Algebra distance word problems

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In summary, there is a conversation discussing how to solve a distance word problem involving a jogger and a car. The jogger is running at 4 miles per hour and takes 45 minutes longer than the car, which is traveling at 40 miles per hour, to complete the course. The conversation involves two different methods of solving the problem, with one method giving a correct answer of 50 minutes for the jogger's time and the other method giving an incorrect answer of 12 minutes. Upon further review, it is determined that the incorrect value was obtained due to a calculation error. Both methods use the equation D=RT, but the second method has a mistake in the calculations. Ultimately, the correct answer for the jogger's
  • #1
bergausstein
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A jogger running at the rate of 4 miles per hour takes 45 minutes more than a car traveling at 40 miles per hour to cover a certain course. How long does it take the jogger to
complete the course and what is the length of the course?

i tried to solve this using this method

i let
$x=$ joggers time to complete the course

$x-\frac{3}{4}=$ car time to complete the course

then,

$4x=40\left(x-\frac{3}{4}\right)$

$x=\frac{5}{6}$ or $50$ minutes.

now when i used another method

i let

$x=$time for car to complete the course

$x+\frac{3}{4}=$ jogger's time to complete the course

$40x=4\left(x+\frac{3}{4}\right)$

i get a different answer $x=$12

can you tell me what's the difference between the two solutions i used and which one is correct?

thanks!
 
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  • #2
Re: algebra distance word problems

Hello.

You observes.

bergausstein said:
i tried to solve this using this method

i let
$x=$ joggers time to complete the course

$x-\frac{3}{4}=$ car time to complete the course

Well.

But, then, you apply badly his resolution:

bergausstein said:
then,

$4x=40\left(x+\frac{3}{4}\right)$

$x=\frac{5}{6}$ or $50$ minutes.

You calculate:

[tex]x-\dfrac{3}{4}[/tex]

And, then you use, in the resolution:

[tex]x+\dfrac{3}{4}[/tex]

Regards.
 
  • #3
Re: algebra distance word problems

yes, but that's just a typo. which I edited now.

my question is, what's the difference in my two solutions and which one is correct?

did I set them up correctly?
in my first solution i chose $x$ to represent the jogger's time so I have $x-\frac{3}{4}=$ for the car's time.

in my second solution i chose $x$ to represent car's time taken to complete the course. so i can have $x+\frac{3}{4}$ to represent jogger's time.

the first solution give $x$ to be $50$mins or $\frac{5}{6}$ of an hour. but the second one give me $x$ to be $12$ which I'm not sure what unit of time should i attached.

please somebody help. this is confusing me.
 
  • #4
Re: algebra distance word problems

In your second method, you obtained an incorrect value for $x$. solving it correctly will give you the same value for the jogger's time.
 
  • #5
Re: algebra distance word problems

bergausstein said:
$x=$time for car to complete the course

$x+\frac{3}{4}=$ jogger's time to complete the course

$40x=4\left(x+\frac{3}{4}\right)$

i get a different answer $x=$12

You will have calculated badly.

Solution:

[tex]x=\dfrac{1}{12} \ h \ = 5 \ minutes[/tex]

Regards.
 
  • #6
Re: algebra distance word problems

MARKFL yes, i also noticed that.

but if review my set up in second solution it seems(for me) that's it is correct.

the time taken for the jogger to complete the course is 45mins more than that of the car to complete the same course.

now if i let $x=$ car's time taken to complete the course

then the time taken for jogger should be $x+\frac{3}{4}$

since the travel the same distance i set them equal to one each other.

by $D=RT$

$40x=4(x+\frac{3}{4})$ what's wrong here?
 
  • #7
Re: algebra distance word problems

MarkFL said:
In your second method, you obtained an incorrect value for $x$. solving it correctly will give you the same value for the jogger's time.

Excuse me, MarkFL.:eek:
 
  • #8
Re: algebra distance word problems

bergausstein said:
$40x=4(x+\frac{3}{4})$ what's wrong here?

This is correct, but you have realized badly the calculations.

Regards.
 

FAQ: Algebra distance word problems

What are algebra distance word problems?

Algebra distance word problems involve using algebraic equations to solve problems related to distance, time, and speed.

How do I set up an algebra distance word problem?

To set up an algebra distance word problem, you need to identify the unknown variables, such as distance, time, and speed, and create equations using known information to solve for the unknown variable.

What is the formula for solving algebra distance word problems?

The formula for solving algebra distance word problems is distance = speed x time, or d = st.

What are some common scenarios that use algebra distance word problems?

Some common scenarios that use algebra distance word problems include calculating the distance traveled by a moving object, determining the time it takes to travel a certain distance at a given speed, and finding the speed of an object based on its distance and time of travel.

What are some tips for solving algebra distance word problems?

Some tips for solving algebra distance word problems include carefully reading the problem to identify the known and unknown variables, setting up equations using the formula d = st, and checking your answer by plugging it back into the original problem.

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