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Re: At a business school conference with 100 attendees, are ther [#permalink]
22 Nov 2011, 23:22

enigma123 wrote:

At a business school conference with 100 attendees, are there any students of the same age (rounded to the nearest year) who attend the same school? (1) The range of ages of the participants is 22 to 30, inclusive (2) Participants represent 10 business schools.

For me its clearcut A. Can someone please let me know if you think it not correct? OA is not provided unfortunately.

I believe the answer should be C.

S1: Only the range of age is given. But there may be 100 different or only 1/2 colleges. In that case the answer in insufficient. S1: Only #of B schools are given.We don't have the range of age. Insufficient

S1+S2 = we have all the data. Sufficient. hence IMO D.

Re: At a business school conference with 100 attendees, are ther [#permalink]
23 Nov 2011, 02:05

4

This post received KUDOS

Expert's post

enigma123 wrote:

At a business school conference with 100 attendees, are there any students of the same age (rounded to the nearest year) who attend the same school? (1) The range of ages of the participants is 22 to 30, inclusive (2) Participants represent 10 business schools.

For me its clearcut A. Can someone please let me know if you think it not correct? OA is not provided unfortunately.

It's not A, because you don't know how many schools are represented. It might be that each of the 100 students if from a different school, in which case the answer is 'no', or they may all be from the same school, in which case the answer is 'yes'. Similarly Statement 2 is not sufficient, because we don't know how many ages are represented.

Using both Statements, we know that there are only 10 schools at the conference, and only 9 different ages (from 22 to 30 inclusive). Certainly it's possible that there are two, say, 28 year-olds from the same school, so the answer can be 'yes'. Can the answer be 'no'? Then we'd need every person of the same age to attend a different school. That means we could have at most ten 22 year olds, at most ten 23 year olds, and so on, and so at most 9*10 = 90 people. But we have 100 people, so it's impossible that the answer is 'no', and there must be at least two people of the same age at the same school, and the answer is C. _________________

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Re: At a business school conference with 100 attendees, are ther [#permalink]
09 Aug 2014, 09:41

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Re: At a business school conference with 100 attendees, are ther [#permalink]
07 Dec 2014, 21:43

Expert's post

4

This post was BOOKMARKED

enigma123 wrote:

At a business school conference with 100 attendees, are there any students of the same age (rounded to the nearest year) who attend the same school?

(1) The range of ages of the participants is 22 to 30, inclusive (2) Participants represent 10 business schools.

For me its clearcut A. Can someone please let me know if you think it not correct? OA is not provided unfortunately.

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.

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