At a business school conference with 100 attendees, are ther

Responding to a pm:

Here is how you can think:

(1) The range of ages of the participants is 22 to 30, inclusive
There could be 100 schools represented by 100 students so no two students will have the same age-school combination.
All students could be from the same school so there would be multiple same age-school combinations.
Not sufficient.

(2) Participants represent 10 business schools.
The age of the students could range from 20 to 80 so we may or may not have the same age-school combinations. Not sufficient.

Now let's consider both statements:
Ages are 22, 23 ...30 - 9 different figures
Schools are A, B, C,..., J - 10 different schools

How many unique age school combinations can we make? A22, A23, ... A30, B22, B23, ..., J22, J23, ...J30
A total of 9*10 = 90 combinations. So we can have 90 unique age-school combinations for 90 students.
Now what about the remaining 10? They must also have age between 22 to 30 and must represent schools A to J. So say for the 91st student, we pick age 25 and school C. But note that we already have a student C25 since we accounted for all combinations in our 90 combinations. So the rest of the 10 students will need to repeat the age-school combination. Hence there must be students (at least 10) who have the same age and represent the same school.

Answer (C)