Employee Ratings Definitely False Analysis

XAT 2025 · Quantitative Ability & Data Interpretation · Hard · Statistics

Eight employees of an organization have been rated on a scale of 1 to 50 for their performance. All ratings are integers. The overall average rating of the eight employees is 30. While the five employees with the highest ratings average 38, the five employees with the lowest ratings average 25.

Which of the following, about the ratings obtained by the eight employees, is DEFINITELY FALSE?

  1. A.

    The second highest rating obtained is 38.

  2. B.

    The median of the eight ratings is 37.5.

  3. C.

    The lowest rating obtained is 1.

  4. D.

    The highest rating obtained is 40.

  5. E.

    The third lowest rating obtained is 37.

Answer

D

Explanation

Let ratings in increasing order be a1≤a2≤a3≤a4≤a5≤a6≤a7≤a8a_1 \le a_2 \le a_3 \le a_4 \le a_5 \le a_6 \le a_7 \le a_8. Sum of all 8 = 8×30=2408 \times 30 = 240. Sum of top 5 (a4+a5+a6+a7+a8a_4 + a_5 + a_6 + a_7 + a_8) = 5×38=1905 \times 38 = 190. Sum of bottom 5 (a1+a2+a3+a4+a5a_1 + a_2 + a_3 + a_4 + a_5) = 5×25=1255 \times 25 = 125.

From overall sum: a1+a2+a3=240−190=50a_1 + a_2 + a_3 = 240 - 190 = 50. a6+a7+a8=240−125=115a_6 + a_7 + a_8 = 240 - 125 = 115. a4+a5=190−115=75a_4 + a_5 = 190 - 115 = 75.

If highest rating a8=40a_8 = 40: Then a6+a7=115−40=75a_6 + a_7 = 115 - 40 = 75. Since a5≥a4a_5 \ge a_4 and a4+a5=75a_4 + a_5 = 75, max possible a4≤37a_4 \le 37. Since a6≤a7a_6 \le a_7, and a6+a7=75a_6 + a_7 = 75, min possible a6≥38a_6 \ge 38. This requires a5≤a6a_5 \le a_6, which gives a5=37.5a_5 = 37.5, impossible since ratings are integers. Hence, a8a_8 cannot be 40. Option D is DEFINITELY FALSE.

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