Math Problem Help

[kagmypts]

Here is a math problem from a post test for AoPS Prealgebra (http://www.artofproblemsolving.com/Store/products/prealgebra/posttest.pdf):
 

"A northbound train from Miami to Jacksonville made the 324-mile journey at an average

speed of 50 miles per hour. On its southbound return trip, it made the journey at an average

speed of 40 miles per hour. To the nearest tenth of a mile per hour, what was the train’s

average speed for the 648-mile roundtrip journey?"

 

The answer given is 44.4 miles per hour.

 

Why is the average speed not 45 mph?  Since the journey is the same distance in both directions, why wouldn't a weighted average computation work?

 

X = .5(50) + .5(40)

x = 25 + 20

x = 45 mph

 

What am I missing?

 

ETA - The computation should be weighted by time and not distance.

 

[scbusf]

For Average Speed problems, you need to divide the Total Distance by the Total Time.

 

So first, use d = r x t to figure out how long each leg of the trip took.  Add those together, and then divide 648 by your total time.

 

For the WHY ......... it's because the train spends more time going 40 mph, so you have to take the time into account.

 

 

[Arcadia]

Because the time taken for northbound and southbound are different.

 

Southbound took 8.1 hrs.

Northbound took 6.48 hrs

 

Total time is 14.58 hrs

Total distance is 648 miles

Average speed is 44.4 mph

It might be easier to see if you think about a similar problem, but have the train go 324 mph north, then 1 mph south.

[kagmypts]

Thank you for the replies.  While the computation makes sense, I was wondering conceptually why the math didn't work.  As many of have pointed out, it's because the speed should be weighted by time and not distance.  

 

Since the train spent 44% of the trip going 50 mph and 56% of the trip going 40 mph, the weighted average computation should look like this:

 

X = .44(50) + .56(40)

x = 22.2 + 22.2

x = 44.4 mph