Ricky drove from town A to town B in 3 hours because he drove 15 miles per hour slower on the return trip. How fast did Ricky drive on the trip from town A to town B?
I know the answer is 37.5mph but can you please explain it and break it down so that I am able to work similar problems like this one. Thank you
outbound at speed v
back at speed (v-15)
say distance d
a to b time = 3 = d/v
b to a time = ? = d/(v-15)
You did not give me all the information
Sorry I didn't type the question correctly.
Ricky drove from town A to town B in 3 hours. His return trip from town B to town A took 5 hours because he drove 15 miles per hour slower on the return trip. How fast did Ricky drive on the trip from town A to town B?
Just put a 5 where I had a ? mark :)
say distance d
a to b time = 3 = d/v
b to a time = 5 = d/(v-15)
so
d = 3 v
and
d = 5(v-15)
3 v = 5 v - 75
2 v = 75
v = 37.5
To find out how fast Ricky drove on the trip from town A to town B, we can set up a simple equation using the information given.
Let's assume that Ricky's speed from town A to town B is represented by "s" miles per hour. Since Ricky drove 15 miles per hour slower on the return trip, his speed on the return trip can be represented as "s-15" miles per hour.
Given that Ricky drove from town A to town B in 3 hours, we know that the distance between the two towns is equal to the speed multiplied by the time: Distance = Speed * Time.
So, for the trip from town A to town B:
Distance = s * 3.
On the return trip from town B to town A:
Distance = (s - 15) * 3.
Since the distance traveled in both trips is the same (since Ricky went from town A to town B and then back to town A), we can set up an equation:
s * 3 = (s - 15) * 3.
Simplifying the equation, we get:
3s = 3s - 45.
If we subtract 3s from both sides of the equation, we get:
0 = -45.
This equation has no solution, which means there is an error in the information provided. In this case, it seems like the given answer of 37.5 mph is incorrect.
However, if we adjust the information slightly and assume that Ricky's speed on the return trip is 15 miles per hour slower, we can set up a new equation:
s * 3 = (s - 15) * 3.
Simplifying the equation, we get:
3s = 3s - 45.
If we subtract 3s from both sides of the equation, we get:
0 = -45.
This equation is also incorrect because it yields no solution. So, unfortunately, we cannot determine Ricky's speed from the information given.