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# M09-11

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:39
2
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Difficulty:

45% (medium)

Question Stats:

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

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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.

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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.

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05 Jan 2015, 08:46
1
I think this question is good and helpful.
6/15 would be 6/14.
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Hasan Mahmud
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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.

Hi Bunuel,

Is there any other way (using algebra) to prove the statement 1 insufficient apart from choosing different numbers ?
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Joined: 22 Jun 2016
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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.

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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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.

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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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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Vidit
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23 Sep 2017, 04:16
1
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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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.

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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.

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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07 May 2018, 02:33
I think this is a high-quality question and I agree with explanation.
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Joined: 20 Jun 2018
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20 Jun 2018, 20:12
I think this is a high-quality question and I agree with explanation.

Posted from my mobile device
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05 Dec 2018, 03:52
I think this is a high-quality question and I agree with explanation.
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25 Nov 2019, 05:47
hello Bunnel,

Thanks
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25 Nov 2019, 05:54
Re: M09-11   [#permalink] 25 Nov 2019, 05:54
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# M09-11

Moderators: chetan2u, Bunuel