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Created 2 cases,
for one mean is lowest as per conservative multiple of 3 given mode of 27. Min value of mean is 21 corresponding to which x is 27

Similarly for one when mean is highest given same conditions, it is 30 corresponding to which x is 81

Hence range is 81-27=54

(B) is the answer
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So I am not sure much

but mode is 27
median = (18+21)/2 = 19.5 but to be consecutive multiple of 3 value should be either 24, 30 30 not possible so I am assuming median they have taken (21+27)/2 = 24

Or it is possible we got x as 27 so all 6 values become 6, 18, 21, 27, 27, 27

Overall seems like multiple possible ways so I will guess.
27 seems to small and 108 too much

54 57 are quite close options so will pick one of the two. C is generally a trap so will go with B

Ans - B
Bunuel
x, 6, 18, 21, 27, 27

If the average (arithmetic mean), median, and unique mode of the six integers shown above, when arranged in increasing order, are consecutive multiples of 3, what is the range of all possible values of x?

A. 27
B. 54
C. 57
D. 81
E. 108


 


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Here, x can be

1. Case 1.
<21, Not possible. x has to divisible by both 3 & odd. Since median has to be divisible by both 2 & 3.
can not be 21<x<27 for the same reason.
Notice unique. can not be any number other than 27.
Only one possible option 27
x=27
Ans. A.27
Bunuel
x, 6, 18, 21, 27, 27

If the average (arithmetic mean), median, and unique mode of the six integers shown above, when arranged in increasing order, are consecutive multiples of 3, what is the range of all possible values of x?

A. 27
B. 54
C. 57
D. 81
E. 108


 


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IMO answer is B 54 as a range so the logic I have solved this is again brute force. I get 27 as first x which solve the constraint of 3 consecutive 21 24 27 and then use the answer choice so then x =81 solved that constraint part so the anwser choice is 54 as range in my opinion
Bunuel
x, 6, 18, 21, 27, 27

If the average (arithmetic mean), median, and unique mode of the six integers shown above, when arranged in increasing order, are consecutive multiples of 3, what is the range of all possible values of x?

A. 27
B. 54
C. 57
D. 81
E. 108


 


This question was provided by GMAT Club
for the GMAT World Cup Competition

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> Unique Mode = 27
> When arranged, they are consecutive multiples of 3
> {21,24,27}
> {24,27,30}
> {27,30,33}
> Median min = $\frac{18+21}2=19.5$
> Median max = $\frac{21+27}2=24$
> So {27,30,33} not possible
> Median which are multiples of 3 : 21,24
> 21 not possible because unique mode
> Median = 24
>
> Avg = $\frac {x+99}6$
>
> When Avg = 21
> $x+99=126$
> $x=27$
>
> When Avg = 30
> $x+99=180$
> $x=21$
>
> Possible values of x = 27,81
> Range = 81 - 27 = 54
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