math problem solved by amok, hail wise one amok! your powers are beyond belief. and perry gets credit for 2nd place.

[RayEarth]

cosmas won the 1995 boston marathon (a 26 mile race) in (about) .8 of an hour less time than the winner of the first boston marathon in 1897, john. find each runner's speed, if john's speed was (about) .73 times cosmas speed. (hint: don't round until the end of the calculations.)

using the d=rt formula i can't figure the darn thing out, stuck for hours, anyone think they can solve this in a few mins?

 

[DesignDawg]

7. Duh.

Ricky
DesignDawg

 

[DesignDawg]

BTW, I could most definitely figure that out in a few mins.
Edit: And since it has been quite a few years since my last math class, and I am rendering something at work right now, I think I will!

Ricky
DesignDawg

 

[RayEarth]

I would like more info on how you got 7?

in the back of the math book, cosmas miles per hour is 12.02 and john miles per hour 8.78

your answer is closer than I've ever gotten to the books answer, but there's suppose to be 2 answers, one speed for each person.

all i got so far is
let x= cosmas rate
cosmas d=26 miles r=x t=?
john d=26 r=.73x t=?
don't know how to get time, 26/x would be right because it says cosmas finished in 0.8 of an hour less time than john meaning that cosmas time should = 26/.73x(john's time) - 0.8?

 

[DesignDawg]

I was just being a smartass when I said 7. And DAMMIT! Why did you have to post the answer! uggh. I was just about to set it up.

Ricky
DesignDawg

 

[Zoltar]

E=MC²

 

[RayEarth]

i posted the answers in the back of the book, but i can't figure out how they got it.

 

[shawnmos]

a marathon is a 26.2 mile race. BTW both of my parents are flying up to boston this weekend to run it. no joke.

 

[Zoltar]

It appears you Dad runs faster than your mom. You're around.

 

[shawnmos]

oh and I am going to be home alone

 

[Zoltar]

Ummm. No thanks. I'm married. To a woman BTW.

 

[shawnmos]

hehe yes actually.

 

[shawnmos]

that was a continuation of the first post.

 

[Zoltar]

Just playing with ya!

 

[Turkey]

You have 4 unknowns, so you need four equations. The equations are simpler than you are making them. John's time is related to Cosmas's time, John's speed is related to Cosmas's speed, and the race is 26.2 miles.

 

[Moonbeam]

.8hr=48min so you get t for the slower runner and t-48 for the faster

The slor runner has a rate r and the fast one 1.23r

t x r = 26 =(t-48) x 1.23r NO?

 

[amok]

x=cosmas speed (I would prefer to use subscript and individually label an xsubc and xsubj and then relate them to be more precise, but if you can use subscripts on this forum i don't know how..)
t=john's time

So, you have:

John: d=0.73xt
Cosmas: d=x(t-0.8)

And since d is 26 in both cases, you can set the two equations equal to one another to solve for t:

x(t-0.8)=0.73xt

Once you get t, you can figure out Cosmas' speed, and proceed to plug that into the equation for John and get his speed. I already checked it, and the answers correspond to those in the book.

 

[amok]

Lol, even when I give somebody an answer to a math problem, I still make them work most of it out themselves. Old habits die hard .

 

[perry]

<< x=cosmas speed (I would prefer to use subscript and individually label an xsubc and xsubj and then relate them to be more precise, but if you can use subscripts on this forum i don't know how..)
t=john's time

So, you have:

John: d=0.73xt
Cosmas: d=x(t-0.8)

And since d is 26 in both cases, you can set the two equations equal to one another to solve for t:

x(t-0.8)=0.73xt
>>



I always hated setting up problems like this. They really got on my nerves, and I'd just skip em if they were on my homework, or usually get em wrong if they were on a test. Substituting the same variables that are related with some sort of proportion into different equations and crap. I don't mind doing differential equations and triple integrals (and other big words), I just hate these, whatever they're called.

 

[amok]

If I were guessing, I would say this is from a low level algebra class. Lol, I understand what you are saying relating to preferring higher level math to this. I have a MS in math, and the only two math classes I ever got below an A in were linear algebra classes that I was required to take (and I got C's in both of them!) .