Interesting GMATFix Problem-33

arora007 wrote:In a room of 25 people, the median age is 50 years and the average age is 42 years. Which of the following statements must be true?
1) At least one person is older than 50 years old.
2) The number of people over 50 years old is equal to the number of people under 50 years old.
3) At least one person is younger than 34 years old.

a)none of the above
b) 1only
c) 3 only
d)1 and 2
e)1,2 and 3

this is just a guess... is it B?

I also thought it was B, but B is the wrong answer!

pretty simple
answer is (E)
1,2,3 all must be true.

Reasoning:-

It says that 50 is the median of the age of the 25 people. For an odd number median is the central value with equal number of values greater than and less than it. This implies that there are 12 people above 50 yrs and 12 people below 50 years of age.
Hence (1) & (2) should be true.

for statement (3)

let us assume that the 12 people above 50 years of age are from 51 to 62 old (one year age difference) and 12 people below 50 years are from 38 to 49 years old.

hence the average age becomes
(38+39+....+50+....+61+62)/25 = 50
or total age = 1250
but originally it is given that average age is 42
or total age = 1050

difference in ages = 1250 - 1050 = 200
this means tat on an average aleast 4 years (200/25) of age should be less for each person in our assumption.
This implies 38-4 = 34 hence atleast one person should be less than 34 years of age.
(3) must be true as well.

Therefore the answer is (e)

I hope i am correct.

Ishaan

IMO : C
I'll plug in some example to prove the problem:
1. 13 people are 50, 11 are 33, 1 is 37.
Average = (13*50+11*33+37)/25 = 42
Median is 50, however there's no one above 50 years old. 1 is not a "must be".
2. Take the example in 1 to prove how 2 is not a "must be" too
3. In this case we will try to allocate as many as possible year old to the smallest ones so see if they all can be equal or bigger than 34.
Because the total sum of year old is fixed: 42*25 =1050
There must be 13 no. equal or larger than 50 (as 50 is median), let's take 50 for all these 13 no. to create the smallest sum of them.
Hence the largest value for the remain 12 no. is: 1050 - 50*13 = 400
Average of those 12 no. :400/12 = 33.3
As 33.3 <34 , then there's must be at least one no. in those 12 no. smaller than 34. Thus, 3 is a "must be"
Pick C

limestone wrote:IMO : C
I'll plug in some example to prove the problem:
1. 13 people are 50, 11 are 33, 1 is 37.
Average = (13*50+11*33+37)/25 = 42
Median is 50, however there's no one above 50 years old. 1 is not a "must be".
2. Take the example in 1 to prove how 2 is not a "must be" too
3. In this case we will try to allocate as many as possible year old to the smallest ones so see if they all can be equal or bigger than 34.
Because the total sum of year old is fixed: 42*25 =1050
There must be 13 no. equal or larger than 50 (as 50 is median), let's take 50 for all these 13 no. to create the smallest sum of them.
Hence the largest value for the remain 12 no. is: 1050 - 50*13 = 400
Average of those 12 no. :400/12 = 33.3
As 33.3 <34 , then there's must be at least one no. in those 12 no. smaller than 34. Thus, 3 is a "must be"
Pick C

hey limestone u are correct.
i didnt think it in that sense. yes c is the correct answer