In a certain set of five numbers the median is 200. Is the range greater than 80?

(1) The average (arithmetic mean) of the numbers is 240
(2) Three of the numbers in the set are equal

why b is clearly insufficient ?

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.
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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
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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
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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
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