The table below provides the percentage of marks obtained by students, A, B, C, D and E in each of the six semesters. Study the table and answer the following question.

Students	1st Sem	2nd Sem	3rd Sem	4th Sem	5th Sem	6th Sem
A	74	79	73	78	72	86
B	55	51	68	53	72	69
C	40	43	50	52	60	66
D	59	59	58	57	59	57
E	66	76	71	81	89	92

Which semester shows the highest improvement in percentage of marks considering average of marks obtained by all the students for the semester?

Correct option is D

Given:

The table provides the percentage of marks obtained by students A, B, C, D, and E across six semesters.

Concept Used:

​$$\text{Percentage Improvement} = \left( \frac{\text{New Average} - \text{Old Average}}{\text{Old Average}} \right) \times 100$$​​

Solution:

​$$\text{1st Sem} = \frac{74 + 55 + 40 + 59 + 66}{5} = \frac{294}{5} = 58.8$$​​

​$$\text{2nd Sem} = \frac{79 + 51 + 43 + 59 + 76}{5} = \frac{308}{5} = 61.6$$​​

​$$\text{3rd Sem} = \frac{73 + 68 + 50 + 58 + 71}{5} = \frac{320}{5} = 64$$​​

​$$\text{4th Sem} = \frac{78 + 53 + 52 + 57 + 81}{5} = \frac{321}{5} = 64.2$$​​

​$$\text{5th Sem} = \frac{72 + 72 + 60 + 59 + 89}{5} = \frac{352}{5} = 70.4$$​​

​$$\text{6th Sem} = \frac{86 + 69 + 66 + 57 + 92}{5} = \frac{370}{5} = 74$$​​

​$$\text{2nd Sem from 1st Sem} = \frac{61.6 - 58.8}{58.8} \times 100 = \frac{2.8}{58.8} \times 100 \approx 4.76\%$$​​

​$$\text{3rd Sem from 2nd Sem} = \frac{64 - 61.6}{61.6} \times 100 = \frac{2.4}{61.6} \times 100 \approx 3.9\%$$​​

​$$\text{4th Sem from 3rd Sem} = \frac{64.2 - 64}{64} \times 100 = \frac{0.2}{64} \times 100 \approx 0.31\%$$​​

​$$\text{5th Sem from 4th Sem} = \frac{70.4 - 64.2}{64.2} \times 100 = \frac{6.2}{64.2} \times 100 \approx 9.66\%$$​​

​$$\text{6th Sem from 5th Sem} = \frac{74 - 70.4}{70.4} \times 100 = \frac{3.6}{70.4} \times 100 \approx 5.11\%$$​​

The 5th Semester shows the highest percentage improvement of 9.66%.

Option (d) is right.

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