Xavier Aptitude Test 2013 Paper

(e) In normal case, Sara has to invite 24 contacts of each of her 5 friends along with those 5 friends as Sara is one among these 25 friends.
∴ 5 + 5 × 24 = 125
But, it is given that,
‘At least two of Sara’s friends are connected with each other.’
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 remaining 3 of Sara’s friends has got 24 contacts.
∴ There are 5 + 2 × 23 + 3 × 24 = 5 + 46 + 72 = 123 invites.
At least two also hints us that at maximum, all the 5 friends of Sara has the other 4 friends in their contacts.
Then each of these 5 friends has 20 contacts (excepting Sara and the other 4).
∴ There are 5 + 5 × 20 = 5 + 100 = 105 invites.