GATE | GATE-CS-2016 (Set 2) | Question 38

Consider a set U of 23 different compounds in a Chemistry lab. There is a subset S of U of 9 compounds, each of which reacts with exactly 3 compounds of U. Consider the following statements:

I.  Each compound in U \ S reacts with an odd number of compounds.
II.  At least one compound in U \ S reacts with an odd number of compounds.
III.  Each compound in U \ S reacts with an even number of compounds. 

Which one of the above statements is ALWAYS TRUE?

(A) Only I
(B) Only II
(C) Only III
(D) None


Answer: (B)

Explanation: Answer is B. This a graph theory question.

“\” is the set difference operation. Same as U – S.

Since U is universal set, U\S would give complement of S = S’
Let S contains Compounds numbered {1,2,3…8, 9} so U\S contains Compounds {10, 11, 12…. 22, 23}



Consider these compounds to be vertices of a graph.
An edge b/w two vertices indicate that the compounds react with each other.

This graph has NO multiple edge cause that doesn’t make sense.
There are no directed edges cause if one compound reacts with other it also means other reacts with it too. Single edge represent reaction b/w both.
It has NO Loops cause compound don’t react with itself.

Hence graph is simple undirected graph.

We know that “An undirected graph has even number of vertices of odd degree”

9 vertices of this graph have degree 3 (odd degree) cause 9 compounds react with 3 other compounds.
Hence there must be at LEAST 1 more vertex which must have an odd degree.
This extra compound must belong to U\S cause 9 compounds in S have already been accounted for.
This implies statement II in the question is TRUE.

Other 2 statements are False.
Statement III – Consider that all 9 compounds in S react with the same 3 compounds in U\S say with {20, 21, 22} hence all compounds in U\S either react with Zero or 9 compounds but not Even number.
Similarly Statement I can be proved wrong by visualizing the graph.


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