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M09-11

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A bowl contains green and blue chips only. If two chips are drawn from the bowl (without replacement) what is the probability that both chips will be blue?


(1) The ratio of blue chips to green chips is 3:4.

(2) There are 5 more green chips than blue chips.
[Reveal] Spoiler: OA

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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
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I think this question is good and helpful.
6/15 would be 6/14.
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New post 26 Jul 2015, 04:13
I don't agree with the explanation. For statement 1) when B=6 and G=8, the probability calculated is as show -
P = (6/15) * (5*13).
Shouldn't the 15 in the denominator be 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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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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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
viditm4 wrote:
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.


If B=3 and G=4 then probability that both chips will be blue is P = (blue)/(total) * (blue - 1)/(total - 1)= 3/7∗2/6
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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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M09-11   [#permalink] 10 Oct 2017, 23:04
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