The Correct Option is D
Solution and Explanation
Approach: Rather than testing each option, find the smallest palindromic starting odometer the trip allows — the smaller the start, the longer the distance, hence the faster the average speed.
Step 1: Let the average speed be $s$ km/h. The reading just before the trip is $26862-8s$, and this must itself be a palindrome.
Step 2: Since $s\le 110$, the trip is at most $8\times110=880$ km, so the start lies in $[26862-880,\ 26862]=[25982,\ 26862]$.
Step 3: A 5-digit palindrome in this range has the form $\overline{2\,b\,c\,b\,2}$; to be at least $25982$ we need $b=6$, giving $26062,26162,\dots,26862$.
Step 4: Greatest speed needs the greatest distance, i.e. the smallest valid start $=26062$. Then $8s=26862-26062=800\Rightarrow s=100$.
Answer: The greatest possible average speed is $100$ km/h.