Math brain teasers require computations to solve.
A car travels at a speed of 30 mph over a certain distance, and then returns over the same distance at a speed of 20 mph. What is the average speed for the total trip?
24 mph. Most people are quick to answer 25 mph, but this is not correct, as we will see. Let the distance traveled be 60 miles each way. Then, the trip out = 60/30 = 2 hours, and the trip back = 60/20 = 3 hours. Therefore: the 120 mile trip = 120/5 = 24 mph.
Mar 12, 2005
|It's weird, there is something about these kind of teasers I find hard. Yet the really difficult rated teasers I do fine on. |
Jun 09, 2005
|I dont understand what time has to do with it. Also would you reach the same answer if the traveled distance was 30 miles? I really hope you can explain these to me so I can get really smart |
Jun 09, 2005
|it does work with 30! Now why is time important?|
May 23, 2006
|You are trying to find the average speed in mpH (which is time). It would work for any amount. If you say it is a 60 mi length, then it would be 120 mi total (there and back) and it would take 5 hours (2h for 30mph + 3h for 20mph), and mph=m/h so 120mi/5h simplifies to 24mi/1h. Get it now? Great teaser, also!|
Jun 23, 2006
|i dont get it |
Feb 07, 2009
|The answer is correct. But the rule for getting it--directly, without using a virtual distance and calculating from it the average velocity based on total distance per total time--is not stated.|
The rule is: [2 (V1.V2) / (V1+V2)] without regard to distance.
Here's how I derived it.
V average = D total/T total
D total is twice D always. (you go and return, covering twice the same distance)
So, V average = 2.D/T total
T total is the sum of T1 (time going) and T2 (time coming back)
So, V average = 2.D/(T1+T2)
Since V = D/T, so T = D/V
V average = 2.D divided by [(D/V1)+(D/V2)]
Take the common factor D
V average = 2 divided by [(1/V1)+(1/V2)]
V average = 2 divided by [(V1+V2)/(V1.V2)]
So, V average = 2(V1.V2)/(V1+V2)
That's it! Good question...
Mar 31, 2009
|Hell that's a complicated way to do a simple problem. A much more interesting question is if he travels there at 30 mph , what speed must he do on the return trip to average 60 mph for the round trip? |
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