Finding Rate with unknown time.

[NobodysHero]

Chuck and Dana agree to meet in Chicago for the weekend. Chuck travels 174 miles in the same time that Dana travels 159 miles. If Chuck's rate of travel is 5mph more than Dana's, and they travel the same length of time, at what speed does Chuck travel? Rate=Distance/Time. For Chuck I have x+5=174/? and for Dana I have x=159/?. I don't know what the time is so I see no way to solve the solution. Any help would be greatly appreciated. Thanks for your time.

 

[Deleted member 4993]

Chuck and Dana agree to meet in Chicago for the weekend. Chuck travels 174 miles in the same time that Dana travels 159 miles. If Chuck's rate of travel is 5mph more than Dana's, and they travel the same length of time, at what speed does Chuck travel? Rate=Distance/Time. For Chuck I have x+5=174/? and for Dana I have x=159/?. I don't know what the time is so I see no way to solve the solution. Any help would be greatly appreciated. Thanks for your time.

Asking - what speed does Chuck travel?

Let Chuck's speed = C mph

Chuck's rate of travel is 5mph more than Dana's, → Dana's speed = D = C - 5

Length of time Chuck traveled = d/r = 174/C

Length of time Dana traveled = d/r = 159/(C-5)

they travel the same length of time, → continue.....

 

[NobodysHero]

So the equation should be 174/c=159/(c-5)? The time is still confusing me.

 

[Deleted member 4993]

So the equation should be 174/c=159/(c-5)? The time is still confusing me.

Why are you worrying about time?

The question asks you to calculate Chuck's speed 'C' - not time.

After calculating 'C' you may choose to calculate respective time/s.

 

[NobodysHero]

I thought you needed the time to find the rate. Isn't the formula Rate=Distance/Time?

 

[NobodysHero]

I haven't been able to find any examples like this question in the textbook that goes with the course but for all the distance questions I've always used Distance=Rate multiplied by Time. In this question i'm trying to find the Rate so the formula should Rate=Distance divided by time. If Chucks rate of travel = C then the equation should be C=174/Time, but the time isn't specified in the question. Maybe i'm over analyzing the question but it feels like I;m missing information needed to solve it.

 

[HallsofIvy]

I thought you needed the time to find the rate. Isn't the formula Rate=Distance/Time?

In this problem you have two equations. Typically, you can solve two equation for two unknowns so you can solve for both time and the speed. Starting from your x+5=174/? and for Dana I have x=159/?, since the problem does not ask for "?", the time, I would eliminate it first by dividing one equation by the other:
\(\displaystyle \frac{x+ 5}{x}= \frac{\frac{174}{?}}{\frac{159}{?}}= \frac{174}{159}\)

Now multiply on both sides to eliminate the denominators: 159(x+ 5)= 174 x.

 

[NobodysHero]

In this problem you have two equations. Typically, you can solve two equation for two unknowns so you can solve for both time and the speed. Starting from your x+5=174/? and for Dana I have x=159/?, since the problem does not ask for "?", the time, I would eliminate it first by dividing one equation by the other:
\(\displaystyle \frac{x+ 5}{x}= \frac{\frac{174}{?}}{\frac{159}{?}}= \frac{174}{159}\)

Now multiply on both sides to eliminate the denominators: 159(x+ 5)= 174 x.

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Ok so I would multiply 159 by x and 5 which gives me 159x+795=174x. Then I subtract 159 from both sides giving me 795=15x. Finally I divide by 15 and get the answer 53=x. So Dana's speed is 53 and Chucks speed is 58. That seems correct to me. Thanks Subhotosh Khan, Denis, and HallsofIvy for all you're help.

 

[lookagain]

\(\displaystyle \dfrac{x+ 5}{x}= \dfrac{\dfrac{174}{?}}{\dfrac{159}{?}}= \dfrac{174}{159}\)

HallsofIvy, if you replace the "\frac" characters with the "\dfrac" characters, you get the larger images of the fractions that you posted for easier readability (as I highlighted above).

 

[HallsofIvy]

HallsofIvy, if you replace the "\frac" characters with the "\dfrac" characters, you get the larger images of the fractions that you posted for easier readability (as I highlighted above).

Thanks. I did not know that! ("d" is not very intuitive!)