Answer : Option A

Explanation :

Total number of students passed in annual exams in a section

= [ (No. of students failed in half-yearly but passed in annual exams + (No. of students passed in both exams)

   ] in that section

 Success rate in annual exams in Section A

= [ \( \frac { no of students of section A passed in annual exams } { Total number of students i section A } \) x 100 ]%

= [ \(\frac { (14+64) } { (28+14+6+64) } \) x 100]%

 = [\(\frac { 78 } {11 2 } \) x 100]%

  = 69.64%

Similarly, success rate in annual exams in:

section B [ \(\frac { (12+55)} { (23+12+17+55) } \) x 100]% = [ \(\frac { 67 } { 107}\) x 100] % =62.62%

section C [ \(\frac { (8+46) } { (17+8+9+46) } \) x 100]% = [ \(\frac { 54} { 80}\)  x 100]% = 67.5%

section D [ \(\frac { (13+76) } { (27+13+15+76) } \) X 100]% = [ \(\frac { 89} { 131}\) X 100]% = 67 .94%

Clearly, the success rate in annual examination is maximum for Section A.