So, since averages are defined as:
[tex] \frac{\sum_{k=1}^{P} P_k}{P}=70 [/tex]
So, since P are the total number of elements and P_k is the P_kth student. This is saying if we sum over each student's score and divide it by the number of students, we should get P, which is true.
So, using that logic, the other class can be represented as:
[tex] \frac{\sum_{k=1}^{N} N_k}{N}=70 [/tex]
We can take both of these equations and multiply them by N:
[tex] \sum_{k=1}^{P} P_k=70P [/tex]
[tex] \sum_{k=1}^N N_k=92N [/tex]
So, if we want to find the average of this we should add both our equations then divide by P+N, which is the number of all the students.
[tex] \frac{\sum_{k=1}^{P} P_k+\sum_{k=1}^{N}N_k}{N+P}=\frac{70P+92N}{N+P} [/tex]
To make this simpler we can replace our LHS with 86, since that's the average of both classes combined.
[tex] 86=\frac{92N+70P}{N+P} \implies\\
86N+86P=92N+70P \implies \\
16P=6N \implies \\
\frac{16P}{N}=6 \implies \\
\frac{P}{N}=\frac{6}{16} [/tex]
Simplified we would have P/N=3/8.
