C
1) Since Kenji has $600 to spend and the classes cost 20x and 15y we can write out the following inequality:
So now, let's test which point is one possible solution by plugging them and verifying whether it is true or false.
![\begin{gathered} (25,20)\rightarrow20(25)+15(20)\leq600\rightarrow800\leq600\:F \\ (10,35)\operatorname{\rightarrow}20(10)+15(35)\leqslant600\operatorname{\rightarrow}725\leqslant600F \\ (10,20)\rightarrow20(10)+15(20)\leq600\Rightarrow500\leq600True! \\ (30,20)\operatorname{\rightarrow}*20(30)+15(20)\leqslant600\operatorname{\Rightarrow}900\leqslant600F \end{gathered}]()
Algebraically testing we can tell that the ordered pair which is a possible solution is
C) (10,20)