Solve the problem.
-Two cars leave a city and head in the same direction. After 7 hours, the faster car is 28 miles ahead of the slower car. The slower car has traveled 343 miles. Find the speeds of the two cars.
A)45 mph and 49 mph
B)49 mph and 53 mph
C)40 mph and 44 mph
D)51 mph and 55 mph

Question

Quizplus · Accepted Answer

Let the speed of the slower car be x mph.
Let the speed of the faster car be y mph.

We know that the slower car travels 343 miles in 7 hours. Therefore:

x = 343/7
x = 49 mph

We also know that after 7 hours, the faster car is 28 miles ahead of the slower car. This means that in 7 hours, the faster car has traveled 343 + 28 = 371 miles. Therefore:

y = 371/7
y = 53 mph

Therefore, the speeds of the two cars are 49 mph and 53 mph. So, the best choice is B. Let the speed of the slower car be x mph
Then, the speed of the faster car = (x+28)/7 mph (since in 7 hours, the faster car is 28 miles ahead of the slower car)
Distance travelled by the slower car = 343 miles
Using the formula distance = speed × time, we get:
343 = x × 7
x = 49 mph
Therefore, the speed of the faster car = (49+28)/7 = 77/7 = 11 mph
Hence, the speeds of the two cars are 49 mph and 11 mph. Therefore, option B) is the correct answer.

Solve the Problem. -Two Cars Leave a City and Head in the Same