1. ## Distance,Time,Rate Story Problem

Hello,

Another fine day for math isn't it? Well thank you all for helping me ahead of time!

Story Problem:

You can jog 8 miles in the same time that it takes to travel 16 miles by car. IF the car's rate is 9 miles per hour faster than your jogging rate, find the average rate of each.
Things we know:

Jogging rate = X
Cars Rate = X+9
Distance jogging = 8 Miles
Distance Car = 16

We don't know over all time and general rate...

I was thinking I could take the time and set it equal to 1 and go from there so:

Jogging Rate:

1=(1/8) / X : So X=1/8

Cars Rate:

(1/8) + 9 = 9.125

Is that correct? .125 and 9.125?

2. ## Re: Distance,Time,Rate Story Problem

It REALLY helps to write complete definitions.

Distance is in Miles
Time is in Hours

X = Jogging Rate in mph -- This would be a little better.

X mph = Jogging Rate -- I would prefer this

X mph + 9 mph = Car Driving Rate

Now, WRITE IT DOWN!!

Distance = Rate * Time

8 Mile = X mph * Time
16 Mile = (X mph + 9 mph) * Time

We know already that "Time" is the same in each.

Now, what on earth is that 1/8 mph business? First, it takes 64 hours to get 8 miles at this speed. Second, the units aren't even mph. What are they? It is all very confusing. Never do that. Third, you're going to get a lot more than 16 miles at 9.125 mph for 64 hours. You do need to make sure your answers make sense.

3. ## Re: Distance,Time,Rate Story Problem

Originally Posted by TKHunny
It REALLY helps to write complete definitions.

Distance is in Miles
Time is in Hours

X = Jogging Rate in mph -- This would be a little better.

X mph = Jogging Rate -- I would prefer this

X mph + 9 mph = Car Driving Rate

Now, WRITE IT DOWN!!

Distance = Rate * Time

8 Mile = X mph * Time
16 Mile = (X mph + 9 mph) * Time

We know already that "Time" is hte same in each.

Now, what on earth is that 1/8 mph business? First, it takes 64 hours to get 8 miles at this speed. Second, the units aren't even mph. What are they? It is all very confusing. Never do that. Third, you're going to get a lot more than 16 miles at 9.125 mph for 64 hours. You do need to make sure your answers make sense.

Okay..... Well......... I'm sorry that I didn't reach your expectation on how to format it? I'm not a math instructor? And as for the 1/8 it was taking what I knew from another problem and trying to apply it to this problem.
So if you don't mind; There is no need for the all CAPS! Because as fair as I can see- I'm back at square one because I don't know what time is, nor the X in rate to figure it out.

If I take

Dist = Rate * Time

That gets me to (Let X = Rate for Jogging)

8= X * T or ( Let X = Rate for Jogging + 9 for Car )

16 = (x+9) * T

So now I'm really confused....

4. ## Re: Distance,Time,Rate Story Problem

Problem: You can jog 8 miles in the same time that it takes to travel 16 miles by car. IF the car's rate is 9 miles per hour faster than your jogging rate, find the average rate of each.
Let x be the rate of "you" and y be the rate of the car.
Distance=rate*time
Distance/rate=time
8/x=16/y
x+9=y
-------------------------------
solve:
8y=16x
8(x+9)=15x
8x+72=15x
72=7x
x=72/7 mph
y=1152/56=288/14=144/7mph

5. ## Re: Distance,Time,Rate Story Problem

Originally Posted by cpuboye11
I'm sorry that I didn't reach your expectation on how to format it?
It's not an expectation. It's a helpful hint. LEARN how to structure so that the notation helps you. I am NOT talking about formatting. I'm talking about organizing in a useful way, to your advantage.

as for the 1/8 it was taking what I knew from another problem and trying to apply it to this problem.
That's nice, but why do you believe the 1/8 has any transferability? CLEAR definitions will help you with such decisions.

There is no need for the all CAPS!
Too bad you just ignored me. I guess I should use DOUBLE CAPS!

8= X * T or ( Let X = Rate for Jogging + 9 for Car )

16 = (x+9) * T
Confused about what? You have it! Now solve it.

Seriously, your life will be improved if you carry the units, rather than just hoping that you have them right.

6. ## Re: Distance,Time,Rate Story Problem

Hello, cpuboye11!

You can jog 8 miles in the same time that it takes to travel 16 miles by car.
If the car's rate is 9 mph faster than your jogging rate, find the average rate of each.

Here's a back-door solution . . .

Let $\displaystyle x$ = your jogging speed.
Then $\displaystyle x+9$ = your driving speed.

We are told: you can drive twice as far as you can jog.

There is an equation . . . $\displaystyle x + 9 \:=\:2x$

By the way, don't get your shorts in a knot.
Tkhunny used caps in one brief phase.
It was for emphasis, not to insult or offend.

If you are offended by italics, disregard this post.

7. ## Re: Distance,Time,Rate Story Problem

You guys act like math is nothing. But for some people such as my self its like a knife being shoved in to my head. So I might be over acting but when you sitting in front of a math book for 2 hrs working on 1 out of 50 problems and the book doesn't do crap for you and you think, maybe someone can explain this to you a little bit better then the book. And then when you ask and you get a bunch of smart a responses in an already stressful situation- the human mind tends to react as such.

So my apologies, I will keep my under-formatted and apparently dumb questions to my self. Thank you.