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# From a bag containing 12 identical blue balls, y identical

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Manager
Joined: 30 Jan 2006
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From a bag containing 12 identical blue balls, y identical [#permalink]  06 Apr 2006, 14:32
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61% (01:53) correct 39% (01:12) wrong based on 235 sessions
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
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Sep 2013, 02:54, edited 1 time in total.
Renamed the topic, edited the question, added the answer choices and the OA.
Manager
Joined: 09 Feb 2006
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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: 4
Location: Bay Area
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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: 136
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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
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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
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Re: PS - probability [#permalink]  09 Apr 2006, 11:38
(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
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Re: [#permalink]  21 Sep 2013, 22:46
It is 18
P(blue) = 12 / (12+Y)
P(blue) < 2/5
solving this: 12 /12+Y = 2/5
y = 18
Math Expert
Joined: 02 Sep 2009
Posts: 27168
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Kudos [?]: 40906 [3] , given: 5576

Re: Re: [#permalink]  22 Sep 2013, 02:57
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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.
_________________
SVP
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 710 Q48 V39
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Kudos [?]: 293 [0], given: 354

Re: From a bag containing 12 identical blue balls, y identical [#permalink]  25 Nov 2013, 12:52
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
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Kudos [?]: 0 [0], given: 19

Re: From a bag containing 12 identical blue balls, y identical [#permalink]  22 Nov 2014, 02:39
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 ....

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
Joined: 02 Sep 2009
Posts: 27168
Followers: 4224

Kudos [?]: 40906 [1] , given: 5576

From a bag containing 12 identical blue balls, y identical [#permalink]  22 Nov 2014, 05:50
1
KUDOS
Expert's post
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.
_________________
From a bag containing 12 identical blue balls, y identical   [#permalink] 22 Nov 2014, 05:50
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