2014 AMC 10B Problem 8

The problem mixes feet with yards and seconds with minutes, so first get everything into the units the question wants.

1. Time in seconds. $3$ minutes is $3 \times 60 = 180$ seconds.
2. How many "$t$-second" blocks? The truck goes $\frac{b}{6}$ feet in every block of $t$ seconds. In $180$ seconds there are $\frac{180}{t}$ such blocks.
3. Distance in feet. Multiply:
$$\frac{b}{6} \times \frac{180}{t} = \frac{180b}{6t} = \frac{30b}{t} \text{ feet}.$$
4. Change to yards. Every $3$ feet make one yard, so divide by $3$:
$$\frac{30b}{t} \div 3 = \frac{10b}{t} \text{ yards}.$$

The answer is $\boxed{\textbf{(E)}\ \frac{10b}{t}}$.

Distance equals rate times time, but only once the units match. Convert the time to match the rate, then convert the final distance to the unit asked for. A quick sense check also helps: a bigger $b$ should mean more yards and a bigger $t$ (a slower truck) fewer yards, so $b$ belongs on top and $t$ on the bottom. That already rules out (B) and (D).

Pick easy numbers. Let $b = 6$ and $t = 1$, so the truck goes $1$ foot every second. In $180$ seconds it goes $180$ feet, which is $60$ yards. Now test the choices with $b = 6$, $t = 1$: only $\frac{10b}{t} = 60$ matches.