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Re: Problem Solving for 780+ Aspirants [#permalink]

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27 Mar 2009, 00:17
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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 $$|{P_s - P_d}|= \frac{1}{2001}$$
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Re: Problem Solving for 780+ Aspirants [#permalink]

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27 Mar 2009, 21:32
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Folks, here comes the next one..This is from Geometry....

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Re: Problem Solving for 780+ Aspirants [#permalink]

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27 Mar 2009, 23:07
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)
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Re: Problem Solving for 780+ Aspirants [#permalink]

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01 Apr 2009, 02:49
Here is the solution....

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Re: Problem Solving for 780+ Aspirants [#permalink]

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01 Apr 2009, 03:16
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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

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Re: Problem Solving for 780+ Aspirants [#permalink]

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01 Apr 2009, 10:31
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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
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Re: Problem Solving for 780+ Aspirants [#permalink]

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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

B. 6

C. 8

D. 10

E. Cannot be determined

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Re: Problem Solving for 780+ Aspirants [#permalink]

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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

B. 6

C. 8

D. 10

E. Cannot be determined

b + g = c + 1
b/2 + g = c - 3

b = 8
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Re: Problem Solving for 780+ Aspirants [#permalink]

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11 Apr 2009, 19:12
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.
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Re: Problem Solving for 780+ Aspirants [#permalink]

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12 Apr 2009, 06:16
=> (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
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Re: Problem Solving for 780+ Aspirants [#permalink]

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12 Apr 2009, 06:36
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
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Re: Problem Solving for 780+ Aspirants [#permalink]

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16 Apr 2009, 03:13
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-1

Seats occupied in the second case = x

Given that 2x-1-x = 3

So x = 4

Hence the number of boys 2x = 8.
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Re: Problem Solving for 780+ Aspirants   [#permalink] 16 Apr 2009, 03:13

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