M46-42

Question

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

Bunuel · Answer

Official Solution:

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

In the list  6, 18, 21, 27, 27, x , the number 27 appears twice, and each other number appears once. So 27 is already a mode. However, for the list to have a  unique  mode,  x  cannot be 6, 18, or 21, because any of those values would then also appear twice, and 27 would no longer be the unique mode.
 
Since the average, median, and mode are consecutive multiples of 3, and the unique mode is 27, the three values could be:
 
 \{21, 24, 27\} 
 
 \{24, 27, 30\} 
 
 \{27, 30, 33\} 
 
Now consider the median. Since the list has an even number of terms, the median is the average of the two middle terms.
 
If  x < 27 , then the median is less than 24. Among the possible sets above, the only value less than 24 that could be the median is 21. But a median of 21 would require  x = 21 , which is not allowed because then 21 and 27 would both appear twice.
 
So  x  cannot be less than 27.
 
When  x \geq 27 , the two middle terms are 21 and 27, so the median is:
 
 \frac{21 + 27}{2} = 24 
 
Thus, only the following two sets remain possible:
 
 \{21, 24, 27\} 
 
 \{24, 27, 30\} 
 
So the average must be either 21 or 30.
 
The sum of the six integers is:
 
 x + 6 + 18 + 21 + 27 + 27 = x + 99 
 
So the average is:
 
 \frac{x + 99}{6} 
 
 Case 1:  \{21, 24, 27\}  
 
 \frac{x + 99}{6} = 21 
 
 x + 99 = 126 
 
 x = 27 
 
 Case 2:  \{24, 27, 30\}  
 
 \frac{x + 99}{6} = 30 
 
 x + 99 = 180 
 
 x = 81 
 
So the possible values of  x  are 27 and 81.
 
The range of all possible values of  x  is:
 
 81 - 27 = 54

Answer: B

nulladicta · Answer

But how is 81 a consecutive multiple of 3? Isnt it a major jump from 27 if we put x=81 after 27?

Bunuel · Answer

The 81 itself does  not  need to be a consecutive multiple of 3 with 27.

The condition applies to the  average, median, and mode , not to the six numbers in the list.

When x = 81:

Average = 30
 Median = 24
 Mode = 27

These three values are 24, 27, 30, which are consecutive multiples of 3.

So x = 81 is valid even though 81 is far from 27.

M46-42

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