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# Problem Solving for 780+ Aspirants

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VP
Joined: 08 Apr 2009
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Re: Problem Solving for 780+ Aspirants [#permalink]

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11 May 2009, 00:13
GMAT TIGER wrote:
No of factors = (3+1)(1+1)(1+1)(1+1) = 32

Hi GMAT TIGER, I don't quite understand your logic behind this.

Here's what I did to find the total number of factors of 840

840 can be divided into 6 multiplications (2X2X2X3X5X7)

Total factors would be the number 1 and (6c1)+(6c2)+...(6c6)=64

I agree with the number of factors divisible by 15, which is 16

16/64 = 1/4

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

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11 May 2009, 20:25
walker wrote:
GMAT TIGER wrote:
But walker how did you get 64 and 16? Why you multiplied by 2 in both cases?

I remember that GMAT always says "number of positive factors" to avoid ambiguity. So, I used "2" to count both negative and positive set of factors. Anyway, it would be better to use word "positive" in the question.

Thanks. Thats real lol......
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Re: Problem Solving for 780+ Aspirants [#permalink]

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11 May 2009, 22:35
840 = 2^3 * 3 ^1 * 5^1 * 7 ^ 1

The No of factors = 4 *2*2 *2 = 32

No of factor which are divible by 15 in 840 = 15 * ( 2 ^3 * 3 ^1 ) = 4* 2 = 8

Probability = 8/32 = 1/4
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Re: Problem Solving for 780+ Aspirants [#permalink]

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12 May 2009, 17:26
asimov wrote:
GMAT TIGER wrote:
No of factors = (3+1)(1+1)(1+1)(1+1) = 32

Hi GMAT TIGER, I don't quite understand your logic behind this.

Here's what I did to find the total number of factors of 840

840 can be divided into 6 multiplications (2X2X2X3X5X7)

Total factors would be the number 1 and (6c1)+(6c2)+...(6c6)=64

I agree with the number of factors divisible by 15, which is 16

16/64 = 1/4

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

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07 Oct 2009, 07:15
840=2^3*3*5*7

no of factors=(3+1)(1+1)(1+1)(1+1)=32

No of factor which are divisible by 15= 3*5,3*5*2,3*5*2*2,3*5*2*2*2,
3*5*7,3*5*2*7,3*5*2*2*7,3*5*2*2*2*7=8 factors

Probability = 8/32 = 1/4

how come others getting 16 factors.can someone explain
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Re: Problem Solving for 780+ Aspirants [#permalink]

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07 Oct 2009, 13:04
cicerone wrote:
Folks, here is the next question..

From the set of positive factors of 840 one factor is chosen at random. What is the probability that it is divisible by 15?

A. 1/4

B. 7/32

C. 1/2

D. 1/56

E. None of these

my approach :
total no of factors =(2^3)*3*5*7=refer powers (3+1)x(1+1)x(1+1)x(1+1)=32 factors
we want to select factors divisible by 15 ie 3 x 5 should be part of all divisors..we can ensure that by neglecting 3^0 & 5^0 in each factor i.e.
= (2^0+2^1+2^2+2^3) * (3^1) * (5^1) * (7^0+7^1) = thus no of factors factors 4x1x1x2 = 8 factors [15 , 15x7, 15x2, 15x4, 15x8, 15x7x2, 15x7x4, 15x7x8]

prob = 8/32 = 1/4 OA A
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Re: Problem Solving for 780+ Aspirants [#permalink]

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05 May 2011, 02:43
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Re: Problem Solving for 780+ Aspirants [#permalink]

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05 May 2011, 08:09
Psame = 1001x1000/(2002x2001) + 1001x1000/(2002x2001) = 2002x1000/(2002x2001) = 1000/2001
Pdifferent = 1001x1001/(2002x2001) = 2002x1001/(2002x2001)=1001/2001

|Ps - Pd| = |1000/2001 - 1001/2001 | = 1/2001

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

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04 Nov 2013, 09:25
1
KUDOS
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

$$|Ps - Pd| = |\frac{1000}{2001} - \frac{1001}{2001}| = \frac{1}{2001}$$

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)

And the final answer would be 2/2001
Re: Problem Solving for 780+ Aspirants   [#permalink] 04 Nov 2013, 09:25

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