Question

Difficulty: Very hardMeasures of Central Tendency (Mean, Median, Mode)

A dataset DD consists of 1111 distinct positive integers. The median of DD is 4040, and the arithmetic mean of DD is 4545. A new dataset D′D' is created by increasing each of the 55 largest integers in DD by 1010 and decreasing each of the 55 smallest integers in DD by a positive integer xx. If the median of D′D' is strictly less than the arithmetic mean of D′D', what is the maximum possible integer value of xx?

  1. 2020Answer
  2. B
    2121
  3. C
    1919
  4. D
    2525
  5. E
    1010

Answer

The maximum possible integer value of xx is 2020.
The sum of the original 1111 values is 11×45=49511 \times 45 = 495. When the 55 largest integers are each increased by 1010, the sum increases by +50+50. When the 55 smallest integers are each decreased by xx, the sum decreases by −5x-5x. The new sum is 545−5x545 - 5x, making the new mean 545−5x11\frac{545 - 5x}{11}. Because DD has 1111 elements, the median is the 6th6^{\text{th}} element. Changing the smallest 55 and largest 55 elements does not alter the value of the 6th6^{\text{th}} element, so the median remains 4040. Requiring the median to be strictly less than the new mean gives 40<545−5x11  ⟹  440<545−5x  ⟹  5x<105  ⟹  x<2140 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21. The greatest integer less than 2121 is 2020.

Step-by-Step Solution

1
Calculate the total sum of the original dataset DD.
Since DD has 1111 elements with a mean of 4545, the sum of elements is 11×45=49511 \times 45 = 495.
The mean formula is Mean=Sumn\text{Mean} = \frac{\text{Sum}}{n}, so Sum=n×Mean\text{Sum} = n \times \text{Mean}.
2
Determine the median and sum of the modified dataset D′D'.
The median remains the 6th6^{\text{th}} element, which is 4040. The sum of D′D' is 495+5(10)−5(x)=545−5x495 + 5(10) - 5(x) = 545 - 5x.
Modifying only the 55 smallest and 55 largest elements leaves the 6th6^{\text{th}} central element unchanged.
3
Set up and solve the inequality comparing the median to the mean of D′D'.
Solve 40<545−5x11  ⟹  440<545−5x  ⟹  5x<105  ⟹  x<2140 < \frac{545 - 5x}{11} \implies 440 < 545 - 5x \implies 5x < 105 \implies x < 21.
The problem specifies that the median must be strictly less than the mean.
4
Identify the maximum integer value satisfying the inequality.
The largest integer strictly less than 2121 is 2020.
xx must be an integer.

Key Concept

Effect of data modifications on mean and median
Estimated Time:2m 30s
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