Examkin

Soru

Zorluk: Çok zorMeasures of Central Tendency (Mean, Median, Mode)

A dataset consists of 99 distinct positive integers a1,a2,a3,a4,a5,a6,a7,a8,a9a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9 listed in strictly increasing order. The mean of the 99 integers is 2828, and the median is 2424. A 10th10\text{th} positive integer xx, where x>a9x > a_9, is added to the dataset, causing the new mean of the 1010 integers to become 3131. If LL represents the minimum possible value of a9a_9, what is the value of x−Lx - L?

  1. A
    22
  2. B
    66
  3. 2121Cevap
  4. D
    3737
  5. E
    5858

Cevap

The value of x−Lx - L is 2121.
The sum of the original 99 elements is 9×28=2529 \times 28 = 252, and the sum of the 1010 elements is 10×31=31010 \times 31 = 310, giving x=58x = 58. To minimize a9a_9 (LL), the sum of all other elements must be maximized. The median a5=24a_5 = 24. The maximum possible values for the first four distinct elements below 2424 are 20,21,22,2320, 21, 22, 23 (summing to 8686). To minimize a9a_9, the upper four elements must be consecutive integers (a9−3,a9−2,a9−1,a9)(a_9 - 3, a_9 - 2, a_9 - 1, a_9). Setting the total sum equation 86+24+(4a9−6)=25286 + 24 + (4a_9 - 6) = 252 yields 4a9=1484a_9 = 148, so L=37L = 37. Consequently, x−L=58−37=21x - L = 58 - 37 = 21.

Adım Adım Çözüm

1
Calculate the sum of the original 9 integers and determine the value of the 10th integer xx.
The sum of the original 99 integers is 9×28=2529 \times 28 = 252. The sum of the 1010 integers after adding xx is 10×31=31010 \times 31 = 310. Therefore, x=310−252=58x = 310 - 252 = 58.
The sum of a set of numbers equals the number of elements multiplied by the mean.
2
Identify the median of the ordered 9-element set.
In an ordered set of 99 elements, the median is the 5th5\text{th} element, so a5=24a_5 = 24.
For an odd number of ordered elements, the median is the exact middle element.
3
Maximize the sum of the first 4 elements a1,a2,a3,a4a_1, a_2, a_3, a_4 to minimize a9a_9.
Since all integers are distinct and strictly increasing, a4<24a_4 < 24. To maximize a1+a2+a3+a4a_1 + a_2 + a_3 + a_4, choose a4=23,a3=22,a2=21,a1=20a_4 = 23, a_3 = 22, a_2 = 21, a_1 = 20. Their sum is 20+21+22+23=8620 + 21 + 22 + 23 = 86.
Maximizing lower elements leaves the smallest possible remainder of the total sum for the upper elements.
4
Express a6,a7,a8a_6, a_7, a_8 in terms of a9a_9 to minimize a9a_9.
To make a9a_9 as small as possible, a6,a7,a8a_6, a_7, a_8 should be as large as possible relative to a9a_9, meaning they are consecutive integers: a8=a9−1a_8 = a_9 - 1, a7=a9−2a_7 = a_9 - 2, a6=a9−3a_6 = a_9 - 3.
Making elements above the median consecutive integers directly below a9a_9 minimizes a9a_9 for a fixed sum.
5
Set up the sum equation for the 9 elements to solve for L=min(a9)L = \text{min}(a_9).
(a1+a2+a3+a4)+a5+(a6+a7+a8+a9)=252  ⟹  86+24+(a9−3+a9−2+a9−1+a9)=252  ⟹  104+4a9=252  ⟹  4a9=148  ⟹  a9=37(a_1 + a_2 + a_3 + a_4) + a_5 + (a_6 + a_7 + a_8 + a_9) = 252 \implies 86 + 24 + (a_9 - 3 + a_9 - 2 + a_9 - 1 + a_9) = 252 \implies 104 + 4a_9 = 252 \implies 4a_9 = 148 \implies a_9 = 37. Thus, L=37L = 37.
Solving the algebraic equation derived from the total sum yields the minimum integer value for a9a_9 while satisfying a6=34>24a_6 = 34 > 24.
6
Calculate x−Lx - L.
x−L=58−37=21x - L = 58 - 37 = 21.
Subtracting the minimum bound LL from xx fulfills the target question requirement.

Anahtar Kavram

Optimization of Dataset Values using Central Tendency Constraints
Tahmini Süre:2m 30s
Bu soruyu puanla

Konular

Measures of Central Tendency (Mean, Median, Mode)

Bu konudan daha fazla soru