Dave walked to his friend's house at a rate of 4 mph returned back biking at a rate of 10 mph. If it took him 18 minutes longer to walk than to bike, then what was the total distance of the round trip?

Answer 1

Let the distance to his friend's house be d.
The time it took Dave to walk to his friend's house is d/4 hours.
The time it took Dave to bike back is d/10 hours.
Since it took him 18 minutes longer to walk than to bike, then d/4 = d/10 + 18/60, or 3d/5 - d/2 = 18/60.
Multiplying through the equation by 60 gives 36d - 30d = 18, or 6d = 18.
Hence, the total distance of the round trip is 2d = 2 × (18/6) = <<2*18/6=6>>6 miles. Answer: \boxed{6}.

Answer 2

Try again

Answer 3

Let the total distance of the round trip be d miles.
The time it took Dave to walk to his friend's house is d/4 hours.
The time it took Dave to bike back is d/10 hours.
Since it took him 18 minutes longer to walk than to bike, we can write the equation d/4 = d/10 + 18/60.
Multiplying through the equation by 20 gives 5d = 2d + 6.
Combining like terms gives 3d = 6.
Dividing both sides by 3 gives d = 2.
Therefore, the total distance of the round trip is 2 miles. Answer: \boxed{2}.

Answer 4

No!

Answer 5

I apologize for the mistake. Let's try again.

Let the distance to his friend's house be d.
The time it took Dave to walk to his friend's house is d/4 hours.
The time it took Dave to bike back is d/10 hours.
Since it took him 18 minutes longer to walk than to bike, we can write the equation d/4 = d/10 + 18/60.
Multiplying through the equation by 20 gives 5d = 2d + 6.
Combining like terms gives 3d = 6.
Dividing both sides by 3 gives d = 2.
Therefore, the distance to his friend's house is 2 miles.
The round trip distance is 2 * 2 = 4 miles. Answer: \boxed{4}.