Question

In a set of 5 observations {3, 5, 9, 12, 14, one more observation is added, such that the new mean of the set is equal to the new median. If the added observation is not equal to any of the previous observations, how many positive integral values of the new observation are possible?}

• A. 5
• B. 4
• C. 3
• D. 2
• E. 1

Hint:

When a problem involves finding a value that affects the median, you must break the problem down into cases based on where the unknown value could fall within the sorted list. Be systematic to ensure you cover all possibilities.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This problem combines concepts of mean and median. We need to analyze how adding a new value \(x\) affects both measures and find the values of \(x\) that make them equal.
Step 2: Detailed Explanation:
The original set of 5 observations is {3, 5, 9, 12, 14}. The sum of these observations is \(3+5+9+12+14 = 43\).
A new observation, \(x\), is added. The new set has 6 observations.

New Mean: The new sum is \(43+x\). The new mean is \( \frac{43+x}{6} \).
New Median: The new set has 6 observations. The median will be the average of the 3rd and 4th terms once the set is sorted. The value of the median depends on where \(x\) falls in relation to the original numbers.
We need to solve the equation: New Mean = New Median, or \( \frac{43+x}{6} = \text{New Median} \).
Let's test the possible cases for the position of \(x\). Case 1: \(x\) is smaller than the middle values (x<9). The original sorted set is {3, 5, 9, 12, 14}. If we add an \(x<9\), the two middle terms of the new sorted set will be 5 and 9.

New Median = \( \frac{5+9}{2} = 7 \).
Set mean equal to median: \( \frac{43+x}{6} = 7 \).
Solve for x: \( 43+x = 42 \implies x = -1 \).
This value is not a positive integer, so it is not a valid solution.
Case 2: \(x\) is between the two original middle values (9<x<12). If we add an \(x\) in this range, the new sorted set is {3, 5, 9, x, 12, 14}. The two middle terms are 9 and \(x\).

New Median = \( \frac{9+x}{2} \).
Set mean equal to median: \( \frac{43+x}{6} = \frac{9+x}{2} \).
Solve for x: \( 43+x = 3(9+x) \implies 43+x = 27+3x \).
\( 16 = 2x \implies x = 8 \).
This value \(x=8\) contradicts the condition for this case (9<x<12). So, there is no solution in this range.
Case 3: \(x\) is larger than the middle values (x>12). If we add an \(x>12\), the two middle terms of the new sorted set will be 9 and 12.

New Median = \( \frac{9+12}{2} = 10.5 \).
Set mean equal to median: \( \frac{43+x}{6} = 10.5 \).
Solve for x: \( 43+x = 6 \times 10.5 \implies 43+x = 63 \).
\( x = 20 \).
This value is a positive integer, it's not equal to any of the original observations, and it satisfies the condition for this case (\(x>12\)). This is a valid solution.
Step 3: Final Answer:
We have analyzed all possible cases and found only one possible positive integral value for the new observation, which is 20. Therefore, there is only 1 possible value.