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

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

by imane81 » Wed Feb 17, 2010 12:25 pm
Thank you for advising on this one...

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
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by ajith » Wed Feb 17, 2010 12:38 pm
imane81 wrote:Thank you for advising on this one...

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
12 is less than 2/5 of the total
30 is less than the total content

so the least no of yellow balls is 31-12 = 19
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by cat er ina » Wed Feb 17, 2010 2:21 pm
HEY AJITH IS GREAT BY I UNDERSTOOD FROM THIS LINK, GIVES A VERY GOOD EXPLANATION

www.manhattangmat.com/forums/from-a-bag ... t8493.html
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by harsh.champ » Thu Feb 18, 2010 2:10 pm
imane81 wrote:Thank you for advising on this one...

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
Formal Approach:-
P(Removed ball is blue) = 12/12+y which is less than 2/5
Thus,we get that 60 < 24 + 2y
i.e. y>18.
So we can rule out option A and B.
Also,we have to find the least no. hence [spoiler]y=19 C should be the answer.[/spoiler]

I guess the detailed explanation makes it clear.
For shortcut approach,you can adopt ajith's method.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by bonetlobo » Wed May 13, 2015 6:13 am
Ajith's explanation is brilliant. I did it the "conventional" way:(.

Number of blue balls = 12
Number of yellow balls = y
=> Total number of balls in the bag = 12+y

=> probability that one ball randomly removed is blue ball = 12/(12+y)

However, it is given that:

12/(12+y) < 2/5

Since both sides are positive, we can do cross multiplication

Solving, we get y > 18

So, minimum value of y = 19
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by Brent@GMATPrepNow » Wed May 13, 2015 6:56 am
imane81 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
If you're not sure where to begin here, TESTING the answer choices will reveal the correct answer in no time.

A) 17
If there are 17 yellow balls, then the TOTAL number of balls = 12 + 17 = 29
So, P(selected ball is blue) = 12/29
12/29 is GREATER THAN 2/5, so we can eliminate A

B) 18
If there are 18 yellow balls, then the TOTAL number of balls = 12 + 18 = 30
So, P(selected ball is blue) = 12/30
12/30 is EQUAL TO 2/5, so we can eliminate B

At this point, we can see that adding 1 more yellow ball (i.e., 19 yellow balls) will make P(selected ball is blue) LESS THAN 2/5
So, the correct answer is C

Cheers,
Brent
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by GMATGuruNY » Wed May 13, 2015 6:57 am
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 leat no. of yellow balls that must be in a bag?

A)17
B)18
C)19
D)20
E)21
Let T = the total number of marbles.
Since the probability of selecting a blue marble is less than 2/5, the 12 blue marbles in the bag must constitute less than 2/5 of the total:
12 < (2/5)T
30 < T.

Since the total number of marbles must be at least 31, the least number of yellow marbles that must be added to the 12 blue marbles is 19.

The correct answer is C.
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