3T+11M=100
M | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
pts for Multiple choice | 11 | 22 | 33 | 44 | 55 | 66 | 77 | 88 | 99 |
Points left | 89 | 78 | 67 | 56 | 45 | 34 | 23 | 12 | 1 |
Is left over divisable by 3 | N | Y | N | N | Y | N | N | Y | N |
How many T/F | 26 | 15 | 4 |
So if could be
2 Multiple choice and 26 T/F check 2*11+26*3 = 100
5 Multiple choice and 15 T/F check 5*11+15*3 = 100
8 Multiple choice and 4 T/F check 8*11+4*3 = 100
Incidentally this is a diophantine equation because all solutions must be integers.
There is a more formal way of solving so that all solutions are found. (including negative solutions)
Of course you problems does not have any negative solutions :))