Calculating Median in a Class with B, D, A, and C Scores: Findings and Solutions

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Discussion Overview

The discussion revolves around determining the median score of four students (Budi, Doni, Adi, and Coki) based on a set of inequalities relating their scores. The scope includes mathematical reasoning and problem-solving related to median calculation.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant states the inequalities: B > D, A + D > B + C, and D > 2B - A, and seeks to determine the median.
  • Another participant interprets the inequalities and suggests that B is less than the average of A and D, and provides possible score arrangements.
  • A third participant summarizes that the median can be expressed as $\frac{1}{2}(B + \max(C, D))$, indicating different cases based on the relative positions of C, D, and A.
  • A later reply expresses gratitude for the assistance but notes that multiple forums have confirmed the impossibility of determining the exact numerical value of the median.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact value of the median, with some suggesting possible expressions for it while acknowledging the limitations in determining a specific number.

Contextual Notes

The discussion highlights limitations in deriving a unique solution due to the nature of the inequalities and the dependence on relative score positions, which remain unresolved.

Monoxdifly
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In a class, Budi's score is greater than Doni's. The sum of Adi's and Doni's scores is greater than the sum of Budi's and Coki's scores. Meanwhile, Doni's score is greater than two times Budi's score substracted by Adi's score. Determine the median of those four students' scores.

All I know, was, by using their initials that:
B > D
A + D > B + C
D > 2B - A

And by using the second and third info I got that their score from lowest to highest is either C, D, B, A or D, B, C, A. However, I met a dead-end after that. Please someone help me.
 
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I sketched a number line ... $B>D$ is obvious. The last inequality states $B < \dfrac{A+D}{2}$, or $B$ is less than the average of $A$ and $D$, equal to the midpoint of segment $AD$.

Finally, the second inequality states $C < A+D-B$. Note the possible positions for $C$ represented by the open-ended blue ray in the diagram.

If $B < C< A$ or $D < C < B$, the the median of the four scores is $\dfrac{B+C}{2}$

If $C < D$, then the median is $\dfrac{B+D}{2}$
 
Yep. Put more concisely, the median is $\frac 12 (B+\max(C,D))$.
 
Thank you everyone who has helped me with this question. I have asked this question in 4 different forums including this one and all confirm that it is impossible to find the exact number of the median's value.
 

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