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The only way the range is equal to twice the difference betwenn the greatest number and the median is when the three numbers are equal.
Consider some ex.
[123] Median=2 Range=1, => out [125] M=2 R=4 => out [222] M=2 R=0 fits
Sufficient
2) [-1,0,1] fits [222] also [125]nope [-2,0,3]nope Sufficient
This is a stupid method and it's inaccurate too, maybe I've overseen a set. Hope you provide a more elegant way to answer this.
allabout, The question asks whether median = mean.
It should be A?
Let {x1,x2,x3} be the numbers
Q: Is x2 = (x1+x2+x3)/3 ? or
is x2 = (x1+x3)/2 ?
(1) Given x3-x1 = 2(x3-x2)
or x2 = (x1+x3)/2
SUFFICIENT.
(2) INSUFFICIENT, since the sum can equal any of the numbers x1,x2 or x3.
Hence A. _________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
The only way the range is equal to twice the difference betwenn the greatest number and the median is when the three numbers are equal.
Consider some ex.
[123] Median=2 Range=1, => out [125] M=2 R=4 => out [222] M=2 R=0 fits
Sufficient
2) [-1,0,1] fits [222] also [125]nope [-2,0,3]nope Sufficient
This is a stupid method and it's inaccurate too, maybe I've overseen a set. Hope you provide a more elegant way to answer this.
allabout, The question asks whether median = mean.
It should be A?
Let {x1,x2,x3} be the numbers
Q: Is x2 = (x1+x2+x3)/3 ? or is x2 = (x1+x3)/2 ?
(1) Given x3-x1 = 2(x3-x2) or x2 = (x1+x3)/2 SUFFICIENT. (2) INSUFFICIENT,since the sum can equal any of the numbers x1,x2 or x3.
Hence A.
Don't know; the sum of the three numbers is equal to three times one number. Do you know a set that fullfills this condition (2) but in which the mean isn't equal to the median?
Don't know; the sum of the three numbers is equal to three times one number. Do you know a set that fullfills this condition (2) but in which the mean isn't equal to the median?
Hmm. I understand it now sir/madam(?) THANK YOU!!
It will work only if the numbers are of the form you mentioned.
(-n,0,n) or (n,n,n). Hence SUFFICIENT.
Therefor D.
Another classic case of "Answer Biasing" as duttsit would like to call it. I just rejected (2) because I didn't like it _________________
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."
Say numbers are x,y,z and these are in ascending order, So median here is y
Range = z-x
St1:
Range = 2 * (z-y)
so z-x = 2z-2y
-x = z-2y
0 = z+x -2y
y = z+x -2y + y
i.e 3y = x+y+z
y = (x+y+z)/3
So median = mean SUFF
St2:
this have three cases
1. x+y+z = 3x
x + (x +a) + (x+b) = 3x
so a + b = 0 (Remember that a and b are +ve)
This means all must be equal to x i.e mean = median
2. x+y+z = 3y
(y-a) + y + y(+b) = 3y
so b-a = 0 i.e a = b (Remember that a and b are +ve)
This means all must be equal to y i.e mean = median
3. x+y+z = 3z
(z-a) + (z-b) + z = 3z
so -a-b = 0 i.e a +b = 0 (Remember that a and b are +ve)
This means all must be equal to z i.e mean = median