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 June 2nd, 2016, 01:48 PM #1 Newbie   Joined: Jun 2016 From: Toronto Posts: 2 Thanks: 2 Problem solving help John and Jane walk for 15 minutes at a constant speed. At the 15 minute mark of their walk, John runs back to their starting point to get something he forgot and then runs back to rejoin Jane. While he's gone, Jane walks twice as slow as usual. Once John and Jane are back together they continue at their regular speed and arrive at their destination 6 minutes later than usual. How many times faster than their normal walking speed did John run? Thanks from manus
 June 2nd, 2016, 03:31 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 Let their walking speed be "v" miles per hour and let the John run at "n" times their regular speed. In 15 minutes, 1/4 of an hour, they will have walked v/4 miles. While John runs back at nv miles per hour, Jane walks on at v/2 miles per hour. If she walks at that speed for T hours, she will have walked an additional Tv/2 miles for a total of v/4+ Tv/2= (v/2)(1/2+ T) miles. In that time, John will have run back v/4 miles, requiring (v/4)/(nv)= 1/(4n) hours. He then has T- 1/(4n) hours to catch up with Jane. In that time, he will have run (T- 1/(4n))(nv)= Tnv- v/4 miles which must equal Jane's (v/2)(1/2+ T) miles: Tnv- 1/4= (v/2)(1/2+ T). Thanks from manus
 June 2nd, 2016, 03:38 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond Hint: how long did Jane walk at a reduced speed? Thanks from manus
 June 2nd, 2016, 03:40 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 let $r$ = normal walking speed in distance units per minute $T$ = normal time in minutes to complete their walk $t$ = time period in minutes John was gone $T+6$ = delayed time in minutes to complete the walk $(T+6) - 15 - t$ = time to finish the walk after John catches up for Jane ... $r \cdot 15 + \dfrac{r}{2} \cdot t + r[(T+6) - 15 - t] = rT$ solving for $t$ ... $t = 12$ minutes Jon is gone. During that 12 minutes, Jane covers an additional distance of $6r$ walking forward at half-speed. while John is gone, he covers a distance of $15r$ backwards and $15r+6r$ forwards for a total distance of $36r$ in 12 minutes ... a rate of $3r$, three times as fast. Thanks from manus and MsHalifax
 June 3rd, 2016, 07:16 AM #5 Newbie   Joined: Jun 2016 From: Toronto Posts: 2 Thanks: 2 Grade level? Thanks for the clear explanation! What would be an appropriate grade level for a problem like this? Thanks from manus
June 3rd, 2016, 08:03 AM   #6
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 Originally Posted by MsHalifax Thanks for the clear explanation! What would be an appropriate grade level for a problem like this?
Over the past year I have been tutoring in person two quite bright and hardworking 9th graders studying second year algebra. I doubt either could solve this problem without considerable help. One reason is that kids find solving for numbers more intuitive than solving to express one variable in terms of another: their sense of abstraction and generalization has not yet been exercised much.

The other reason is that skeeter's explanation implies a good deal of mental analysis not shown. There are two unknown rates, four unknown distances, and three unknown times. A good student will follow skeeter's analysis, but may despair of replicating it. I am writing on an IPad and can't use a real computer, but I am going to think about how you might show a child how to think his or her way through this problem in a way that would make them say "Hey, I could do that."

 June 3rd, 2016, 08:35 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond As the walk took 6 minutes longer than usual to complete Jane must have walked at half speed for 12 minutes. That means that John covered a distance that would require 15 + 15 + 6 = 36 minutes walking at regular speed in 12 minutes, so John runs at 3 times normal walking speed. Thanks from skeeter, manus, JeffM1 and 1 others
 June 3rd, 2016, 08:46 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,559 Thanks: 2561 Math Focus: Mainly analysis and algebra One way to easily see Skeeter's approach is to draw a speed/time graph. The area outlined in red is Jane's trip. The shaded area is the normal trip. The distance travelled on a speed/time graph is the area underneath the graph. So the two areas must be equal. This means that the area of A is equal to the area of B. Thus $\frac12rt = 6r$, giving us $t=12$. We can then see that John ran the forst rectangle twice (to home and back again) and the second rectangle once or $15r + 15r + 12\cdot \frac12r = 36r$. And he did it in those $12$ minutes. Thanks from greg1313, skeeter, manus and 1 others
June 3rd, 2016, 11:42 AM   #9
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Quote:
 Originally Posted by greg1313 As the walk took 6 minutes longer than usual to complete Jane must have walked at half speed for 12 minutes. That means that John covered a distance that would require 15 + 15 + 6 = 36 minutes walking at regular speed in 12 minutes, so John runs at 3 times normal walking speed.
Bravo.

June 3rd, 2016, 07:13 PM   #10
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Quote:
 Originally Posted by v8archie One way to easily see Skeeter's approach is to draw a speed/time graph.
@V8archie, did you just use paint to draw that? Or something better.

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