sandy wrote:
Jeff and Ali race each other at the Tentleytown Speedway. Ali’s car travels at 300 feet per second, and Je?’s car travels at 250 feet per second. If one lap around the track is 3000 feet long, and each car travels at a constant rate, how many laps will it take Ali to overtake Jeff?
A. 1
B. 5
C. 6
D. 10
E. 60
Drill 4
Question: 2
Page: 531
Since the question asks for a certain number of LAPS, let's first rewrite the given information in terms of laps.
300/3000 = 1/10, which means 300 ft = 1/10 laps. So, Ali’s speed of 300 feet per second is equal to a speed of
1/10 laps per second
250/3000 = 1/12, which means 250 ft = 1/12 laps. So, Jeff’s speed of 250 feet per second is equal to a speed of
1/12 laps per second
Since each person travels for the same amount of time, we can start with the following word equation:
Ali's travel time (in seconds) = Jeff's travel time (in seconds)When Ali laps Jeff, Ali has completed 1 lap more than Jeff has.
Let
x = the number of laps Ali completes (i.e., Ali's travel distance)
So,
x - 1 = the number of laps Jeff completes (i.e., Jeff's travel distance)
time= distance/ratePlug our values into our word equation to get:
(x)/(1/10) = (x - 1)/(1/12)Simplify:
(10)(x) = (12/1)(x - 1)Simplify:
10x = 12x - 12Solve:
x = 6Answer: C