In the first case, if d is the number of days, and B and C are the amounts Benedict and Calvin each have, then after d days, Benedict has B-36d and Calvin has C-12d. But Benedict has no money left after d days, so B-36d=0, therefore d=B/36. But Calvin still has $260, so C-12d=260, and we can substitute for d:
C-12(B/36)=260, C-B/3=260, so C=260+B/3.
In the second case, after time t days, B-12t=0, so t=B/12. And C-36t=20. Substitute for t:
C-36(B/12)=20, C-3B=20, so C=20+3B, because Calvin has $20 left when Benedict spends all his money.
Therefore C=260+B/3=20+3B. Therefore,
260+B/3=20+3B, 240=8B/3, B=(3/8)240=$90.
C=20+3B=20+270=$290.
So Calvin has $290 and Benedict has $90.
Note, however, that d and t are not whole numbers of days. In the first case, d=2½ days and in the second case, t=7½ days. It’s assumed that in half a day they each spend half their daily amount. Benedict would take 2½ days to spend all his $90 in the first case and 7½ days in the second case: 2½×36=7½×12=90.