Student Marks Range Optimization

Let marks of 5 students in ascending order be $A < S_2 < S_3 < S_4 < S_5$. Total marks = $5 \times 38 = 190$. Exactly 3 students got $> 32$, so $S_3, S_4, S_5 > 32$, which means $S_3 \ge 33$. And $A, S_2 \le 32$.

To find highest possible value of $A$: To maximize $A$, we should minimize $S_5, S_4, S_3$ and make $S_2$ close to $A$. Min values for $S_3, S_4, S_5$ with distinct integers $> 32$ are 33, 34, 35. So $A + S_2 + 33 + 34 + 35 = 190 \implies A + S_2 = 88$. Since $S_2 \le 32$, $A = 88 - S_2 \ge 88 - 32 = 56$, which exceeds 32 (contradiction). So $S_2$ can be at most 32. But $A < S_2 \le 32$. Wait, to maximize $A$, we make $S_5 = 50, S_4 = 49, S_3 = 48$? No, to maximize $A$, we MINIMIZE $S_3, S_4, S_5$. Min $S_3, S_4, S_5$: if $S_3=33, S_4=34, S_5=35$, sum = 102. $A + S_2 = 88$. Since $A < S_2 \le 32$, $A + S_2 \le 31 + 32 = 63 < 88$. So $S_3, S_4, S_5$ must sum to MORE to leave a smaller sum for $A + S_2$. To get $A + S_2 \le 63$, sum of $S_3+S_4+S_5 \ge 190 - 63 = 127$. To maximize $A$, we set $S_2 = 32$, then $A + 32 = 190 - (S_3+S_4+S_5)$. To maximize $A$, we minimize $S_3+S_4+S_5$ subject to $A + S_2 \le 63$. $A + S_2 = 190 - (50 + 49 + 48) = 190 - 147 = 43$. With $S_2 = 32$, $A = 43 - 32 = 11$. Wait, if $S_5=50, S_4=49, S_3=33$, sum = 132 $\implies A+S_2 = 58$. If $S_2 = 32, A = 26$. Check if $A=26, S_2=32, S_3=33, S_4=49, S_5=50$: sum = $26+32+33+49+50 = 190$. All conditions met! So Max $A = 26$.

To minimize $A$: Maximize $S_2, S_3, S_4, S_5$. Max $S_5=50, S_4=49, S_3=48, S_2=32$. Sum = $50+49+48+32 = 179$. Then Min $A = 190 - 179 = 6$ or check $S_2 < 32$ giving even smaller $A$. Wait, $A \ge 0$, but if $A=6, S_2=32, S_3=48, S_4=49, S_5=50$, sum = 190. Max $A - \text{Min } A = 26 - 6 = 20$.