# Math word problem

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**1**Senior Member

Thread Starter

Join Date: Feb 2011

Location: Pacific NW coast~!

Posts: 2,289

**Math word problem**

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?

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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**6**Senior Member

Join Date: Oct 2007

Posts: 9,807

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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**11**Senior Member

Join Date: Oct 2007

Posts: 9,807

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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Posts: 5,406

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.

***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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**13**Senior Member

Join Date: Apr 2009

Location: Michigan

Posts: 3,893

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.

- 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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**14**Senior Member

Join Date: Sep 2017

Location: Scottsdale, AZ

Posts: 390

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.

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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**15**Senior Member

Join Date: Oct 2007

Posts: 9,807

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

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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**17**Senior Member

Join Date: Apr 2009

Location: Michigan

Posts: 3,893

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.

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.

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**18**Senior Member

Join Date: Sep 2017

Location: Scottsdale, AZ

Posts: 390

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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**19**Senior Member

Join Date: Aug 2009

Location: CT

Posts: 4,024

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.

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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Location: NSB, Florida, USA, earth (mostly)

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