karovd wrote:
A certain list consists of 3 different numbers. Does the median of the 3 numbers equal the average (arithmetic mean) of the 3 numbers?
(1) The range of the 3 numbers is equal to twice the difference between the greatest number and the median.
(2) The sum of the 3 numbers is equal to 3 times one of the numbers.
Solution:
Question Stem Analysis:We need to determine whether the median of the 3 numbers of a certain list is equal to the average of the 3 numbers, given that the 3 numbers are different. We can let the 3 numbers be a, b, and c such that a < b < c. Therefore, we need to determine whether b = (a + b + c)/3.
Statement One Alone:We see that c - a = 2(c - b). Simplifying, we have:
c - a = 2c - 2b
2b - a = c
Now, let’s substitute this for c in b = (a + b + c)/3:
b = (a + b + (2b - a))/3 ?
b = 3b/3 ?
b = b ?
We can see that the answer is yes. Statement one alone is sufficient.
Statement Two Alone:If the sum of the three numbers is equal to 3 times one of the numbers, then the average of the three numbers is equal to one of the three numbers. The average cannot equal the smallest or the largest integer; thus the average must equal the median.
To see why the average cannot equal the smallest integer, notice that a < b and a < c. Adding these two inequalities, we obtain:
2a < b + c
Adding a to each side, we have:
3a < a + b + c
Dividing each side by 3, we see that a < (a + b + c)/3, which shows that the average is strictly greater than the smallest integer. It can be shown using similar arguments that the average is strictly less than the largest integer. Since we know the average is equal to one of the three numbers and since the average cannot equal the smallest or the largest number, it is equal to the median. Statement two alone is sufficient.
Answer: D
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