A school has four sections A, B, C, D of Class IX students. Look at the following result table:



Which section exhibits the highest pass percentage in either of the two examinations?

To find the highest pass percentage in either of the two examinations, let's determine the total number of students in each section first:
$$\text{Total Students}=\text{Failed Both}+\text{Failed Half-Yearly, Passed Annual}+\text{Passed Half-Yearly, Failed Annual}+\text{Passed Both}$$

* Section A: $28+14+6+64=112$
* Section B: $23+12+17+55=107$
* Section C: $17+8+9+46=80$
* Section D: $27+13+15+76=131$

Now, let's compute the overall pass percentage (students who passed at least one of the exams, i.e., total students minus those who failed in both exams):

$$\text{Pass Percentage}=\frac{\text{Total Students}-\text{Failed in both}}{\text{Total Students}}\times 100$$

* Section A: $\frac{112-28}{112}\times 100=\frac{84}{112}\times 100=75.00\%$
* Section B: $\frac{107-23}{107}\times 100=\frac{84}{107}\times 100\approx 78.50\%$
* Section C: $\frac{80-17}{80}\times 100=\frac{63}{80}\times 100=78.75\%$
* Section D: $\frac{131-27}{131}\times 100=\frac{104}{131}\times 100\approx 79.39\%$

Comparing the pass percentages, Section D has the highest pass percentage ($79.39\%$).