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A set contains five numbers: 11, 23, 14, 20, and x. How many different values can x take if the mean (average) and median are equal?

Correct Answer:

**D**The sum of any two sides of a triangle must be larger than the third side.

- To find the median of a set, it is often helpful to arrange the numbers from smallest to largest, and the median is the number in the middle. Since we do not know the value of x, we are unable to do this. However, there are only five possible orders for our numbers, depending on the value of x:

Position of x Median of set x, 11, 14, 20, 23 14 11, x, 14, 20, 23 14 11, 14, x, 20, 23 x 11, 14, 20, x, 23 20 11, 14, 20, 23, x 20 - We can see that in all of these situations, neither 11 nor 23 can be the median. So we have three cases: the median is 14, the median is 20, or the median is x.

Case 1: The median is 14. If the mean equals the median, then we want the mean to equal 14. The mean can be calculated to be . For the mean to be 14, we have (68+x)/5 = 14, which has a solution of x = 2. If x = 2, then the mean and the median are both 14. Thus 2 is our first possible value for x.

Case 2: The median is 20. Now we want the mean to equal 20. Thus (68+x)/5 = 20, which has x = 32 as a solution. If x = 32, then the mean and the median are both 20. Thus 32 is our second possible value for x.

Case 3: The median is x. Now we want the mean to equal x. Thus (68+x)/5 = x, which has x = 17 as a solution. Since 17 is between 14 and 20, setting x equal to 17 would make the median equal to 17. Thus 17 is our third and last possible value for x. - Since these were the only three possible values for the median of the set, we know that we have found all possible solutions. Thus there are three values that x can equal so that the mean and the median of the set are equal. The correct answer is D.