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A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls

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Math Revolution GMAT Instructor
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A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls  [#permalink]

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New post 19 Feb 2019, 01:52
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E

Difficulty:

  65% (hard)

Question Stats:

49% (01:28) correct 51% (01:50) wrong based on 57 sessions

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[GMAT math practice question]

A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls, 9 white balls, and 11 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 7 balls of the same color are drawn?

A. 7
B. 14
C. 15
D. 27
E. 28

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Re: A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls  [#permalink]

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New post 19 Feb 2019, 02:03
1
MathRevolution wrote:
[GMAT math practice question]

A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls, 9 white balls, and 11 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 7 balls of the same color are drawn?

A. 7
B. 14
C. 15
D. 27
E. 28



total balls = 36
so as to have atleast 7 balls of same color ; we need to remove balls which are max in no of same color ;
11+5+9+3= 28 balls
so now we are left with 7 balls of same color i.e blue and 1 of red..
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A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls  [#permalink]

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New post 19 Feb 2019, 04:43
1
MathRevolution wrote:
[GMAT math practice question]

A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls, 9 white balls, and 11 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 7 balls of the same color are drawn?

A. 7
B. 14
C. 15
D. 27
E. 28


Determine the WORST-CASE-SCENARIO -- for each color of ball, the MAXIMUM number that can be removed WITHOUT removing 7 of the same color:
1 red
3 green
5 yellow
6 blue
6 white
6 black
Sum = 1+3+5+6+6+6 = 27

Implication:
It is possible to remove 27 balls without selecting 7 of the same color.
Thus, to GUARANTEE that 7 of the same color are removed, we must remove ONE MORE ball:
27+1 = 28

.
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Re: A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls  [#permalink]

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New post 19 Feb 2019, 19:11
Archit3110 wrote:
MathRevolution wrote:
[GMAT math practice question]

A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls, 9 white balls, and 11 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 7 balls of the same color are drawn?

A. 7
B. 14
C. 15
D. 27
E. 28



total balls = 36
so as to have atleast 7 balls of same color ; we need to remove balls which are max in no of same color ;
11+5+9+3= 28 balls
so now we are left with 7 balls of same color i.e blue and 1 of red..


Archit3110

Could you please explain to me why do we have to keep the red one?

Kind regards!
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Re: A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls  [#permalink]

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New post 21 Feb 2019, 01:14
=>

The maximum number of draws without \(7\) balls of a single color is \(1 + 3 + 5 + 6 + 6 + 6 = 27\), obtained by drawing \(1\) red ball, \(3\) green balls, \(5\) yellow balls, \(6\) blue balls, \(6\) white balls and \(6\) black balls. If we draw one more ball, then we must have \(7\) balls of one color.
Thus, we need to draw \(28\) balls to ensure that \(7\) balls of the same color are drawn.

Therefore, E is the answer.
Answer: E
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Re: A box contains 1 red ball, 3 green balls, 5 yellow balls, 7 blue balls   [#permalink] 21 Feb 2019, 01:14
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