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In a certain laboratory, chemicals are identified by a color-coding sy

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In a certain laboratory, chemicals are identified by a color-coding sy  [#permalink]

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12 Dec 2018, 03:24
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55% (hard)

Question Stats:

58% (01:41) correct 42% (01:48) wrong based on 113 sessions

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In a certain laboratory, chemicals are identified by a color-coding system. There are 20 different chemicals. Each one is coded with either a single color or a unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

A. 5
B. 6
C. 7
D. 20
E. 40

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Re: In a certain laboratory, chemicals are identified by a color-coding sy  [#permalink]

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05 Jun 2019, 05:14
Bunuel wrote:
In a certain laboratory, chemicals are identified by a color-coding system. There are 20 different chemicals. Each one is coded with either a single color or a unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

A. 5
B. 6
C. 7
D. 20
E. 40

Shouldn't the approach to solve this question be the following:

Let the number of colours be 'n', and we have n + n(n-1)/2 >= 20

We get 2n + n^2 - n = 40, simplifying further we get n^2 + n = 40

Now, putting in the values (since we only require the minimum no. of colors), we see that n cannot be equal to 5 and 6 is the first (and the minimum) value that satisfies the equation.

Hence, option (B)
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In a certain laboratory, chemicals are identified by a color-coding sy  [#permalink]

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12 Dec 2018, 03:33
Bunuel wrote:
In a certain laboratory, chemicals are identified by a color-coding system. There are 20 different chemicals. Each one is coded with either a single color or a unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

A. 5
B. 6
C. 7
D. 20
E. 40

6c1 +6c2 : 21

IMO 6 should be correct.. IMO b
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Re: In a certain laboratory, chemicals are identified by a color-coding sy  [#permalink]

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12 Dec 2018, 10:07
Let say there are 2 colors : A & B. All possible combination would be A , B , AB.
If there are 3 : A, B & C. Combinations would be A , B , C , AB, AC , BC.

n(n+1)/ 2 > 20
n(n+1) > 40

Lowest value which fits is 6.

Ans B
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In a certain laboratory, chemicals are identified by a color-coding sy  [#permalink]

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04 Jan 2019, 03:19
In a certain laboratory, chemicals are identified by a color-coding system. There are 20 different chemicals. Each one is coded with either a single color or a unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

5 colours

A,B,C,D,E
BC, BD, BE
CD, CE
DE
falls short of 20.

6 colours

all the above + a new colour F
F, FA, FB, FC,FD, FE
totals 21.

B is the correct option.
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Re: In a certain laboratory, chemicals are identified by a color-coding sy  [#permalink]

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06 Jun 2019, 18:01
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Bunuel wrote:
In a certain laboratory, chemicals are identified by a color-coding system. There are 20 different chemicals. Each one is coded with either a single color or a unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

A. 5
B. 6
C. 7
D. 20
E. 40

If there are 5 different colors, the number of codes that can be formed is 5C1 + 5C2 = 5 + 10 = 15, which is not sufficient for the 20 different chemicals.

If there are 6 different colors, the number of codes that can be formed is 6C1 + 6C2 = 6 + 15 = 21, which IS sufficient for the 20 different chemicals. Therefore, we need at least 6 different colors.

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Re: In a certain laboratory, chemicals are identified by a color-coding sy   [#permalink] 06 Jun 2019, 18:01
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