Question 1189622
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A girl starts a hike up the mountain at 6 am. 2 hours into the hike she is passed up by a group of people on their way up. 
At 10 am, the same group of people passes her again on the way down. She finally reached the summit at noon. 
If the girl and the group of people traveled at constant speeds, at what time did the group reach the summit?
sorry I asked this question yesterday but missed an important part (that she reached the summit at noon).
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<pre>
Below is my illustration to the solution.  
Let me measure all the distances in kilometers and the rates in kilometers per hour.


                                            
      --|---------------|---------------|---------------|
                                            <-----
            ----->          ----->          ----->  
        A               B               C               D
      girl           passing         meeting         summit
    started           point          point          (the peak)
    at 6 am          at 8 am        at 10 am        12 am for the girl
                                                    unknown time for the group


So, according to the problem, the girl and the group started SIMULTANEOUSLY from point B at 8 am

with different rates.  Let "u" be the rate of the girl and "v" be the rate of the group.

OBVIOUSLY,  u < v.     Let d be the distance from B to D, in kilometers.


The girl   moved from B to C for 2 hours. So, the distance BC is 2u kilometers.
The group  moved from B to D and then back to C for 2 hours, too (the same time).
So, the group covered, in total, the distance 2v in 2 hours, i.e. 2v kilometers.


    +---------------------------------------------------------+
    |    Now it is very important moment for understanding.   |
    |           Track attentively my reasoning.               |
    +---------------------------------------------------------+


The distance covered by the girl in 2 hours from 8 am to 10 am (BC)  PLUS  the distance covered 
by the group in 2 hours from 8am to 10am (BD + DC) is  2d,

    so we can write our first equation  2u + 2v = 2d.    (1)


From the other side,  d = 4u                             (2)
since the girl covered this distance BD in 4 hours.


Thus we have two equations, (1) and (2).


To solve the problem, replace 2d in equation (1) by 8u, based on equation (2).  Then you get

    2u + 2v = 8u,

or

    2v = 8u - 2u = 6u.

Hence

    v = 6u/2 = 3u.


Thus we obtain that the rate of the group is three times the rate of the girl.


Next, we know that the girl covered BD in 4 hours, from 8 am to 12 am.

Hence, the group spent 1/3 of 4 hours to get the peak, or 1 hour and 20 minutes.


It means that the group reached the peak at 9:20 am.
</pre>

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Solved.


It is very nice Travel & Distance problem.



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<U>Comment from student</U>: &nbsp;&nbsp;hi, thank you for the solution, but the problem does not say that they started out at the same time. 
it only says that the girl got passed up by the group 2 hours after she started. we only know about the group's 
time and distance from that point on.




<U>My response</U> : &nbsp;&nbsp;Your comment told me that you did not understand &nbsp;BOTH &nbsp;the problem and my solution.


The problem &nbsp;SAYS &nbsp;that the group passed the girl at &nbsp;8 am.


It &nbsp;MEANS &nbsp;that at &nbsp;8 am they were at the same location at the same time;

hence, &nbsp;we can say that they &nbsp;STARTED &nbsp;from this point simultaneously at &nbsp;8 am.



What happened before &nbsp;8 am, &nbsp;DOES &nbsp;NOT &nbsp;MATTER &nbsp;in this problem.


Learn to read, &nbsp;to think and to understand the things correctly . . . 



..........................



The level of this problem is much higher than regular problem in &nbsp;Physics at high school.


It is even higher than the level of problems in any college/university for students who are not majoring in &nbsp;Physics 
(Chemistry, &nbsp;Biology, &nbsp;Engineering, &nbsp;Psychology, &nbsp;Business administration etc.)


It is the level of leading universities for students majoring in &nbsp;Physics;

it is also the level of &nbsp;Physics circles/(Olympiads in Physics) &nbsp;at high school/college/leading universities - - - 
- - - for those students who really want to study &nbsp;Physics at the advanced/professional level.


I know it very well, &nbsp;because I solved &nbsp;MANY &nbsp;similar problems in my time, &nbsp;from more than &nbsp;10 &nbsp;first-class problems books.


So, &nbsp;the problem is intended for advanced students, &nbsp;who &nbsp;WANT &nbsp;to know &nbsp;MORE - as much as possible, 

who see the beauty of Physics and ready to work &nbsp;(to study) &nbsp;on their own, &nbsp;applying as much efforts as it is required.


THEREFORE, &nbsp;when I explain the solution, &nbsp;I see such a student in front of me . . . , and my explanation is intended for such a student.


If I will explain any microscopic detail - such teaching will be &nbsp;USELESS.



So, if you do not understand the solution presented in my post, &nbsp;it simply means that you are in wrong class.



Nevertheless, &nbsp;I am ready to answer any your question, &nbsp;if you have it.


Do not hesitate to ask - - - and happy learning &nbsp;(&nbsp;!&nbsp;)



Which high school / college / university / thinking center are you from ?


Which textbook in Physics do you use ?



If you respond &nbsp;(pointing your level, too), &nbsp;I can recommend you some textbooks . . .