Download Problem of the Week Problem D and Solution Just Average

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Problem of the Week
Problem D and Solution
Just Average
Problem
The mean, median and mode of the seven numbers in the list 10, 2, 5, 2, 4, 2, x are distinct. The
mean, median and mode are calculated and then listed in order from smallest to largest. The
differences between adjacent numbers in this new list are equal. Determine all possible values
of x.
Solution
Since there are at least three 2’s, the mode will be 2 regardless the value of x.
x + 25
10 + 2 + 5 + 2 + 4 + 2 + x
=
.
The mean of the numbers is
7
7
The median of the numbers will depend on the value of x compared to the other numbers.
Since there are seven numbers in the list, the median will be the fourth number in the ordered
list of numbers. We will break the problem into cases.
Case 1: x ≤ 2 and the ordered list is x, 2, 2, 2, 4, 5, 10.
The fourth number in the list is 2 and it follows that the median is 2. But the mode is also 2.
Since the mean, median and mode are all distinct, there is no possible value of x ≤ 2 which
produces a valid solution.
Case 2: 2 < x ≤ 4 and the ordered list is 2, 2, 2, x, 4, 5, 10.
The fourth number in the list is x and it follows that the median is x. We also know that the
mode is 2. We do not, however, know which is larger, the median or the mean so the mean,
median and mode, in numerical order could be
2, x,
x + 25
7
or 2,
x + 25
, x
7
Case 2a: Let the median be smaller than the mean. Then the ordered list is 2, x,
Since the differences between adjacent numbers are equal, we have
x + 25
−x
7
7x − 14 = x + 25 − 7x
13x = 39
x=3
x + 25
.
7
x−2=
(Multiply both sides of the equation by 7.)
So x = 3 and the mode, median and mean are 2, 3, 4, respectively. Each adjacent pair of
numbers in the list differs by 1.
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Case 2b: Let the mean be smaller than the median. Then the ordered list is 2,
Since the differences between adjacent numbers are equal, we have
x + 25
x + 25
−2=x−
7
7
x + 25 − 14 = 7x − x − 25
36 = 5x
36
x=
= 7.2
5
x + 25
, x.
7
(Multiply both sides of the equation by 7.)
But x ≤ 4, so there is no value of x that satisfies the conditions in this case.
Case 3: x > 4 and the first 4 numbers in the list, in numerical order, are 2, 2, 2, 4.
Since the fourth number in the list is 4, then it follows that the median is 4. The mode is 2.
x + 25
29
Since x > 4, the mean
>
> 4. It follows that the mean is greater than the median.
7
7
x + 25
The ordered list of numbers is then 2, 4,
.
7
Since the differences between adjacent numbers are equal, we have
x + 25
−4
7
x + 25
2=
−4
7
x + 25
6=
7
42 = x + 25
17 = x
4−2=
In this case, x > 4 and therefore x = 17 is a valid solution.
Therefore, there are two values of x that satisfy the conditions of the problem.
When x = 3, the ordered list is 2, 2, 2, 3, 4, 5, 10. The mean is 4, the median is 3 and the mode
is 2. When listed from smallest to largest, the three numbers are 2, 3, 4 and the difference
between adjacent terms is 1.
When x = 17, the ordered list is 2, 2, 2, 4, 5, 10, 17. The mean is 6, the median is 4 and the
mode is 2. When listed from smallest to largest, the three numbers are 2, 4, 6 and the
difference between adjacent terms is 2.
Note: 2, 3, 4 and 2, 4, 6 are both examples of arithmetic sequences. An arithmetic sequence is a
list of numbers with the difference between adjacent numbers being constant.