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Re: If there are different numbers of red, blue and white balls, is the nu [#permalink]

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28 Oct 2017, 04:55

Is it C? St 1: b^2 =rw r can be prime or composite ex: b= 10, r can be 20 or 5 St 2: R can be anything again.

1+2 B is prime , b^2= rw It means either b=r=w or r= 1, w= b^2 or r= b^2, w= 1 Given the numbers are different, so r= 1 or b^2 In either case it is not prime.

Re: If there are different numbers of red, blue and white balls, is the nu [#permalink]

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28 Oct 2017, 05:51

saicharan1191 wrote:

Is it C? St 1: b^2 =rw r can be prime or composite ex: b= 10, r can be 20 or 5 St 2: R can be anything again.

1+2 B is prime , b^2= rw It means either b=r=w or r= 1, w= b^2 or r= b^2, w= 1 Given the numbers are different, so r= 1 or b^2 In either case it is not prime.

Re: If there are different numbers of red, blue and white balls, is the nu [#permalink]

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28 Oct 2017, 05:53

Statement 1 & 2 are alone insufficient

taking both option together we get :

r = b^2/w and we know that r is an integer. hence, since b is not equal to w and b^2/w is an integer. we, can conclude that r = 1, which is not a prime number.

If there are different numbers of red, blue and white balls, is the nu [#permalink]

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28 Oct 2017, 05:54

sevenplusplus wrote:

saicharan1191 wrote:

Is it C? St 1: b^2 =rw r can be prime or composite ex: b= 10, r can be 20 or 5 St 2: R can be anything again.

1+2 B is prime , b^2= rw It means either b=r=w or r= 1, w= b^2 or r= b^2, w= 1 Given the numbers are different, so r= 1 or b^2 In either case it is not prime.

Re: If there are different numbers of red, blue and white balls, is the nu [#permalink]

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28 Oct 2017, 05:56

kunalsinghNS wrote:

sevenplusplus wrote:

saicharan1191 wrote:

Is it C? St 1: b^2 =rw r can be prime or composite ex: b= 10, r can be 20 or 5 St 2: R can be anything again.

1+2 B is prime , b^2= rw It means either b=r=w or r= 1, w= b^2 or r= b^2, w= 1 Given the numbers are different, so r= 1 or b^2 In either case it is not prime.

If there are different numbers of red, blue and white balls, is the nu [#permalink]

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28 Oct 2017, 06:30

chetan2u wrote:

If there are different numbers of red, blue and white balls, is the number of red ball equal to a prime number?

(1) The ratio of red to blue ball is same as ratio of blue to white. (2) The number of blue ball is equal to a prime number .

source-self made

Statement 1: given \(\frac{r}{b}=\frac{b}{w} => r=\frac{b^2}{w}\)

Case 1: if, \(w=1\), then \(r=b^2\) i.e a perfect square hence cannot be prime

Case 2: if, \(w≠1\), then for \(r\) is prime if \(\frac{b^2}{w}\) is prime and if \(\frac{b^2}{w}\) is a composite no, then \(r\) is not prime. Insufficient

Statement 2: nothing mentioned about \(r\). Insufficient

Combining 1 & 2, given \(b\) is prime so for \(r=\frac{b^2}{w}\) to be an integer \(w=1\), hence \(r\) is not prime. Sufficient

Option C

Last edited by niks18 on 28 Oct 2017, 07:14, edited 1 time in total.

Re: If there are different numbers of red, blue and white balls, is the nu [#permalink]

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28 Oct 2017, 06:36

kunalsinghNS wrote:

Statement 1 & 2 are alone insufficient

taking both option together we get :

r = b^2/w and we know that r is an integer. hence, since b is not equal to w and b^2/w is an integer. we, can conclude that r = 1, which is not a prime number.

r = b^2/w and we know that r is an integer. hence, since b is not equal to w and b^2/w is an integer. we, can conclude that r = 1, which is not a prime number.

Re: If there are different numbers of red, blue and white balls, is the nu [#permalink]

Show Tags

28 Oct 2017, 06:50

chetan2u wrote:

niks18 wrote:

kunalsinghNS wrote:

Statement 1 & 2 are alone insufficient

taking both option together we get :

r = b^2/w and we know that r is an integer. hence, since b is not equal to w and b^2/w is an integer. we, can conclude that r = 1, which is not a prime number.