SOLUTION: An executive would arrive 10.0 min early for an appointment if traveling at 60.0 mi/h, or 5.0 min early if traveling at 45.0 mi/h. How much time is there until the appointment? I

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: An executive would arrive 10.0 min early for an appointment if traveling at 60.0 mi/h, or 5.0 min early if traveling at 45.0 mi/h. How much time is there until the appointment? I       Log On

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Question 1128025: An executive would arrive 10.0 min early for an appointment if traveling at 60.0 mi/h, or 5.0 min early if traveling at 45.0 mi/h. How much time is there until the appointment?
I am having trouble setting up an equation for this problem.
Found 2 solutions by ikleyn, Boreal:
Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let D be the distance to travel, in miles.


Then the time traveling at the speed of 60 mph is  D%2F60 hours,

while the time traveling at the speed of 45 mph is  D%2F45 hours. 


We are given that the difference  D%2F45  - D%2F60   is 10 minutes - 5 minutes = 5 minutes = 1%2F12 of an hour:


    D%2F45  - D%2F60  = 1%2F12.


It is your basic equation, and the setup is just DONE.


To solve the equation, multiply its both sides by  180.  You will get


    4D - 3D = 15,

    D = 15    miles.


The appointment is scheduled at  15%2F60 hours + 10 minutes = 15 minutes + 10  minutes = 25 minutes counting from the time

the executive starts his journey.

Solved.

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To see other similar solved problems,  look into the lesson
    - How far do you live from school?
in this site.

Your problem is non-standard and is different from that usually are offered in typical Math classes.

Since you are interested in such kind of problems,  I recommend you to look into my other lessons on Travel & Distance
that you will find in accompanied references to that lesson.

You will find there  A  LOT  of  unique  and  interesting   non-standard  Travel & Distance problems.


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
time to appointment is x
x-10 minutes is 10 minutes early, x-5 is 5 minutes early
at 60 mph, he travels 1 mile a minute, and in x-10 minutes will travel x-10 miles
at 45 mph, he travels 3/4 a mile a minute, and in x-5 minutes will travel (3/4)(x-5) or (3/4)x-3.75
those two are equal since the starting and ending points are the same.
x-10=(3/4)x-3.75
(1/4)x=6.25
x=25 minutes ANSWER
10 minutes early means he travels 15 min at 60 mph and that is 15 miles
5 minutes early means he travels 20 min at 45 mph and that is 15 miles