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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.

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 _________________

Bhushan S. If you like my post....Consider it for Kudos

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

gmatclubot

Re: Problem Solving for 780+ Aspirants
[#permalink]
04 Nov 2013, 09:25

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