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From a bag containing 12 identical blue balls, y identical [#permalink]

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06 Apr 2006, 15:32

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From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?

The denominator must be 30. You have 12 already which means you need 18 yellow balls. This makes the equation equal to each other. However, you want less than 2/5 so add one ball to make it 19.

It is 18 P(blue) = 12 / (12+Y) P(blue) < 2/5 solving this: 12 /12+Y = 2/5 y = 18

The answer is 19, not 18. Notice that we have an inequality sign there.

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag? A. 17 B. 18 C. 19 D. 20 E. 21

Re: From a bag containing 12 identical blue balls, y identical [#permalink]

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25 Nov 2013, 13:52

Matador wrote:

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?

A. 17 B. 18 C. 19 D. 20 E. 21

It goes like this.

12 Blue y yellow

12 + y total

12/x = 2/5

So x = 30, x being the total number of ballos Then yellow> 30-12 = 18

Re: From a bag containing 12 identical blue balls, y identical [#permalink]

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22 Nov 2014, 03:39

Can u pls help in explaining the shift in the inequilaties symbol .

I mean 12/12+y < 2/5 so cross multiplying

12*5 < (12+y)2 , am i missing out on reversing the sign .... thanks in advance

Bunuel wrote:

alexgomez2013 wrote:

It is 18 P(blue) = 12 / (12+Y) P(blue) < 2/5 solving this: 12 /12+Y = 2/5 y = 18

The answer is 19, not 18. Notice that we have an inequality sign there.

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag? A. 17 B. 18 C. 19 D. 20 E. 21

Can u pls help in explaining the shift in the inequilaties symbol .

I mean 12/12+y < 2/5 so cross multiplying

12*5 < (12+y)2 , am i missing out on reversing the sign .... thanks in advance

Bunuel wrote:

alexgomez2013 wrote:

It is 18 P(blue) = 12 / (12+Y) P(blue) < 2/5 solving this: 12 /12+Y = 2/5 y = 18

The answer is 19, not 18. Notice that we have an inequality sign there.

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag? A. 17 B. 18 C. 19 D. 20 E. 21

Re: From a bag containing 12 identical blue balls, y identical [#permalink]

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26 Dec 2015, 19:07

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From a bag containing 12 identical blue balls, y identical [#permalink]

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10 Jan 2016, 03:53

Other way of doing is substitute the value.

The question says it probability of blue should be less than 2/5 means it should be less than 40%

Also For a constant numerator and if the denominator is increasing the value of the output will be decreasing.

In our scenario, the numerator is constant which is 12

Based on the given the choice the denominator will be 12/29, 12/30, 12/31, 12/32, 12/33

Since numerator is same and denominator doesn't vary we can conclude as 12/29 > 12/30 > 12/31 > 12/32 > 12/33

Lets choose the number which is easy to solve. Hence choosing 12/30 first which is equal to 2/5

Straight away I can go for "C" from the above answer since 12/29 must be greater than 2/5 and 12/31 should hold the least value of yellow balls which will give the probability as less than 2/5

gmatclubot

From a bag containing 12 identical blue balls, y identical
[#permalink]
10 Jan 2016, 03:53

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