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# There are 12 red balls, 10 blue balls, 15 red balls and 9

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Joined: 28 Mar 2012
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There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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Updated on: 13 Jun 2012, 03:15
1
7
00:00

Difficulty:

55% (hard)

Question Stats:

63% (01:03) correct 38% (01:27) wrong based on 184 sessions

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There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 23
C. 43
D. 40
E. 16

Originally posted by cyberjadugar on 12 Jun 2012, 23:08.
Last edited by cyberjadugar on 13 Jun 2012, 03:15, edited 3 times in total.
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Status: ISB 14...:)
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Re: There are 12 red balls, 10 blue balls, 15 red balls..  [#permalink]

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12 Jun 2012, 23:26
1
Total balls = 46

If you leave 3 balls in the bag, even if they are of different colors, you can be sure that you have all balls of the fourth color.

So, leaving 3 balls, you need to pick 43 balls.

This is the minimum number I suppose.

Ans. C
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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13 Jun 2012, 03:08
4
3
There are 12 red balls, 10 blue balls, 15 red balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 23
C. 43
D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color.

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Current Student
Joined: 28 Mar 2012
Posts: 312
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Re: There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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13 Jun 2012, 03:15
2
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 red balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 23
C. 43
D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color.

Edited and updated the question.
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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13 Jun 2012, 03:18
1
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 red balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 23
C. 43
D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color.

Edited and updated the question.

Thank you! Now it's better. +1.
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Joined: 28 Feb 2012
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WE: Marketing (Other)
Re: There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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14 Jun 2012, 23:15
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 red balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 23
C. 43
D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color.

Sorry, maybe i am missing something, but i cannot understand how we come up to this. So the question is how many balls we need to pick min to make sure that we have on hand at least one colour of each ball! So if we pick all the reds, blues and greens then we have 37 total, but this is not enough since we don't have black, that means if we pick 38, no matter in what way, we will have on hand at least one colour of each ball. So the min number of balls is 38, Could you please tell me where i went wrong in my logic? (most probably i misunderstand the question)
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Current Student
Joined: 28 Mar 2012
Posts: 312
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Re: There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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15 Jun 2012, 00:31
1
ziko wrote:
Sorry, maybe i am missing something, but i cannot understand how we come up to this. So the question is how many balls we need to pick min to make sure that we have on hand at least one colour of each ball! So if we pick all the reds, blues and greens then we have 37 total, but this is not enough since we don't have black, that means if we pick 38, no matter in what way, we will have on hand at least one colour of each ball. So the min number of balls is 38, Could you please tell me where i went wrong in my logic? (most probably i misunderstand the question)

Hi Ziko,

Only flaw in the reasoning is that you are choosing the balls you are picking. If you are lucky, then in 9 chances you may get 9 black balls, but it will not happen every time.

Thus to be sure, considering the worst case scenario, after picking 11 red balls, 9 blue balls, 14 green balls and 8 black balls (total = 42 balls) the next ball you pick would be of any one of these colors. You will definitely have balls of all colors in your $$43^{rd}$$draw.

Regards,
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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15 Jun 2012, 01:43
ziko wrote:
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 red balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 23
C. 43
D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color.

Sorry, maybe i am missing something, but i cannot understand how we come up to this. So the question is how many balls we need to pick min to make sure that we have on hand at least one colour of each ball! So if we pick all the reds, blues and greens then we have 37 total, but this is not enough since we don't have black, that means if we pick 38, no matter in what way, we will have on hand at least one colour of each ball. So the min number of balls is 38, Could you please tell me where i went wrong in my logic? (most probably i misunderstand the question)

The point here is that we don't want to have at least one ball of each color, we need ALL balls of some color, so all red balls, or all blue balls, and so on.

Anyway not a good question because of the wording, so I wouldn't worry about it at all.

Check similar questions to practice:
DS: search.php?search_id=tag&tag_id=42
PS: search.php?search_id=tag&tag_id=63

Hope it helps.
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Posts: 9162
Re: There are 12 red balls, 10 blue balls, 15 red balls and 9  [#permalink]

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12 Aug 2018, 11:07
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Re: There are 12 red balls, 10 blue balls, 15 red balls and 9 &nbs [#permalink] 12 Aug 2018, 11:07
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