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If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 05:34
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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 . sourceself made
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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 05:55
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. So C Sent from my Moto G (5) Plus using GMAT Club Forum mobile app



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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 06:51
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. So C Sent from my Moto G (5) Plus using GMAT Club Forum mobile appE Why can’t r = b = w = 3? Sent from my iPhone using GMAT Club Forum mobile app



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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 06:53
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.
Answer = C



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If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 06:54
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. So C Sent from my Moto G (5) Plus using GMAT Club Forum mobile appE Why can’t r = b = w = 3? Sent from my iPhone using GMAT Club Forum mobile appQuestion has mentioned that no. of red, blue and white balls are different. hence, we can't consider it as equal.



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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 06:56
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. So C Sent from my Moto G (5) Plus using GMAT Club Forum mobile appE Why can’t r = b = w = 3? Sent from my iPhone using GMAT Club Forum mobile appQuestion has mentioned that no. of red, blue and white balls are different. hence, we can't consider it as equal. Missed that. Thanks. Sent from my iPhone using GMAT Club Forum mobile app



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If there are different numbers of red, blue and white balls, is the nu
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Updated on: 28 Oct 2017, 08:14
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 .
sourceself 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. InsufficientStatement 2: nothing mentioned about \(r\). InsufficientCombining 1 & 2, given \(b\) is prime so for \(r=\frac{b^2}{w}\) to be an integer \(w=1\), hence \(r\) is not prime. SufficientOption C
Originally posted by niks18 on 28 Oct 2017, 07:30.
Last edited by niks18 on 28 Oct 2017, 08:14, edited 1 time in total.



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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 07:36
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.
Answer = C Hi kunalsinghNS, Can you explain your reason for rejecting statement 1?



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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 07:41
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.
Answer = C Hi kunalsinghNS, Can you explain your reason for rejecting statement 1? Hi... r:B:w is 20:10:5.....ans NO ratio is 2:1 r:b:w is 5:10:20......ans YES ratio is 1:2
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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 07:50
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.
Answer = C Hi kunalsinghNS, Can you explain your reason for rejecting statement 1? Hi... r:B:w is 20:10:5.....ans NO ratio is 2:1 r:b:w is 5:10:20......ans YES ratio is 1:2 I completely missed that can you explain what is missing in my assumption?



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If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 08:03
niks18 wrote: 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 .
sourceself 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 \(r\) to be prime \(\frac{b^2}{w}\) has to be prime. let \(\frac{b^2}{w}=p\), where \(p\) is any prime no so \(b^2=p*w => b=\sqrt{p*w}\) so for \(b\) to be an integer \(p=w\) which in turn will mean that \(b=w=p\) which is not possible. Hence \(r\) is not prime. SufficientStatement 2: nothing mentioned about \(r\). InsufficientOption Ahi, you have gone wrong in the coloured portion.. \(b^2=P_1*w => b=\sqrt{P_1*w}\) here w can be easily \(P_2^2*P_1\) say p*w, p is any prime number say 2, w could be \(3^2*2\) so \(p*w= 2*3^2*2=36\)
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Re: If there are different numbers of red, blue and white balls, is the nu
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28 Oct 2017, 08:06
chetan2u wrote: niks18 wrote: 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 .
sourceself 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 \(r\) to be prime \(\frac{b^2}{w}\) has to be prime. let \(\frac{b^2}{w}=p\), where \(p\) is any prime no so \(b^2=p*w => b=\sqrt{p*w}\) so for \(b\) to be an integer \(p=w\) which in turn will mean that \(b=w=p\) which is not possible. Hence \(r\) is not prime. SufficientStatement 2: nothing mentioned about \(r\). InsufficientOption Ahi, you have gone wrong in the coloured portion.. \(b^2=P_1*w => b=\sqrt{P_1*w}\) here w can be easily \(P_2^2*P_1\) say p*w, p is any prime number say 2, w could be \(3^2*2\) so \(p*w= 2*3^2*2=36\) Yes agreed completely missed the point



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If there are different numbers of red, blue and white balls, is the nu
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21 Dec 2017, 12:49
Statement 1: \(R/B = B/W\) \(B^2 = RW\) => tempting to say it as sufficient to say R is not prime number => But not so.., R can be 2, W = can be odd power of 2, say W = 8 => B = 4 => R is prime => insufficient Statement 2: clearly insufficient (1) + (2) => \(B^2 = R * W\) => \(Prime^2 = R * W\) since R cannot be same as W, either \(R = prime^2\) and \(W = 1\) or R = 1, \(W = prime^2\) => either ways, R is not prime number, => (C) Excellent question chetan2u




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