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)
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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.

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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imo C.

Stmt 1 does not tell you how many schools there are. So all though you have a range of 9 years for 100 attendees, there could be 100 schools.

Stmt 2 is not enough for the same reason. We know that there are 10 schools but the age range could be anything.

Stmt 1 & 2: suff

very hard.

at one school, the maximum student diferent at age is 9 (there are 9 ages).

we have 10 schools,so, the number of students with different ages at different schools is 90

some school must have more than 9 students. those schools must contain student at the same age

Statement 1: Not sufficient
It doesn't mention the number of schools which had their students participate in the conference.

Statement 2: Not sufficient
We don't know the range of the age of the students attending the conference. There can be only one student from the participating colleges or 100 students from the same college.

Together: Sufficient
Range: 22-30 = 9 age group
Number of participating schools: 10
Total number of students attending the conference: 100
Min number of students from each age group: 9*10 = 90
Therefore, there has to be at least one repeatation (actually 10 repeatation) in the age group.


Note: Statement 1 can be mistaken to be the answer if we don't read the last few words of the question ("who attend the same school"). Happened with me :D

Regards
Pratik

hard one.

look at both 1 and 2.

there are at least 10 student in one business school
there are at most 9 ages. so if 10 student are different most, at least 2 student have the same age.

C

(rounded to the nearest year)
what is the importance of this part in the question?
I thought like 21 years 6 months would become 22 years...
so we have range from 21 years to 30 years

You are given that range of ages is 22 to 30. So you do not have a 21 year (after rounding) old. So there is no one whose age is 21 years and 3 months or 20 years and 8 months etc. You might have a 21 year 7 months old (something that will get rounded to 22 years).
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100 attendees.
A) Maybe attendees are all from different schools. or maybe same Insufficient
B) on average, 10 attendees per school. there age can be 21,22...30 or 21,21,...21
Insufficient

A&B) range=22 to 30: 22,23...30 (total 9 ages..)
also 10 business schools. therefore atleast one school will have more than 9 attendees. otherwise total will be only 90 max. Let that school be A.
Now A has 10 students and we have only 9 different ages. Thus atleast 2 students will have same age. Sufficient

without using pen and paper:

Stmt 1: we can't do anything with a range without knowing the number of schools the students belong to. Not sufficient
Stmt 2: again, without information of ages/range of age we cannot solve this.

combining 1 & 2: we know that there are 10 B Schools and 9 (from 22-30 exclusive) age groups, therefore some have to belong to the same school, i.e, the least number of age repetitions within the same school is 1 (can be more but we don't need that to answer the question)

therefore, C stands.
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