BTGmoderatorDC wrote: ↑

Sat Oct 09, 2021 6:50 pm

Two coal carts, A and B, started simultaneously from opposite ends of a 400-yard track. Cart A traveled at a constant rate of 40 feet per second; Cart B traveled at a constant rate of 56 feet per second. After how many seconds of travel did the two carts collide? (1 yard = 3 feet)

(A) 75
(B) 48
(C) 70/3
(D) 25/2
(E) 25/6


OA D

Source: Manhattan Prep

Let's start with the word equation that features both travelers (i.e., Cart A and cart B).

Since the track is 400 yards long, we can write: (distance travelled by cart A) + (distance travelled by cart B) = 400 yards

Since the speeds are given in FEET per second, we should you convert 400 yards to 1200 feet

So our word equation becomes: (distance travelled by cart A) + (distance travelled by cart B) = 1200 feet
Let t = the travel time (in seconds) of cart A.
So, t = the travel time (in seconds) of cart B

Distance = (rate)(time)

Plug the relevant values into the word equation to get: 40t + 56t = 1200 feet
Simplify: 96t = 1200
Solve: t = 1200/96 = 600/48 = 100/8 = 25/2 (seconds)

Answer: D