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Comment on Finding the Average Speed
Avg Speed = ( speed1 + speed2
= ( 40 + 40 + 60 + 60 + 60)/5 = 260/5 = 52 Miles/Hour
You're referring to the first
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.
Brent,
I solved each trip separately and the results of each trip I added and divided by 2 in order to get the average but I got 50.
R=80/2 and R=180/3. (40+60 )/2 = 50
Hi Sthepany,
Hi Sthepany,
The first part of the video explains why that approach won't work.
To find average speed, we must use the following formula:
Average speed = (total distance traveled)/(total travel time)
Cheers,
Brent
Hi! I had the equations 20t
You're referring to the
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.
Andrew drove from A to B at
The easiest approach is to
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
Hi, I have difficulty to
Two cars started from the same point and traveled on a straight course in opposite directions for 2 hours, at which time they were 208 miles apart. If one car traveled, on average, 8 miles per hour faster than the other car, what was the average speed of each car for the 2-hour trip?
Here says they traveled opposite direction for 2 hours, so the total time spent 4 hours, correct? I always don't understand something faster/greater/more/order than other should add the difference on which subject. Could you explain? Many thanks.
The speed of the FAST car is
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
That does help, now I
Hi Mr hanneson.
A question for you.
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. if one car traveled, on average, 8 miles per hour faster then the other car, what was the average speed of each car for the 2-hour trip?
Let S = speed (in miles per
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
https://gre.myprepclub.com/forum
This question might be unrelated to this topic but what types of conversions should I know for the GRE. I wasn't able to answer this question right even though its an easy one.
Thanks
Question link: https:/
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.
An airplane has a 3600 mile
for this question i used the equation Avg speed = Total distance/ Total time. and set it up as the following. 450 = (3600)/(4.5 +1800/x)) and solve for x. Is that the correct approach. at this point with the equation, I could either solve or plug in values into x from the AC. Please let me know your thoughts. thanks.
Hi Ravin,
Hi Ravin,
That equation is exactly how I'd approach it.
Nice work.
Hi Brent, how did you get to
Good question.
Good question.
In calculating the total travel time, Ravin first calculated the time it took the airplane to travel the first 1800 miles (at 400 miles per hour).
This time = distance/rate = 1800/400 = 4.5 hours