Casey went on a road trip that covered 100 miles, stopping for a lunch break along the way. The trip took 3 hours in total and her average speed while driving was 40 miles per hour (mph). In minutes, how long was the lunch break?
(A) 15 \text{(A)}\;15 (A) 15 (B) 30 \text{(B)}\;30 (B) 30 (C) 40 \text{(C)}\;40 (C) 40 (D) 45 \text{(D)}\;45 (D) 45 (E) 60 \text{(E)}\;60 (E) 60 Hints (3) Hint 1 v = d t v=\frac{d}{t} v = t d t A B t_{AB} t A B is the travel time from A A A to B B B , excluding the lunch break. ⇒ t A B = d A B v A B = 100 40 = 5 2 hr = 5 2 × 60 min = 150 min \Rightarrow t_{AB}=\frac{d_{AB}}{v_{AB}}=\frac{100}{40}=\frac52\text{ hr}=\frac52\times60\text{ min}=150\text{ min} ⇒ t A B = v A B d A B = 40 100 = 2 5 hr = 2 5 × 60 min = 150 min Hint 2 t t o t a l = 3 hr = 3 × 60 min = 180 min t_{total}=3\text{ hr}=3\times60\text{ min}=180\text{ min} t t o t a l = 3 hr = 3 × 60 min = 180 min Hint 3 By Hints 1 and 2, Ans = t t o t a l − t A B = 180 − 150 = 30 min \text{Ans}=t_{total}-t_{AB}=180-150=30\text{ min} Ans = t t o t a l − t A B = 180 − 150 = 30 min 