Notand cu A, B, C, mediile claselor avem relatiile
A=115B/100
A=98C/100
atunci
a)B=100A/115=20A/23=20*8,25/23≈7,17 (aproximare la 2 zecimale si asa vor fi si urmatoarele)
b) A=0,98C=0,98*8,75=8,575≈8,58
c) A=115B/100=98C/100
atunci
23B/20=49C/50⇒C=50*23B/49*20=5*23*8,2/49*2≈9,62