Given:
The average score of students on section A and B = 49
Total number of students in A and B = 90
Number of students in A = 25% more than that of B
The average score of students in B = 20% higher than that of A
Concept used:
Average = sum of the elements/number of elements
Calculation:
Let the number of students in section B be x.
Let the average score of students in A be y.
Nubmer of students in section A = 125% of B = (5/4) × x
We know, x + (5/4) x = 90
Or, x = 90 × 4/9 = 40
Then, the number of students in section A = 50
Total score of students in section A = 50 × y
Total score of students in section B = 40 × 120% × y
Total score of students in section B = 40 × (6y/5) = 48y
Total score of all students of A and B = 90 × 49 = 4410
50y + 48y = 98y = 4410
Or, y = 4410/98 = 45
Thus, the average score of students in class A id 45.
Let the number of students in section B be 4
Number of students in section A = 4 × (5/4) = 5
Total students = 5 + 4 = 9
Average score in mathematics of 9 students = 49
Let the average score of section A be 5x
Average score of section B = 5x × (6/5) = 6x
According to the question
⇒ 5x × 5 + 6x × 4 = 9 × 49
⇒ 25x + 24x = 9 × 49
⇒ 49x = 9 × 49
⇒ x = 9
Average score of section A = 5 × 9 = 45
∴ The average score of students in class A is 45.