Saturday, 12 January 2013

Puzzle-35 (XAT-2013)

Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?
(A)  ≥ 105  (B) ≤ 123        (C) < 125        (D) ≥ 100 and ≤ 125 (E) ≥ 105 and ≤ 123
For solution click on 'Read more' below..
 The key point:
“Each of Sara’s 5 friends has twenty five friends” => Sara is one among these 25 friends.
I call this as ‘Facebook Logic’. If A is in friend-contacts of B, then B is also in the friend-contacts of A.
So, in normal case, Sara has to invite 24 contacts of each of her 5 friends along with those 5 friends (ie., 5 + 5*24 = 125). But this is not end of story. It is also given that,
At least two of Sara’s friends are connected with each other”:          
Here ‘at least two’ is another point to note. If two of 5 friends of Sara are connected with each other, then each one of this pair has 23 contacts (excepting Sara and the other friend in the pair). Each of the remaining 3 of Sara’s friends has got 24 contacts (excepting Sara). So if Sara invites all, there are 5 + 2*23 + 3*24 = 123 invitees in all.
At least two” hints us that -at maximum, all the 5 friends of Sara has the other four friends in their contacts. Then each of these 5 friends has 20 contacts (excepting Sara and the other 4). So if Sara invites all, there are 5 + 5*20 = 105 invitees in all.
Finally we arrived at the answer: The count of invitees ranges from 105 to 123
Ans: E
For easy understanding I depict these images:


  1. according to testfunda and time edu. answer to this problem is <=123 .i.e. minimum 25 max which is correct.what is flaw in that method .