Lesson: Finding the Average Speed

You're referring to the first question we examine in the above video.

Your calculations are correct but, to avoid confusion among others reading this, I want to point out that you aren't really finding the averages of the two speeds in your calculations.

Instead, speed1 represents the DISTANCE traveled in the first hour, and speed2 represents the DISTANCE traveled in the second hour, etc.

So, what you really have is: average speed = (total DISTANCE)/(total TIME), which is correct. However, the terms speed1, speed2, etc are misleading, since those terms really represent distances.

You're referring to the question that starts at 3:35

For questions of this nature (where the two legs of the journey are the same distance and you're told the average speed of each leg), you can choose ANY value for the distance, and you'll get the same answer each time.

You might want to try it with this question. See what happens if you set the distance to 80 miles or 40 miles. You'll get the same overall average speed each time.

The easiest approach is to assign a nice value to the distance from A to B, one that works well with all 3 speeds.

720 miles works perfectly.

Average speed = (TOTAL travel time)/(TOTAL distance)

"Andrew drove from A to B at 60 miles per hour": time = 720/60 = 12 hours

"Then he realized that he forgot something at A, and drove back at 80 miles per hour": time = 720/80 = 9 hours

"He then zipped back to B at 90 mph": time = 720/90 = 8 hours

So, average speed = (720 miles + 720 miles + 720 miles)/(12 hrs + 9 hrs + 8 hrs)
= (2160 miles)/(29 hours)
= some awful number mph

The speed of the FAST car is 8 mph MORE THAN the speed of the slow car.

Let x = the speed (in miles per hour) of the SLOW car
So, x+8 = the speed (in miles per hour) of the FAST car

Each car travels for 2 hours.
Distance = (time)(speed)

After 2 hours, the SLOW car's distance = (2)(x) = 2x
After 2 hours, the FAST car's distance = (2)(x+8) = 2x + 16

We're told that, after 2 hours, the cars are 208 miles apart.

So, we can write: 2x + (2x + 16) = 208
Simplify: 4x + 16 = 208
So: 4x = 192
Solve: x = 48

So, the SLOW car's speed = 48 mph
And the FAST car's speed = 56 mph

Does that help?

Cheers,
Brent

Let S = speed (in miles per hour) of slow car
Let F = speed (in miles per hour) of fast car

GIVEN: One car traveled, on average, 8 miles per hour faster then the other car
So, the speed of the fast car is 8 mph greater than the speed of the slow car
We can write: F = S + 8

GIVEN: Two cars started from the same point and traveled on a straight course in opposite directions for 2 hours, at which they were 208 miles apart
So, after ONE hour, the cars were 104 miles apart
After ONE hour, the fast car traveled F miles
After ONE hour, the slow car traveled S miles

So, we can write: F + S = 104

We now have 2 equations:
F = S + 8
F + S = 104

Several options from here.

One approach is to rewrite the top equation as follows:
F - S = 8
F + S = 104

ADD the two equations to get: 2F = 112
Solve: F = 56

Now that we know F = 56, we can solve for S to get S = 48

So, the speed of the slow car was 48 miles per hour, the speed of the fast car was 56 miles per hour

Question link: https://gre.myprepclub.com/forum/runner-a-ran-4-5-kilometer-and-runner-b...

In MOST cases, the GRE will give you the conversion (e.g., 5280 feet = 1 mile)
However, for this question, it appears that you're expected to know what the prefix KILO refers to.
That is, 1000 meters = 1 kilometer.

Hi Ravin,
That equation is exactly how I'd approach it.
Nice work.