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M09-11

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New post 16 Sep 2014, 00:39
2
11
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A
B
C
D
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  45% (medium)

Question Stats:

58% (01:18) correct 42% (01:06) wrong based on 132 sessions

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New post 16 Sep 2014, 00:39
1
Official Solution:


(1) The ratio of blue chips to green chips is 3:4. If \(B=3\) and \(G=4\) then \(P=\frac{3}{7}*\frac{2}{6}\) but if \(B=6\) and \(G=8\) then \(P=\frac{6}{14}*\frac{5}{13}\). Two different answers, hence not sufficient.

(2) There are 5 more green chips than blue chips. So, \(G=B+5\). Clearly insufficient.

(1)+(2) \(B:G=3:4\) and \(G=B+5\). Solving gives: \(B=15\) and \(G=20\). Sufficient.


Answer: C
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New post 05 Jan 2015, 08:46
1
I think this question is good and helpful.
6/15 would be 6/14.
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New post 03 Jan 2016, 16:57
Bunuel wrote:
Official Solution:


(1) The ratio of blue chips to green chips is 3:4. If \(B=3\) and \(G=4\) then \(P=\frac{3}{7}*\frac{2}{6}\) but if \(B=6\) and \(G=8\) then \(P=\frac{6}{14}*\frac{5}{13}\). Two different answers, hence not sufficient.

(2) There are 5 more green chips than blue chips. So, \(G=B+5\). Clearly insufficient.

(1)+(2) \(B:G=3:4\) and \(G=B+5\). Solving gives: \(B=15\) and \(G=20\). Sufficient.


Answer: C


Hi Bunuel,

Is there any other way (using algebra) to prove the statement 1 insufficient apart from choosing different numbers ?
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New post 07 Oct 2016, 17:27
SD007 wrote:
Bunuel wrote:
Official Solution:


(1) The ratio of blue chips to green chips is 3:4. If \(B=3\) and \(G=4\) then \(P=\frac{3}{7}*\frac{2}{6}\) but if \(B=6\) and \(G=8\) then \(P=\frac{6}{14}*\frac{5}{13}\). Two different answers, hence not sufficient.

(2) There are 5 more green chips than blue chips. So, \(G=B+5\). Clearly insufficient.

(1)+(2) \(B:G=3:4\) and \(G=B+5\). Solving gives: \(B=15\) and \(G=20\). Sufficient.


Answer: C


Hi Bunuel,

Is there any other way (using algebra) to prove the statement 1 insufficient apart from choosing different numbers ?


how did we reach values of 15 and 20
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Re: M09-11  [#permalink]

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New post 08 Oct 2016, 03:03
jonmarrow wrote:
SD007 wrote:
Bunuel wrote:
Official Solution:


(1) The ratio of blue chips to green chips is 3:4. If \(B=3\) and \(G=4\) then \(P=\frac{3}{7}*\frac{2}{6}\) but if \(B=6\) and \(G=8\) then \(P=\frac{6}{14}*\frac{5}{13}\). Two different answers, hence not sufficient.

(2) There are 5 more green chips than blue chips. So, \(G=B+5\). Clearly insufficient.

(1)+(2) \(B:G=3:4\) and \(G=B+5\). Solving gives: \(B=15\) and \(G=20\). Sufficient.


Answer: C


Hi Bunuel,

Is there any other way (using algebra) to prove the statement 1 insufficient apart from choosing different numbers ?


how did we reach values of 15 and 20


Simply by solving the system of equations: \(B/G=3/4\) and \(G=B+5\). Substitute \(G=B+5\) into \(B/G=3/4\), solve for B and then get G.
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Re: M09-11  [#permalink]

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New post 22 Sep 2017, 23:29
Hi Bunnel, How did you arrive for P=3/7∗2/6. What is the significance of 2/6 in this. Appreciate your response. Thanks.
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New post 23 Sep 2017, 04:16
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New post 10 Oct 2017, 23:04
This is how I approached this.

Given: Bowl contains only Blue and Green Chips. Replacement is not allowed. This is Value question and not yes/no

S1: Blue Chips : Green Chip = 3x : 4x

=> Probability of picking two blue =\(\frac{3x}{7x} * \frac{(3x-1)}{(7x-1)}\). This cannot be solved without knowing the value of x. Insufficient.

S2: Clearly insufficient. Blue and Green could be any number. The probability will be different for different cases.


S1+S2 : We have \(4x - 3x = 5 => x = 5\). Knowing x and using S1, we can now find the probablity. Hence sufficient.

Final answer: C
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New post 01 Jan 2018, 17:59
jonmarrow wrote:
SD007 wrote:
Bunuel wrote:
Official Solution:


(1) The ratio of blue chips to green chips is 3:4. If \(B=3\) and \(G=4\) then \(P=\frac{3}{7}*\frac{2}{6}\) but if \(B=6\) and \(G=8\) then \(P=\frac{6}{14}*\frac{5}{13}\). Two different answers, hence not sufficient.

(2) There are 5 more green chips than blue chips. So, \(G=B+5\). Clearly insufficient.

(1)+(2) \(B:G=3:4\) and \(G=B+5\). Solving gives: \(B=15\) and \(G=20\). Sufficient.


Answer: C


Hi Bunuel,

Is there any other way (using algebra) to prove the statement 1 insufficient apart from choosing different numbers ?


how did we reach values of 15 and 20




Cheat Sheet:
For ratio of 3:4, we have denominator less numerator as 1. Since in the resulting ratio (statement 2 of the question), denominator is 5 more than numerator, simply multiply numerator and denominator by 5 to get 15:20.
If the denominator less numerator was 2, then the multiplying factor would have been 2.5 to achieve denominator more than numerator by 5 and so on.
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New post 07 May 2018, 02:33
I think this is a high-quality question and I agree with explanation.
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New post 20 Jun 2018, 20:12
I think this is a high-quality question and I agree with explanation.

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New post 05 Dec 2018, 03:52
I think this is a high-quality question and I agree with explanation.
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New post 25 Nov 2019, 05:47
hello Bunnel,

Can you please share the link of more similar problems.

Thanks
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New post 25 Nov 2019, 05:54
prabsahi06 wrote:
hello Bunnel,

Can you please share the link of more similar problems.

Thanks


Similar questions:
https://gmatclub.com/forum/a-box-contai ... fl=similar
https://gmatclub.com/forum/a-box-contai ... 44131.html
https://gmatclub.com/forum/initially-ho ... fl=similar
https://gmatclub.com/forum/if-a-barrel- ... fl=similar
https://gmatclub.com/forum/a-barrel-con ... fl=similar
https://gmatclub.com/forum/a-certain-bo ... fl=similar
https://gmatclub.com/forum/a-box-contai ... fl=similar
https://gmatclub.com/forum/a-box-contai ... fl=similar
https://gmatclub.com/forum/a-box-contai ... fl=similar
https://gmatclub.com/forum/a-bag-contai ... fl=similar
https://gmatclub.com/forum/a-box-contai ... fl=similar
https://gmatclub.com/forum/a-box-contai ... fl=similar
https://gmatclub.com/forum/a-box-contai ... fl=similar

Hope it helps.
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Re: M09-11   [#permalink] 25 Nov 2019, 05:54
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