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Is there a good site to help me figure out Rate*Time problems...LOF....


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I am working through LOF Beginning Algebra, and quite frankly, i am really having a hard time grasping these rate * time= distance problems. I need help....

 

For example....

Alex and Betty walk @ 4mph for almost an hour. They then travel the same distance they had walked in their car at 16 mph. All together, their walking and traveling by car takes exactly 1 hour. How long did they walk?

 

Now this may seem easy to you...but I swear I have that deer in the headlights look my kids give ME when i ask them to parse a Bible verse:-)

 

Can someone help me get a grasp on these types of problems? Are there particular formulas for different types of these...

OR can someone PLEASE explain all this to me IN ENGLISH??

 

Oh...the frustration...I feel bad for my poor kids....UGH!

Faithe

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I usually start a word problem like this by deciding what I want to know (how long they walked), and assigning this as my variable or unknown (x). If they walked for x hrs, then they must have driven for 1-x hrs because they walked and drove for exactly 1 hr. Also, you know that the distance walked (rate x time) is equal to the distance driven (rate x time).

 

4 mi/hr * x hrs = 16 mi/hr * (1-x) hrs

 

4x = 16 - 16x

20x = 16

x = 16/20 = 4/5 hr

 

They walked for 4/5 hr, and then drove for 1/5 hr. Does that help?

 

Martha

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I started to post some links, then decided it would be more helpful for you to select the ones most understandable for you and your dc.

 

Just go to Google (or your preferred search engine) and type three words:

algebra rate time

(No boolean needed, just the three words)

 

You will receive over a quarter of a million hits, many of which look helpful.

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For example....

Alex and Betty walk @ 4mph for almost an hour. They then travel the same distance they had walked in their car at 16 mph. All together, their walking and traveling by car takes exactly 1 hour. How long did they walk?

 

 

From my 11YO son...

 

First we have to clarify that by "how long" you mean "how far," right? Because we already know how long in time (1 hour).

 

4 mph (4x) + 16 mph (16x) = 60 (mins)

4X + 16X = 60 mins

20X = 60 mins

 

At this point they traveled an average of 20 mph for 60 mins.

 

So, we think (not sure!) that they traveled for 20 miles.

 

Anybody else?

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What Martha said.

 

And also, just so you don't feel TOO bad -- we do this in beginning algebra, we do this in intermediate algebra, and when we get to college algebra, many students still get bogged down. So you're not alone :)

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I usually start a word problem like this by deciding what I want to know (how long they walked), and assigning this as my variable or unknown (x). If they walked for x hrs, then they must have driven for 1-x hrs because they walked and drove for exactly 1 hr. Also, you know that the distance walked (rate x time) is equal to the distance driven (rate x time).

 

4 mi/hr * x hrs = 16 mi/hr * (1-x) hrs

 

4x = 16 - 16x

20x = 16

x = 16/20 = 4/5 hr

 

They walked for 4/5 hr, and then drove for 1/5 hr. Does that help?

 

Martha

 

This is the answer the book gave...but there was (and isn't) many solutions....

 

How do I know whixh x gets the + or minus??

 

 

Your explanation makes sense, but there are so many different types of problems....Is there a simplified way of figuring out HOW to set up the problems??

 

Thanks,

Faithe

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Your explanation makes sense, but there are so many different types of problems....Is there a simplified way of figuring out HOW to set up the problems??

 

Thanks,

Faithe

 

Faithe,

 

Have you emailed Stan? He's always been prompt with his responses to my dd. And, he welcomes calls and/or email.

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From my 11YO son...

 

First we have to clarify that by "how long" you mean "how far," right? Because we already know how long in time (1 hour).

 

4 mph (4x) + 16 mph (16x) = 60 (mins)

4X + 16X = 60 mins

20X = 60 mins

 

At this point they traveled an average of 20 mph for 60 mins.

 

So, we think (not sure!) that they traveled for 20 miles.

 

Anybody else?

 

No, because the walking PLUS the car travel time = 60 minutes. We actually don't know how long, in time, they walked.

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This is the answer the book gave...but there was (and isn't) many solutions....

 

How do I know whixh x gets the + or minus??

 

 

Your explanation makes sense, but there are so many different types of problems....Is there a simplified way of figuring out HOW to set up the problems??

 

Thanks,

Faithe

 

 

Faithe,

 

Take what Martha posted and write it out separately as "defining what you know" prior to attempting to solve the problem. It is the first step that I always make my kids do before anything else.

 

You know that

r*t=d

distance walked (dw)=distance driven (dd)

total time=1 hr

 

next substitute what you have been given into what you know:

 

rate for walking: 4m/hr

time walking: x hr (you don't know)

so distance walked= 4 m/hr * x hr

 

rate for driving= 16 m/hr

time for driving= (total is 1 hr. If dw=x, than 1hr-x=time for driving)= 1-x

 

Go back to the original "what you know list." You know that dw=dd or in complete terms: rate of walking * time walking = rate of driving * time driving

 

so substitute in:

4 m/h *x = 16 m/h * (1-x)

 

and then just solve for x.

 

Does that help?

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Faithe,

 

Take what Martha posted and write it out separately as "defining what you know" prior to attempting to solve the problem. It is the first step that I always make my kids do before anything else.

 

You know that

r*t=d

distance walked (dw)=distance driven (dd)

total time=1 hr

 

next substitute what you have been given into what you know:

 

rate for walking: 4m/hr

time walking: x hr (you don't know)

so distance walked= 4 m/hr * x hr

 

rate for driving= 16 m/hr

time for driving= (total is 1 hr. If dw=x, than 1hr-x=time for driving)= 1-x

 

Go back to the original "what you know list." You know that dw=dd or in complete terms: rate of walking * time walking = rate of driving * time driving

 

so substitute in:

4 m/h *x = 16 m/h * (1-x)

 

and then just solve for x.

 

Does that help?

 

Yes, this does help a lot. I feel like I am learning a new language and that I have to get used to the nuances and jargon. Letting x = something is just a very strange concept to my mode of thinking. I am a bookkeeper and deal with real numbers...I can add columns in my brain...but throw in something abstract in the least and it throws me for a loop....

 

I am determined to struggle through...as long as it takes me. Tell me I am not too old to learn this stuff.....sigh....

 

~~Faithe

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Yes, this does help a lot. I feel like I am learning a new language and that I have to get used to the nuances and jargon. Letting x = something is just a very strange concept to my mode of thinking. I am a bookkeeper and deal with real numbers...I can add columns in my brain...but throw in something abstract in the least and it throws me for a loop....

 

I am determined to struggle through...as long as it takes me. Tell me I am not too old to learn this stuff.....sigh....

 

~~Faithe

 

I don't know what you have used for algebra in the past, but Foerster does an excellent job of teaching students how to set up word problems exactly like I described and there is a complete step by step solutions manual. ;)

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I am determined to struggle through...as long as it takes me. Tell me I am not too old to learn this stuff.....sigh....

 

 

You aren't. I had a student in his 80s in my Beginning Algebra course at the cc once. Also have had some students in their 60s. You can do this! :001_smile:

 

Learning algebra is very much like learning a new language.

 

With word problems, one of the things I recommend is to write out the information you have, then close the book so you don't look back at the problem, hoping something will magically appear. Actually closing the book can be tremendously difficult, but it can help too.

 

A site I like for additional explanations is purplemath. Here's her section on these type of word problems.

 

You've gotten some good explanations, so I don't know if this'll help in any way, but for motion problems (d=rt), setting up the chart is really helpful in organizing the information.

Using words for an in-between step between English and math can help too.

Here, you know that (time walking) + (time driving) = 1hr.

 

Fill in a d = r*t chart (may not type up well)

 

d r t

walking 4

driving 16

 

We know that the DISTANCE traveled is the same.

So now we can fill in the chart two different ways depending on what we set our variable to be.

 

Probably the best way is what's explained by momof7,

Since you're looking for the time spent walking, let that be the variable. (I'll use t for time walking.)

 

We know (time walking) + (time driving) = 1hr.

So now t + (time driving) = 1, so (time driving) = 1-t

 

Our chart now looks like this:

d r t

walking 4 t

driving 16 1-t

 

Since d = rt, we can fill in d

 

d r t

walking 4t 4 t

driving 16(1-t) 16 1-t

 

And we know the distance walking = distance driving, so we can set the distances the same: 4t = 16(1-t)

 

I find using the tables is really helpful for organizing the information in these problems. Hang in there - you can learn this!

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Several things I repeatedly tell my students: watch your units!! and check to see whether your answer makes sense!! make sure you answer the question that was asked!!

 

In his first equation, the 4 and 16 are miles per hour, but his answer is 60 minutes.

 

Even if he used 1 hour instead of 60 minutes, his "x" would need units of hr^2/mi to make the units come out right, but there would be no significance to that equation.

 

It doesn't make sense to say they averaged 20mph for the hour -- part of the time they were going 4mph, the rest 16mph -- their average speed couldn't possibly be higher than either of them. And don't make the mistake of actually averaging the two speeds (4+16)/2 = 10mph, because that doesn't work either. They traveled a much longer time at the slower speed.

 

 

 

From my 11YO son...

 

First we have to clarify that by "how long" you mean "how far," right? Because we already know how long in time (1 hour).

 

4 mph (4x) + 16 mph (16x) = 60 (mins)

4X + 16X = 60 mins

20X = 60 mins

 

At this point they traveled an average of 20 mph for 60 mins.

 

So, we think (not sure!) that they traveled for 20 miles.

 

Anybody else?

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