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Math word problem

Old 06-01-2018, 11:29 PM
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John runs at an average rate of 5MPH while Lucy runs at an average rate of 3.5MPH. If Lucy starts running at 10AM and John begins running the same path at 10:30AM, what time will John catch up with Lucy?

Ok so my son and I do math together (yes I know we're geeks). He thinks the answer is 10:51AM, 21 minutes after. I think he is correct ONLY IF Lucy completely stops running at 10:30AM. Here is his math:

D(lucy) = 3.5H
D(lucy) = 3.5(0.5hr)
D(lucy) = 1.75mi

5mi / 60min = 1.75mi / x
x = 21min

10:30AM + 21min = 10:51AM

I think he forgot to accomdoate for the fact that Lucy is still running while John starts 30 minutes prior. I see the linear equation as:

D(lucy) = 3.5t (where t is in minutes)
D(john) = 5(t - 30min)

At what point does D(lucy) = D(john)
D(lucy) = D(john)
3.5t = 5(t - 30)
3.5t = 5t - 150
-1.5t = -150
t = 100min

John will catch up with Lucy 1hr 40min after Lucy starts which is 11:40AM

What do you think?
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Old 06-02-2018, 03:00 AM
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If John wants to get into Lucy's pants he will start at the same time and run the same speed.
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Old 06-02-2018, 04:14 AM
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11:40
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Old 06-02-2018, 04:22 AM
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100 min correct
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Old 06-02-2018, 05:33 AM
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Well. As soon as John gets insight of Lucy’s butt in her tight shorts he will speed up. So you will have to factor in his speed increase.
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Old 06-02-2018, 05:42 AM
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Question cannot be answered. The speeds are average and not constant. John could catch her at any given time given the variable of speed, but he may stay ahead or fall behind as the speed is only average over a given time. A better question is at what estimated time will John catch her, not an exact of when will he catch her.
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Old 06-02-2018, 05:45 AM
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Maybe a couple generations ago this would be valid. Lucy won't be up at 10 am. John will be playing video games on his phone.
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Old 06-02-2018, 05:49 AM
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11:40 is correct ..100 mins after Lucy starts or 70 after John starts..depending on how you set up your equation
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Old 06-02-2018, 06:01 AM
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Originally Posted by Jersus View Post
Question cannot be answered. The speeds are average and not constant. John could catch her at any given time given the variable of speed, but he may stay ahead or fall behind as the speed is only average over a given time. A better question is at what estimated time will John catch her, not an exact of when will he catch her.
while correct, do you think that’s the lesson? If so, can you provide the mathematical proof that shows there is more than one possible time? If the kid does that, if i’m the teacher, I give him mad props.
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Old 06-02-2018, 06:07 AM
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Old 06-02-2018, 06:42 AM
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Originally Posted by chrispnet View Post

while correct, do you think that’s the lesson? If so, can you provide the mathematical proof that shows there is more than one possible time? If the kid does that, if i’m the teacher, I give him mad props.
The “lesson” is not relevant, you answer the question asked based on the information presented. In this case they asked for a time, not an estimated time, ergo not possible without knowing a constant speed.

If if I am the teacher, the student that explains this to me instead of an estimated answer received the highest grade. Why? Simple, that student or students is/are the critical thinkers amongst the group.
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Old 06-02-2018, 06:48 AM
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Originally Posted by Jersus View Post

The “lesson” is not relevant, you answer the question asked based on the information presented. In this case they asked for a time, not an estimated time, ergo not possible without knowing a constant speed.

If if I am the teacher, the student that explains this to me instead of an estimated answer received the highest grade. Why? Simple, that student or students is/are the critical thinkers amongst the group.
Of course the lesson is relevant. Your verbal assertion is irrelevant in a math class. Proving your assertion mathematically, however, is the whole point. It is definitely provable, and that’s the way to show your ‘critical thinking’.
***if a kid brought me a verbal intuitive assertion in my math class to show me how his clever parsing of the words showed how poorly the question was worded and didn’t provide the math he gets a zero from me

Last edited by chrispnet; 06-02-2018 at 06:57 AM.
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Old 06-02-2018, 07:00 AM
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There are a number of ways to figure this out. I would do it this way using D = Rate * Time

- Lucy runs for 30min at 3.5mph, so she will be 1.75 miles ahead when they start running.

- The question is when does Lucy's distance = John's distance

D(L) = 1.75 miles + 3.5mph * t
D(J) = 5mph * t
When are they equal?
1.75 + 3.5t = 5t
1.75 = 1.5t
t = 1.1667, or 1 hour and 10 min (after they both start running)

To me, this is easier to visualize.
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Old 06-02-2018, 07:04 AM
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Time is independent variable, distance is dependent.

I'm using time in hours, rather than minutes.

So Lucy's linear equation is x=3.5t
John starts 30 minutes later. at 5mph, this is the equivalent of him starting at the same time but 2.5 miles behind Lucy. So John's equation is y=5t-2.5

You want to find where they are the same distance, so set the equations equal.

3.5t=5t-2.5

Solve for t

1.5t=2.5

t=2.5/1.5

t=1.666667 (t is in hours)

convert hours to minutes

1.66667 * 60 = 100

So 100 minutes after 10:00 AM, or 11:40 AM.

Last edited by thunder550; 06-02-2018 at 07:09 AM.
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Old 06-02-2018, 07:05 AM
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Originally Posted by chrispnet View Post

Of course the lesson is relevant. Your verbal assertion is irrelevant in a math class. Proving your assertion mathematically, however, is the whole point. It is definitely provable, and that’s the way to show your ‘critical thinking’.
***if a kid brought me a verbal intuitive assertion in my math class to show me how his clever parsing of the words showed how poorly the question was worded and didn’t provide the math he gets a zero from me
There is no clever parsing of words. As the question is asked the answer is not possible given an almost infinite number of speed possibilities across two variables.

I would much rather teach thinkers than those that think everything is finite. Do not blame me for the problem being very poorly worded. If the “lesson” is so important, then word the problem accordingly. The real lesson here is that there is no answer for the problem as presented, without using broad assumptions that are not given.

Even you admitted in your first reply to me that I am correct, then you proceed to claim that you would give a student a zero for using the same logic. So, which is it?
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Old 06-02-2018, 07:10 AM
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Sure sure. Of course the teacher is trying to teach algebra. There is an algebraic approach to present your assertion. If your critical thinking skills can’t break that code...your skills are lacking. Zero.
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Old 06-02-2018, 07:11 AM
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Originally Posted by thunder550 View Post
Time is independent variable, distance is dependent.

I'm using time in hours, rather than minutes.

So Lucy's linear equation is x=3.5t
John starts 30 minutes later. at 5mph, this is the equivalent of him starting at the same time but 2.5 miles behind Lucy. So John's equation is y=5t-2.5

You want to find where they are the same distance, so set the equations equal.

3.5t=5t-2.5

Solve for t

1.5t=2.5

t=2.5/1.5

t=1.666667 (t is in hours)

convert hours to minutes

1.66667 * 60 = 100

So 100 minutes after 10:00 AM, or 11:40 AM.
We get the same answer, but I like mine better because you are saying john is 2.5 miles behind Judy. She is never that far ahead. Harder to picture...
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Old 06-02-2018, 07:16 AM
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Originally Posted by ericinmich View Post
We get the same answer, but I like mine better because you are saying john is 2.5 miles behind Judy. She is never that far ahead. Harder to picture...
Same answer yeah, personal preference. You are using 10:30 as the x-axis zero value, am using 10:00. I find the 10:00 easier to picture. But answer is the same, so who cares.
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Old 06-02-2018, 07:18 AM
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Stop trying to be so fucking fancy, very basic time distance calculation.


Lucy travels 1.75 miles in 30 minutes and has that lead on John when he gets underway.




Every hour underway John gains 1.5 miles on Lucy.




1.75 miles / 1.5 miles = 1.166 that’s how long it takes John to eliminate Lucy’s 1.75 mile lead.




Convert the .1666 to minutes by multiplying x 60 = 10 minutes.




Takes John 1 hour + 10 minutes to make up Lucy’s 1.75 mile lead. If he takes off at 10:30 he catches her at 11:40.



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Old 06-02-2018, 07:22 AM
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Pics of Lucy?
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