In a certain set of five numbers the median is 200

Lets say the set is a,b,200,c,d as taken by bunuel...

Now we donot know anything about the mean of the numbers.

If the numbers are 200,200,200,220,230 the range will be 30
If the numbers are 200,200,200,220,290 the range will be 40. and so onnnnnn....

So we cannot say desicively what the range is? Whether it is greater than 80 or not..So it is not sufficient

From F.S 1, as the average is 240, we can assume that each of the 5 elements is 240 . However , as the median is 200, the two elements below the median can be at-most 200 each and the two elements above median can be at-least 200 each. Now,Range = Max(element)-Min(element). After maximizing the 2 elements below the median, we can have the following set : 200,200,200,300,300. Now,to have a value of range=80, we would have to have the value of the Max(element) = 280. Notice : 200,200,200,320,280 --> The Max(element) changes to 320 . Thus, any change on the Max(element) will ALWAYS lead to a range>80.Also, any decrease on the Min(element) will only increase the numerical value of Range above 80. Sufficient.

From F.S 2, we can have either 160,180,200,200,200 where range<80 OR 100,120,200,200,200 where range>80. Insufficient.

A.

Because it's easy to construct the sets which give two different answers to the question:
{200, 200, 200, 201, 100000000}
{200, 200, 200, 201, 202}

This is a great Question.
Excellent Solutions have been provided above.
Here is my solution to this one=>

Let the 5 numbers be in increasing order as =>

a
b
c
d
e



Now #=5

Hence median = 3rd term => c
So c= 200

Now we are asked if range is >80 or not.
Here we will calculate the minimum value of range and if the minimum value of range >80 then range will be great than 80


Statement 1=>
Mean =240

There is a very important property of mean => Sum of deviations around the mean is always zero.

Now to minimise the range => a=b=c=200

Total negative deviation occurred =120
Hence to balance out this => d and e must have a 120 positive devotion when added.
To minimise range we must minimise e
for that we she maximise d
d=e=300

Hence minimum range = 300-200=100

Hence sufficient

Statement 2-->
Three values are equal
Notice no mean is provided.
E.g=> a=e=200=> Range =0
a=200 and e=999999 => Range>80
Hence insufficient

Hence A

Bunuel, shouldnt it be minimize A and maximize D?

Hi
You see, if we minimise a and maximise d => The range would be maximum.
we have to check for the minimum value of range.
E.g=>
Lets see an example with 3 numbers => a,b,c
Median = 100 => b=100
a,100,c be the three numbers in increasing order.
a=0
c=200
the range = 200

if a=100
c=100
Range =0


Hence we have to maximise a and minimise d in our original question


Regards
Stone Cold

Given: In a certain set of five numbers the median is 200.

Asked: Is the range greater than 80?

(1) The average (arithmetic mean) of the numbers is 240
Sum of 5 numbers = 240 * 5 = 1200
For minimum range
Let the numbers be {200,200,200,x,x}
2x = 600
x = 300
Minimum Range = 300-200=100>80
SUFFICIENT

(2) Three of the numbers in the set are equal
Case 1: Let the numbers be {200,200,200,x,y}
Range = y-200
But value of y is not provided.
Case 2: Let the numbers be {x,y,200,200,200}
Range = 200-x
But value of x is not provided.
NOT SUFFICIENT

IMO A

TheNona

Let us consider 2 cases

Case 1: {200,200,200,c,d}
Since d may take any value
NOT SUFFICIENT

Case 2: {a, b, 200, 200, 200}
Since a may take any value
NOT SUFFICIENT

OreoShake

We have to prove than minimum possible range > 80
So that all possible ranges will be > 80

To minimise range we have to maximise a and minimise d

Hey Bunuel,
Maximum value of a as well as b is 200 and minimum value of d is c, so our set will be: {200, 200, 200, d, d}
If the minimum value is c, then your set should be: {200, 200, 200, c, c}

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