Employee Ratings False Statement Identification

Let the 8 ratings in ascending order be $a, b, c, d, e, f, g, h$. Sum of all 8 ratings = $8 \times 30 = 240$. Sum of top 5 ratings ($d+e+f+g+h$) = $5 \times 38 = 190$. Sum of bottom 5 ratings ($a+b+c+d+e$) = $5 \times 25 = 125$. Sum of top 3 ratings ($f+g+h$) = $240 - 125 = 115$. Sum of bottom 3 ratings ($a+b+c$) = $240 - 190 = 50$. Sum of 4th and 5th ratings ($d+e$) = $190 - 115 = 75$. If highest rating $h = 40$, then $f + g = 115 - 40 = 75$. Since $e \le f \le g \le h$, if $f+g = 75$, then $g \ge 38$. Also $d+e = 75 \implies e \le 37.5 \implies e \le 37$. So $f \ge 38$ and $e \le 37$, which is consistent, but if $h=40$, $g \le 40$. Wait, if $h=40$ and $f+g=75$, since $g \le h=40$, $f \ge 35$. But $e \ge f$, so $d+e=75$ with $e \ge 35 \implies d \le 40$. However, $c \le d$. Let's test if $h=40$ is possible: If $h=40, g=38, f=37 \implies f+g+h = 115$. Then $e=38, d=37 \implies d+e=75$. Then $e=38 > f=37$, which violates $e \le f$! Thus, if $h=40$, $f+g=75 \implies f \le 37.5$. But $e \le f$, so $e \le 37.5$. Since $d+e=75$, $d = 75-e \ge 37.5 \implies d \ge 38$. But $d \le e$, contradiction ($38 \le d \le e \le 37$)! So $h=40$ is impossible.