ASSUMPTION:
These two groups contain different students, that is, there are no students that belong to both groups and attend both classes, so the total number of students is 120+150=270, and our task is to evaluate the mean attendance of any of two classes by a combined set of students of both groups.
Let x_i represent attendance of the first class (XI-A) by an i^(th) person of the first group of N_1=120 students. It is equal to 1 if this person attended the first class and 0 otherwise.
Since the mean attendance in this group equals to M_1=56.35, we can write the equation:
M_1 = (Sigma x_i)/N_1 = (x_1+x_2+...+x_120)/120 = 56.35
or
Sigma x_i = x_1+x_2+...+x_120 = N_1*M_1 = 120 * 56.35 = 6762
Let y_j represent attendance of the second class (XI-B) by an j^(th) person of the second group of N_2=150 students. It is equal to 1 if this person attended the second class and 0 otherwise.
Since the mean attendance in this group equals to M_2=63.45, we can write the equation:
M_2 = (Sigma y_j)/N_2 = (y_1+y_2+...+y_150)/150 = 63.45
or
Sigma y_j = y_1+x_2+...+y_150 = N_2*M_2 = 150 * 63.45 = 9517.5
Combining both groups, we have N_1+N_2=120+150=270 students, whose attendance of one of two classes is represented by variables x_i (where i in [1,120]) and y_j (where j in [1,150]).
Their mean attendance of any of two classes can be represented as
M = (Sigma x_i + Sigma y_j)/(N_1+N_2) =
= (N_1*M_1+N_2*M_2)/(N_1+N_2) =
= (6762+9517.5)/270 ~~ 60.34
