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

Question

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?

CONSIDER ANSWER CHOICES

5 colours

A,B,C,D,E
AB, AC, AD, AE
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.

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

ScottTargetTestPrep · Accepted Answer

Let’s analyze the answer choices.

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.

Answer: B

RJ7X0DefiningMyX · Answer

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)

Archit3110 · Answer

6c1 +6c2 : 21

IMO 6 should be correct.. IMO b

pandeyashwin · Answer

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

kanakdaga · Answer

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?

CONSIDER ANSWER CHOICES

5 colours

A,B,C,D,E
AB, AC, AD, AE
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.

CONSIDER ANSWER CHOICES

5 colours

A,B,C,D,E 
AB, AC, AD, AE 
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.

Kinshook · Answer

Given: 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.

Asked: 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?

If n=5; Number of chemicals = 5C1 + 5C2 = 5 + 10 = 15 <20 ; Not feasible

If n=6; Number of chemicals = 6C1 + 6C2 = 6 + 15 = 21 > 20; Feasible

IMO B

Posted from my mobile device

luisdicampo · Answer

Deconstructing the Question

If there are  n  colors, we can form codes using either a single color or an unordered pair of colors.

Total possible codes:

n + \binom{n}{2}

We need at least 20 codes.

Step-by-step

Compute:

n + \binom{n}{2} = n + \frac{n(n-1)}{2} = \frac{n^2 + n}{2}

We need:

\frac{n^2 + n}{2} \ge 20

n^2 + n \ge 40

Test values:

n = 5: 5 + 10 = 15  (not enough)

n = 6: 6 + 15 = 21  (sufficient)

So the minimum value is  6 .

Answer B

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

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