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Manager  Joined: 30 Jan 2006
Posts: 110
From a bag containing 12 identical blue balls, y identical  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 66% (01:34) correct 34% (01:54) wrong based on 1117 sessions

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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?

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

Originally posted by Matador on 06 Apr 2006, 15:32.
Last edited by Bunuel on 22 Sep 2013, 03:54, edited 1 time in total.
Renamed the topic, edited the question, added the answer choices and the OA.
Math Expert V
Joined: 02 Sep 2009
Posts: 58469

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16
11
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

$$P(blue) = \frac{12}{12+y}<\frac{2}{5}$$ --> $$y>18$$ --> $$y_{min}=19$$.

Hope it's clear.
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Manager  Joined: 09 Feb 2006
Posts: 89
Location: New York, NY

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12/x = 2/5

x = 30 total number of balls.

There must be at least 18 yellow balls in the bag.
Intern  Joined: 02 Jan 2006
Posts: 3
Location: Bay Area

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1
Agree with NTLancer.

Since the prob of blue balls is LESS than 2/5, you need at least 19 yellow balls in the bag.
Manager  Joined: 27 Mar 2006
Posts: 94

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Agree

The number of yellow balls is atleast 19.

Since yellow balls must be greater than 18...hence least number is 19
Manager  Joined: 21 Dec 2005
Posts: 76

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Agree with 19 too.

Since 2/5=12/x, x = 30 so if probability is less than 2/5 that means x is >30, say 31 then 31-12=19 yellow balls.
Intern  Joined: 12 Feb 2006
Posts: 24
Re: PS - probability  [#permalink]

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(12 blue) / (12 blue + x yellow) = 2/5

12/30 = 2/5

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.
Intern  Joined: 21 Sep 2013
Posts: 1

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It is 18
P(blue) = 12 / (12+Y)
P(blue) < 2/5
solving this: 12 /12+Y = 2/5
y = 18
SVP  Joined: 06 Sep 2013
Posts: 1566
Concentration: Finance
Re: From a bag containing 12 identical blue balls, y identical  [#permalink]

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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?

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

Therefore, yellow must be at least 19

Cheers!
Kudos Rain
J Intern  Joined: 03 Jul 2014
Posts: 13
Re: From a bag containing 12 identical blue balls, y identical  [#permalink]

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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 ....

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

$$P(blue) = \frac{12}{12+y}<\frac{2}{5}$$ --> $$y>18$$ --> $$y_{min}=19$$.

Hope it's clear.
Math Expert V
Joined: 02 Sep 2009
Posts: 58469
From a bag containing 12 identical blue balls, y identical  [#permalink]

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2
2
hanschris5 wrote:
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 ....

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

$$P(blue) = \frac{12}{12+y}<\frac{2}{5}$$ --> $$y>18$$ --> $$y_{min}=19$$.

Hope it's clear.

It goes like this:
$$\frac{12}{12+y}<\frac{2}{5}$$

Cross multiply: $$60< 24+2y$$;

Subtract 24 from both sides: $$36<2y$$;

Divide by 2: $$18<y$$, which is the same as $$y>18$$ (y is greater than 18).

Hope it's clear.
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Manager  Joined: 25 Dec 2012
Posts: 116
From a bag containing 12 identical blue balls, y identical  [#permalink]

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1
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
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Re: From a bag containing 12 identical blue balls, y identical  [#permalink]

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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?

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

We can create the following inequality:

12/(y+12) < 2/5

60 < 2(y + 12)

30 < y + 12

18 < y

Since y is greater than 18, the least number of yellow balls is 19.

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

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_________________ Re: from a bag containing 12 identical blue balls, y identical   [#permalink] 08 Oct 2019, 02:22
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