If currents entering a node are $2\,\text{A}$, $3\,\text{A}$, and $5\,\text{A}$, and two leaving currents are $4\,\text{A}$ and $1\,\text{A}$, what must the third leaving current be?

By KCL, total entering current equals total leaving current. Entering current is $2+3+5=10\,\text{A}$. Known leaving current is $4+1=5\,\text{A}$, so the remaining leaving current must be $10-5=5\,\text{A}$.