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Question

Five people C, D, X, Y and Z each scored different marks in the same test. All scores are in whole numbers only.
D scored the second lowest and D's score was 40. X scored more than D but less than the one who scored 45. 45 was not the highest score. C's score is an even number. C did not score the least. Y scored 6 marks less than Z. Which of the following can be a possible score of C?

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
46

Let the five scores in increasing order be $S_1 < S_2 < S_3 < S_4 < S_5$. We are given the following information:

  • There are five distinct scores for C, D, X, Y, and Z.
  • All scores are whole numbers.
  • $D$'s score is the second lowest, so $S_2 = D = 40$. This implies $S_1 < 40$.
  • $X$'s score is between $D$'s score and 45, so $40 < X < 45$. Since scores are whole numbers, $X$ can be 41, 42, or 43.
  • The score 45 was achieved, but it was not the highest score. So, $45 \ne S_5$.
  • $C$'s score is an even number.
  • $C$ did not score the least, so $C \ne S_1$.
  • $Y$'s score is 6 less than $Z$'s score, so $Y = Z - 6$.

Score Structure Analysis

From $S_2 = 40$ and $45 \ne S_5$, we deduce the possible positions for the score 45:

  • Case 1: 45 is $S_3$. The order would be $S_1 < 40 < 45 < S_4 < S_5$. However, $X$ must be between 40 and 45 ($40 < X < 45$). Since $X$ must be one of $S_3$ or $S_4$, and $S_3 = 45$, $X$ cannot be $S_3$. If $X = S_4$, then $40 < S_4 < 45$. This contradicts $S_3 < S_4$ because $45 < S_4$ is required. Thus, 45 cannot be $S_3$.
  • Case 2: 45 is $S_4$. The order must be $S_1 < 40 < S_3 < 45 < S_5$. Since $40 < X < 45$, and $X$ must be $S_3$ or $S_4$, $X$ cannot be $S_4=45$. Therefore, $X$ must be $S_3$. So $S_3 = X$, and $X \in \{41, 42, 43\}$.

The score structure is $S_1 < 40 < X < 45 < S_5$, where $X \in \{41, 42, 43\}$.

Evaluating Options for C's Score

We know $C$ is even and $C \ne S_1$. $C$ must be one of the scores $S_3, S_4,$ or $S_5$. Since $S_4 = 45$ (odd), $C$ cannot be $S_4$. Thus, $C$ must be $S_3$ or $S_5$. Also, $C$ and $X$ are different people, so they cannot have the same score.

  • Option 1: C = 38. If $C=38$, since $C \ne S_1$, then $S_1 < 38$. This would mean $S_1 < C < D=40$. This contradicts $D=40$ being the second lowest score ($S_2$). So, $C \ne 38$.
  • Option 2: C = 44. If $C=44$, $C$ must be $S_3$ or $S_5$.
    • If $C=S_3=44$. Then $X=S_3$, so $X=44$. This contradicts $40 < X < 45$. Also, $C$ and $X$ are different people, they cannot both be $S_3$.
    • If $C=S_5=44$. This contradicts $S_5 > 45$.
    So, $C \ne 44$.
  • Option 4: C = 42. If $C=42$, $C$ must be $S_3$ or $S_5$.
    • If $C=S_3=42$. Then $X=S_3$, so $X=42$. Since $C$ and $X$ are different people, this is not possible.
    • If $C=S_5=42$. This contradicts $S_5 > 45$.
    So, $C \ne 42$.
  • Option 3: C = 46. If $C=46$, $C$ must be $S_3$ or $S_5$.
    • If $C=S_3=46$. This contradicts $S_3 < 45$ (as $S_3=X$ and $X<45$).
    • If $C=S_5=46$. The structure is $S_1 < 40 < X < 45 < C=46$. $X \in \{41, 42, 43\}$. $C=46$ is even and $C \ne S_1$. This fits. Let's check the $Y=Z-6$ condition.
      • If $X=41$, possible scores are $\{S_1, 40, 41, 45, 46\}$. Let $S_1=35$. Scores: $\{35, 40, 41, 45, 46\}$. People: Y(35), D(40), X(41), Z(45), C(46). Check $Y=Z-6$: $35 = 45 - 6$. This is true. All conditions are met.
      • If $X=43$, possible scores are $\{S_1, 40, 43, 45, 46\}$. Let $S_1=39$. Scores: $\{39, 40, 43, 45, 46\}$. People: Y(39), D(40), X(43), Z(45), C(46). Check $Y=Z-6$: $39 = 45 - 6$. This is true. All conditions are met.
      Since we found valid scenarios where $C=46$, this is a possible score.

Conclusion

Based on the analysis, the only possible score for C among the given options is 46.

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