Adalberto has $\$2.25$ in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?

Let $d$ be the number of dimes. Then $d + 9$ is the number of nickels. Total value: $0.10 d + 0.05 (d + 9) = 2.25$. Distribute: $0.10 d + 0.05 d + 0.45 = 2.25$. Combine: $0.15 d + 0.45 = 2.25$. Subtract $0.45$: $0.15 d = 1.80$. Divide: $d = 12$. So $d = 12$ dimes and $d + 9 = 21$ nickels. Check: $12 (0.10) + 21 (0.05) = 1.20 + 1.05 = \$2.25$ ✓. Trap D swaps the two counts. Trap A and C give totals that miss $\$2.25$.