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December 8th, 2016, 03:34 AM  #1 
Senior Member Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4  Something that's confusing me
I have the following question: If I drive one way at a speed of 40 km/h, what is the speed I should drive back in, in order for the average speed for both ways (back and forth) to be 80 km/h. After looking at the solution, I understood that it's not possible. But at first I thought it is simply to drive back at 120 km/h? so the average speed would be 80 km/h? why that's not true? If you want me to post the solution let me know 
December 8th, 2016, 07:04 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314 
$\overline{spd} = \dfrac{40+x}{2} = 80$ $40+x = 160$ $x=120$ I'd like to know why you think it's not possible to achieve and average speed of 80 km/hr. 
December 8th, 2016, 07:18 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
The simple answer is that division by zero is not a valid arithmetical operation. Remember the basic formula $r * t = d \implies t = \dfrac{d}{r}\ and\ r = \dfrac{d}{t}.$ Let's say that the distance traveled ONE WAY is 160 kilometers. So clearly the distance to and fro is 320. $Let\ t = total\ time\ going\ and\ coming.$ If the average speed to and fro is 80 kilometers per hour, then $80 * t = 320 \implies t = \dfrac{320}{80} = 4\ hours.$ Nothing weird so far. $Let\ u = time\ spent\ going\ and\ v = time\ spent\ coming.$ $\therefore u + v = 4 \implies v = 4  u.$ With me so far? But the trip going was done at 40 kilometers per hour for a distance of 160, right? $\therefore u = \dfrac{160}{40} = 4 \implies v = 4  4 = 0.$ Is there any rate of speed that would allow you to travel 160 miles in no time at all? Obviously not. How does this manifest itself mathematically? The indicated rate on the return trip is $\dfrac{160}{v} = \dfrac{160}{0},\ which\ is\ not\ a\ real\ number.$ Now you can substitute x and 2x in for 160 and 320 and reach the same result. Division by zero is not a mathematical thing. 
December 10th, 2016, 12:26 AM  #4  
Senior Member Joined: Oct 2014 From: Complex Field Posts: 119 Thanks: 4  Quote:
Quote:
Last edited by noobinmath; December 10th, 2016 at 12:33 AM.  
December 10th, 2016, 05:20 AM  #5  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
The average speed is the total distance driven divided by the total time taken. A difficulty here is that the OP does not say what distance is being driven. Let "x" be the distance, in miles, one way. Then the total distance driven is 2x. Since the first leg, distance x, was driven at 40 mph, the time taken was x/40 hours. Let "v" be the speed driving back. Then the time required is x/v so the total time required is x/40+ x/v= x(1/40+ 1/v)= x(v+ 40)/(40v). The average speed then is 2x/(x(v+ 40)/40v)= 2x(40v/x(v+ 40))= 80v/(v+ 40)= 80. Multiplying both sides by v+ 80, 80v= 80(v+ 40)= 80v+ 3200. The "80v" terms cancel leaving 3200= 0 which is not true so there is no solution. Last edited by Country Boy; December 10th, 2016 at 05:23 AM.  
December 10th, 2016, 05:32 AM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,151 Thanks: 875 Math Focus: Wibbly wobbly timeywimey stuff.  
December 10th, 2016, 08:58 AM  #7  
Senior Member Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314  Quote:
I sure wish gasoline consumption worked the way you, Jeff, and the problem writer feel average speeds work.  
December 11th, 2016, 04:12 AM  #8 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 
On average, this was a good thread !

December 11th, 2016, 12:28 PM  #9 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,151 Thanks: 875 Math Focus: Wibbly wobbly timeywimey stuff.  
December 11th, 2016, 12:31 PM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 
Dan, what's also funny is that I couldn't stop laughing when I thought of that! 

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