kp1811 wrote:
The question stem says " assuming there is only one mode" so we need to consider sets where Mode is well defined [ I hope my understanding is correct]
A
stmnt 1 - median is equal to range
lets consider a set { 0,2,2}. here we have median is equal to range. mode[2] is also equal to range
lets consider the set { 2,2,4} here we median is equal to range but mode is equal to range. hence suff
stmnt2 - lets consider a set { 2,2,4}. here 4 = 2* 2(smallest number) and mode[2] is equal to range[2].
lets consider a set { 5,10,10} here 10 = 2* 5(smallest number) but mode[10] is not equal to range[5].
hence insuff
PS: What is the source of this question?
I think that above solution is correct and the answer is A, though there is one case missing.
Set can be of a form:
A. {X,Y,Y}
B. {X,X,Y}
OR
C. {XXX}
Basically telling us that there is only one mode, stem is saying that we
do not have three distinct numbers in the set.
(1) The median equals the range:
A. {X,Y,Y} --> Y=Y-X --> X=0 --> Set: {0,Y,Y}. Mode=Y, Range=Y-0=Y, --> Y=Y. True.
B. {X,X,Y} --> X=Y-X --> 2X=Y --> Set: {X,X,2X}. Mode=X, Range=2X-X=X --> X=X. True.
C. {XXX} --> X=X-X --> X=0 --> Set: {0,0,0}. Mode=0, Range=0 --> 0=0. True.
Sufficient.
(2) The largest number is twice the value of the smallest number:
A. {X,Y,Y} --> Y=2X --> Set: {X,2X,2X}. Mode=2X, Range=2X-X=X, --> 2X equals to X only if X=0 (set: {0,0,0}), but we don't know that. Not always true. For example we can have set: {1,2,2} Mode=2, but range=1, 2#1.
B. {X,X,Y} --> Y=2X --> Set: {X,X,2X}. Mode=X, Range=2X-X=X --> X=X. True.
C. {XXX} --> X=2X --> X=0 --> Set: {0,0,0}. Mode=0, Range=0 --> 0=0. True.
Two different answers (cases B and C always true, A not always).
Not sufficient.
Answer: A.
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