At 10 am two cars start traveling toward each other from 287 miles apart. They passed each other at 1:30 pm. If the rate of the fster car exceeded the rate of the slower car by 6 miles per hour, find the rat (in mph) of the faster car.
A. 38
B. 40
C. 44
D. 48
E. 50
I think it is C, am I correct?
t = 3.5 hours
distance = rate * time
287 = r*t+(r+6)t = (2r+6)t =(2r+6)(3.5)
so
2r+6 = 287/3.5
solve for r and then answer with r+6
To solve this problem, we can use the formula:
Distance = Rate * Time
Let's first calculate the time it took for the two cars to meet. The problem states that they started at 10 am and passed each other at 1:30 pm, which is a total of 3 hours and 30 minutes.
Next, we need to find the combined rate of the two cars. Since their total distance was 287 miles, and they traveled for 3.5 hours, we can substitute these values into the formula:
287 = (Rate of the slower car + Rate of the faster car) * 3.5
Now we can use the information that the rate of the faster car exceeded the rate of the slower car by 6 mph. Let's call the rate of the slower car "x" mph. Therefore, the rate of the faster car would be "x + 6" mph.
Substituting these values into the equation, we have:
287 = (x + x + 6) * 3.5
Simplifying further:
287 = (2x + 6) * 3.5
287 = 7x + 21
7x = 287 - 21
7x = 266
x = 266 / 7
x ≈ 38
So, the rate of the slower car is approximately 38 mph. Since the rate of the faster car is 6 mph higher, the rate of the faster car is approximately 38 + 6 = 44 mph.
Hence, the correct answer is C.