# Story Math Question. Yes, it is homework. **URL:** \ **Category:** In My Humble Opinion\ **Created:** [September 28, 2010, 1:58pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346 "2010-09-28T13:58:46Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Shirley\_Ujest](https://avatars.discourse-cdn.com/v4/letter/s/45deac/32.png) [@Shirley\_Ujest](https://boards.straightdope.com/u/Shirley_Ujest)\ **Post date:** [September 28, 2010, 1:58pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/1 "2010-09-28T13:58:46Z")
Last night three of us could not solve a story question in our son’s 7th grade math book. Person A leaves home at 9am and he is walking 4 mph. Person B leaves home a half hour later in the same direction running 8.5 mph. How long will it take for B to catch A? What would be the algebraic expression to figure this out? (We cheated and went to the back of the book and got the answer, which was 9 something, but we could not figure out how they came up with this answer. ) Our son went to school with this unanswered, so I/we are not violating the TOS stuff. ---
**Author:** ![The\_Lurker\_Above](https://avatars.discourse-cdn.com/v4/letter/t/dc4da7/32.png) [@The\_Lurker\_Above](https://boards.straightdope.com/u/The_Lurker_Above)\ **Post date:** [September 28, 2010, 2:10pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/2 "2010-09-28T14:10:44Z")
> [@Shirley\_Ujest](#): > > (We cheated and went to the back of the book and got the answer, which was 9 something, but we could not figure out how they came up with this answer. ) > > Our son went to school with this unanswered, so I/we are not violating the TOS stuff. You want to set up 2 equations for the positions of the walker/runner. Starting with t=0 when the runner left the house I set up function 1 for the position of the walker wrt time. And then set up function 2 for the position of the runner wrt time. Since when the runner catches up to the walker they will be at the same position you can set function 1 = function 2 and solve for t. Assuming they’re leaving from the same house I got a time around 26 minutes. ---
**Author:** ![Indian](https://avatars.discourse-cdn.com/v4/letter/i/c0e974/32.png) [@Indian](https://boards.straightdope.com/u/Indian)\ **Post date:** [September 28, 2010, 2:13pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/3 "2010-09-28T14:13:46Z")
x multiplied by 4 = (x-0.5) multiplied by 8.5 ---
**Author:** ![Animastryfe](https://avatars.discourse-cdn.com/v4/letter/a/8797f3/32.png) [@Animastryfe](https://boards.straightdope.com/u/Animastryfe)\ **Post date:** [September 28, 2010, 2:14pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/4 "2010-09-28T14:14:38Z")
Here’s a way to think about it. At 9:30, when person B has just left, person A has already traveled 2 miles. The difference in speed between the two people is 4.5 mph. So how long does it take for person B to go 2 miles at 4.5 mph? ---
**Author:** ![tdn](https://avatars.discourse-cdn.com/v4/letter/t/94ad74/32.png) [@tdn](https://boards.straightdope.com/u/tdn)\ **Post date:** [September 28, 2010, 2:32pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/5 "2010-09-28T14:32:42Z")
> [@Animastryfe](#): > > Here’s a way to think about it. At 9:30, when person B has just left, person A has already traveled 2 miles. The difference in speed between the two people is 4.5 mph. So how long does it take for person B to go 2 miles at 4.5 mph? That won’t do it. We can’t assume that person A stopped walking after 30 minutes and is waiting for person B to catch up. ---
**Author:** ![Folacin](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/folacin/32/3195_2.png) [@Folacin](https://boards.straightdope.com/u/Folacin)\ **Post date:** [September 28, 2010, 2:37pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/6 "2010-09-28T14:37:06Z")
He hasn’t stopped walking - but the relative velocity is 4.5 between the two - so if you reduce the runner’s speed to 4.5, then the walker has relatively stopped. I like this solution. Joe ---
**Author:** ![The\_Lurker\_Above](https://avatars.discourse-cdn.com/v4/letter/t/dc4da7/32.png) [@The\_Lurker\_Above](https://boards.straightdope.com/u/The_Lurker_Above)\ **Post date:** [September 28, 2010, 2:38pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/7 "2010-09-28T14:38:54Z")
> [@tdn](#): > > > [@Animastryfe](#): > > > > Here’s a way to think about it. At 9:30, when person B has just left, person A has already traveled 2 miles. The difference in speed between the two people is 4.5 mph. So how long does it take for person B to go 2 miles at 4.5 mph? > > That won’t do it. We can’t assume that person A stopped walking after 30 minutes and is waiting for person B to catch up. This method does work (ie. gives the same answer as my method :D). 4.5 mph is the speed of the runner relative to the walker. ---
**Author:** ![tdn](https://avatars.discourse-cdn.com/v4/letter/t/94ad74/32.png) [@tdn](https://boards.straightdope.com/u/tdn)\ **Post date:** [September 28, 2010, 2:39pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/8 "2010-09-28T14:39:27Z")
Oh yeah, I hadn’t thought of it that way. ---
**Author:** ![Shirley\_Ujest](https://avatars.discourse-cdn.com/v4/letter/s/45deac/32.png) [@Shirley\_Ujest](https://boards.straightdope.com/u/Shirley_Ujest)\ **Post date:** [September 28, 2010, 2:42pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/9 "2010-09-28T14:42:41Z")
My husband and his boss ( an engineer with 30 years experience) both futz over this and came up with this: Brother A leaves at 9:00am at 4 mph Brother B leaves at 9:30am at 8.5 mph in the same direction Distance = Rate x Time We need to find out what time is (T)? 4mph x T = 8.5mph x ( T - 30 min) since the one brother left 30 min later 4T = 8.5T - 255 255 = 4.5T 255/4.5 = T 56.66 = T 56.66 minutes after brother A leaves, brother B will be next to him. At 9:56am they will be side by side! Naturally, I said, it would be sooner if there were a Starbucks in the path of A. ---
**Author:** ![Shirley\_Ujest](https://avatars.discourse-cdn.com/v4/letter/s/45deac/32.png) [@Shirley\_Ujest](https://boards.straightdope.com/u/Shirley_Ujest)\ **Post date:** [September 28, 2010, 2:44pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/10 "2010-09-28T14:44:59Z")
Thanks everyone. ---
**Author:** ![statsman1982](https://avatars.discourse-cdn.com/v4/letter/s/e79b87/32.png) [@statsman1982](https://boards.straightdope.com/u/statsman1982)\ **Post date:** [September 28, 2010, 2:54pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/11 "2010-09-28T14:54:03Z")
\*\*The Lurker Above \*\*is using this equation: 4(t+30) = 8.5t which gives 26.66 minutes (A is 30 minutes ahead of B). **ShirleyUjest** is using this equation: 4t = 8.5(t-30), which gives 56.66 minutes (B is 30 minutes behind A). They give different answers because they’re answering different questions. Using the first equation, you’re answering the question: _once B leaves_\*, \*\*\*how long will it take to catch A? Well, B leaves 30 minutes later, and it will take 26.66 minutes once he does, so 26.66 + 30 = 56.66. The second equation finds the answer directly. ---
**Author:** ![Indian](https://avatars.discourse-cdn.com/v4/letter/i/c0e974/32.png) [@Indian](https://boards.straightdope.com/u/Indian)\ **Post date:** [September 28, 2010, 2:55pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/12 "2010-09-28T14:55:34Z")
> [@Shirley\_Ujest](#): > > My husband and his boss ( an engineer with 30 years experience) both futz over this and came up with this: > Brother A leaves at 9:00am at 4 mph > > Brother B leaves at 9:30am at 8.5 mph in the same direction > > Distance = Rate x Time > > We need to find out what time is (T)? > > 4mph x T = 8.5mph x ( T - 30 min) since the one brother left 30 min later > > 4T = 8.5T - 255 > > 255 = 4.5T > > 255/4.5 = T > > 56.66 = T > > 56.66 minutes after brother A leaves, brother B will be next to him. > > At 9:56am they will be side by side! > > Naturally, I said, it would be sooner if there were a Starbucks in the path of A. This is the same equation in my post . ---
**Author:** ![Shirley\_Ujest](https://avatars.discourse-cdn.com/v4/letter/s/45deac/32.png) [@Shirley\_Ujest](https://boards.straightdope.com/u/Shirley_Ujest)\ **Post date:** [September 28, 2010, 2:58pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/13 "2010-09-28T14:58:23Z")
Sorry! Sorry! ---
**Author:** ![Peter\_Morris](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/peter_morris/32/359_2.png) [@Peter\_Morris](https://boards.straightdope.com/u/Peter_Morris)\ **Post date:** [September 28, 2010, 3:03pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/14 "2010-09-28T15:03:36Z")
> [@Shirley\_Ujest](#): > > Last night three of us could not solve a story question in our son’s 7th grade math book. > > Person A leaves home at 9am and he is walking 4 mph. > > Person B leaves home a half hour later in the same direction running 8.5 mph. > > How long will it take for B to catch A? B can never catch A. It’s been proved. At 9:30, when B sets out, A has walked 2 miles. It will take 0.23529 hours for B to reach that point. However, in that time, A has walked another 0.94196 miles. To run that extra distance will take B another 0.11082 hours. But in that time, A will have walked another 0.22163 miles. And so on. The sequence never converges, so A and B never meet. ---
**Author:** ![The\_Lurker\_Above](https://avatars.discourse-cdn.com/v4/letter/t/dc4da7/32.png) [@The\_Lurker\_Above](https://boards.straightdope.com/u/The_Lurker_Above)\ **Post date:** [September 28, 2010, 3:05pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/15 "2010-09-28T15:05:20Z")
> [@statsman1982](#): > > They give different answers because they’re answering different questions. Using the first equation, you’re answering the question: _once B leaves_\*, \*\*\*how long will it take to catch A? Well, B leaves 30 minutes later, and it will take 26.66 minutes once he does, so 26.66 + 30 = 56.66. > > The second equation finds the answer directly. The problem statement says “How long will it take for B to catch A?”. That’s why I set t=0 for when B left the house. 26.7 minutes after B leaves and 56.7 minutes after A leaves are indeed the same time; but 26.7 minutes is the direct answer to the problem statement as I read it. I’d mark both correct, so long as the answer statement specified which. ---
**Author:** ![statsman1982](https://avatars.discourse-cdn.com/v4/letter/s/e79b87/32.png) [@statsman1982](https://boards.straightdope.com/u/statsman1982)\ **Post date:** [September 28, 2010, 3:14pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/16 "2010-09-28T15:14:57Z")
I suppose if showing our work were important, we should have converted time to hours first. We then would have gotten 0.944 hours, which is 56 minutes. ---
**Author:** ![beowulff](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/beowulff/32/542_2.png) [@beowulff](https://boards.straightdope.com/u/beowulff)\ **Post date:** [September 28, 2010, 3:39pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/17 "2010-09-28T15:39:44Z")
> [@Peter\_Morris](#): > > > [@Shirley\_Ujest](#): > > > > Last night three of us could not solve a story question in our son’s 7th grade math book. > > > > Person A leaves home at 9am and he is walking 4 mph. > > > > Person B leaves home a half hour later in the same direction running 8.5 mph. > > > > How long will it take for B to catch A? > > B can never catch A. It’s been proved. > > At 9:30, when B sets out, A has walked 2 miles. > It will take 0.23529 hours for B to reach that point. > However, in that time, A has walked another 0.94196 miles. > To run that extra distance will take B another 0.11082 hours. > But in that time, A will have walked another 0.22163 miles. > And so on. The sequence never converges, so A and B never meet. Thanks, Zeno. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [September 28, 2010, 4:35pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/18 "2010-09-28T16:35:54Z")
> [@Peter\_Morris](#): > > B can never catch A. It’s been proved. > > At 9:30, when B sets out, A has walked 2 miles. > It will take 0.23529 hours for B to reach that point. > However, in that time, A has walked another 0.94196 miles. > To run that extra distance will take B another 0.11082 hours. > But in that time, A will have walked another 0.22163 miles. > And so on. The sequence never converges, so A and B never meet. Furthermore, at no point are A or B actually moving. ---
**Author:** ![Clothahump](https://avatars.discourse-cdn.com/v4/letter/c/51bf81/32.png) [@Clothahump](https://boards.straightdope.com/u/Clothahump)\ **Post date:** [September 28, 2010, 4:43pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/19 "2010-09-28T16:43:23Z")
> [@Peter\_Morris](#): > > > [@Shirley\_Ujest](#): > > > > Last night three of us could not solve a story question in our son’s 7th grade math book. > > > > Person A leaves home at 9am and he is walking 4 mph. > > > > Person B leaves home a half hour later in the same direction running 8.5 mph. > > > > How long will it take for B to catch A? > > B can never catch A. It’s been proved. > > At 9:30, when B sets out, A has walked 2 miles. > It will take 0.23529 hours for B to reach that point. > However, in that time, A has walked another 0.94196 miles. > To run that extra distance will take B another 0.11082 hours. > But in that time, A will have walked another 0.22163 miles. > And so on. The sequence never converges, so A and B never meet. Good old Zeno… ---
**Author:** ![Yllaria](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/yllaria/32/3452_2.png) [@Yllaria](https://boards.straightdope.com/u/Yllaria)\ **Post date:** [September 28, 2010, 4:51pm UTC](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346/20 "2010-09-28T16:51:04Z")
> [@Peter\_Morris](#): > > . . . B can never catch A. It’s been proved. > > At 9:30, when B sets out, A has walked 2 miles. > It will take 0.23529 hours for B to reach that point. > However, in that time, A has walked another 0.94196 miles. > To run that extra distance will take B another 0.11082 hours. > But in that time, A will have walked another 0.22163 miles. > And so on. The sequence never converges, so A and B never meet. I thought that only worked for tortoises. [Next page](https://boards.straightdope.com/t/story-math-question-yes-it-is-homework/555346.md?page=2)