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XAT 2013 Question 58

This is a detailed step-by-step solution for XAT 2013 Question 58. Understand the concept, common mistakes, and expert tips to solve similar questions in XAT exam.

XAT

Sarah has 5 friends and each of her 5 friends have 25 friends.
The 5 friends of Sarah have Sarah as one of their 25 friends. Therefore, apart from Sarah the five friends will have 24 friends each.
If no 2 friends know each other, then Sarah will have to invite 5*24 + 5 = 120 + 5 = 125 persons.
Sarah knows that at least 2 of her friends are connected to each other.
Let the 5 friends be A, B, C, D, and E. Let us consider that A and B are friends.
A will be counted twice (first as one among the 5 persons and second as one of the friends of B).
B will also be counted twice (first as one among the 5 persons and second as one of the friends of A).
Subtracting these 2 friends, we can infer that the maximum number of persons that Sarah could have invited to the party is 123.

The minimum number of friends that Sarah could have invited is obtained when all the friends are connected to each other and their friends are also the same. In this case, the five friends will have 20 common friends, will be friends with each other (4) and Sarah (1).
The minimum number of persons Sarah could have invited to the party = 20+5 = 25.
Option B is more appropriate that option C.
Therefore, option B is the right answer.

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