Rates & Work

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Rates & Work

by markmosby » Sun Dec 13, 2015 2:47 pm
If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend along the same route later that evening, what was the average speed for the entire trip?

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
(2) The route between Townsend and Smallville is 165 miles long.


*So I've reviewed this problem multiple times, but I want to make sure I'm learning everything I can from this single problem. I understand that in order to find the average speed for the entire trip, I need the total distance and total time for the trip; I also understand that the average speed for each leg of the trip will suffice. Is there anything else I should be gleaning from this problem?

Thank you so much!

-Mark
Source: — Data Sufficiency |

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by DavidG@VeritasPrep » Sun Dec 13, 2015 5:14 pm
markmosby wrote:If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend along the same route later that evening, what was the average speed for the entire trip?

(1) The trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
(2) The route between Townsend and Smallville is 165 miles long.


*So I've reviewed this problem multiple times, but I want to make sure I'm learning everything I can from this single problem. I understand that in order to find the average speed for the entire trip, I need the total distance and total time for the trip; I also understand that the average speed for each leg of the trip will suffice. Is there anything else I should be gleaning from this problem?

Thank you so much!

-Mark
One important thing to note: the relationship between rate and time is reciprocal.

In this case, we know that the time it takes to go from Townsend to Smallville is 50% greater than - or (3/2) of - the time it takes to go from Smallville - Townsend. Therefore, the rate Townsend - Smallville will be 2/3 of the rate from Smallville - Townsend. So 40 = 2/3 (Rate from T-S.) Rate from T-S = 60. If the distance is the same for each leg of the trip, this is clearly enough to find the average overall speed, so statement 1 alone is sufficient.

If you were doing this for Problem Solving you could pick a simple distance, such as 120 miles for each leg. The 40mph leg will take 3 hours. The 60 mph leg will take 2 hours. Total distance = 120 + 120 = 240. Total time: 3 + 2 = 5. Average overall speed: 240/5 = 48mph. (Note that 50mph would be a trap here.) This brings us to another important thing to note: Average Speed = Total Distance/Total Time.
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by MartyMurray » Sun Dec 13, 2015 9:25 pm
markmosby wrote:*So I've reviewed this problem multiple times, but I want to make sure I'm learning everything I can from this single problem. I understand that in order to find the average speed for the entire trip, I need the total distance and total time for the trip; I also understand that the average speed for each leg of the trip will suffice. Is there anything else I should be gleaning from this problem?

-Mark
I just want to add to what David said about picking simple numbers. If you were not sure about the answer to this question, you could try picking some numbers and using them, the information given in the question and the information given in Statement 1 to see what happens.

For instance, you could try 120 miles, then 60 miles, and notice that you get the same average speed.

Using 120 miles, you get 48 miles per hour, as David outlined.

Using 60 miles you get 1.5 hours there, 1 hour back. 2.5 hours to go 120 miles. So the average speed is again 120/2.5 = 48.

So you could see a pattern and determine that Statement 1 is sufficient.

For this relatively simple problem you might not really need to do that, but it's good to be aware of the availability of that method for when you run into problems whose answers are not as simple to get to.
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