Press #1 can print the magazine in $6$ hours alone, and Press #2 can do the same job in $12$ hours alone. If both presses run at the same time on the same job (without interfering with each other), how long does the job take?

**Rate model.** In one hour, Press #1 completes $\tfrac{1}{6}$ of the job; Press #2 completes $\tfrac{1}{12}$. Together in $t$ hours they complete $1$ full job, so their combined rate is $\tfrac{1}{t}$ per hour: $\tfrac{1}{6} + \tfrac{1}{12} = \tfrac{1}{t}$. Find a common denominator ($12$): $\tfrac{2}{12} + \tfrac{1}{12} = \tfrac{3}{12} = \tfrac{1}{4}$. So $\tfrac{1}{t} = \tfrac{1}{4} \Rightarrow t = 4$ hours. - Trap C ($9 = 6 + 3$) or D ($18$) treats times as if they added — but you can't add hours-per-job that way. - Trap A ($3$) is below the faster machine's solo time, which **is** possible but doesn't match the equation. (Two machines working together can finish faster than either alone, but never faster than the time the equation gives.) Intuition check: the answer must be *less than the faster individual time* ($6$ hours) and *more than half of it* ($3$ hours). $4$ hours sits in the right band.