SOLUTION: Claire is away at college and needs something from home, 625 miles away. She arranges to meet her father at a location in between home and college, so he can give her the item. a.

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Question 1209220: Claire is away at college and needs something from home, 625 miles away. She arranges to meet her father at a location in between home and college, so he can give her the item.
a. Claire averages 55 mph, and her part of the trip takes h hours. Represent the miles she drove algebraically.
b. Her father takes 2 more hours than Claire to make his part of the trip, and his average speed is 50 mph. Represent the distance he drove algebraically.
c. Using your answers from (a) and (b), write the sum of the distances driven by Claire and her father. Simplify your expression as much as possible.
d. What is the actual sum of the distances that they both drove?
e. Write an equation that can be used to find the number of hours they traveled.
f. How many hours did it take Claire to drive her part of the trip?
g. How many miles did Claire drive?
h. How many hours did her father drive for his part of the trip?
i. How many miles did her father drive?

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Claire is away at college and needs something from home, 625 miles away.
She arranges to meet her father at a location in between home and college,
so he can give her the item.
a. Claire averages 55 mph, and her part of the trip takes h hours. Represent the miles she drove algebraically.
b. Her father takes 2 more hours than Claire to make his part of the trip, and his average speed is 50 mph.
Represent the distance he drove algebraically.
c. Using your answers from (a) and (b), write the sum of the distances driven by Claire and her father.
Simplify your expression as much as possible.
d. What is the actual sum of the distances that they both drove?
e. Write an equation that can be used to find the number of hours they traveled.
f. How many hours did it take Claire to drive her part of the trip?
g. How many miles did Claire drive?
h. How many hours did her father drive for his part of the trip?
i. How many miles did her father drive?
~~~~~~~~~~~~~~~~~~~~~

(a)  55*h  miles.


(b)  50*(h+2)  miles.


(c)  55h + 50*(h+2) = 55h + 50h + 100 = 105h + 100  miles.


(d)  105h + 100 miles.


(e)  105h + 100 = 625  miles.


(f)  From the equation,  h =  =  = 5  hours


(g)  55*5 = 275  miles.


(h)  5+2  = 7 hours.


(i)  50*7 = 350  miles.

All the parts from (a) to (i) are completed.


=====================


Looking at this problem, I think that it is strange way to teach.

This way determines each step of a student and does not leave a room for him / (for her) to think on his/her own.


The problems like this one are the simplest algebra word problems on Travel and Distance.

I placed several problems on this subject in my lesson
    - Travel and Distance problems
in this site.

To be prepared to solving such problems, a student should be armed with these simple ideas:

        formulas of the type distance = rate * time
and
        understanding that the total distance in oncoming traffic
        is the sum of partial distances.

The rest is for the student's common sense.

Developing his or her common sense is the major goal of these exercises.

At least, my teachers taught me this way during my school years.

This goal is not achievable, if the student's room of thinking
is tightly compressed to minimum possible dimensions.


So, I am strictly opposed to such methods of teaching.



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