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Doublecheck homework percentages, and equations

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data

    A.) Distance, the length of which is 350km, was done within 4 hours.
    Some part of the distance was driven at speed of 60km/h, and some part of the distance was drivent at speed 100km/h. What part of the distance was driven at 60km/h speed

    B.) How many students were at the course?
    If those who fulfilled the course with midterms were 75% of the students.
    The rest took part in the exam, and of those 50% fulfilled the course.
    The ones who got the grade zero, were 7 students.

    (It was assumed by myself that grade zero correspond to failure of the course, because this was ostensibly supposed to be the same grading policy as it is currently at our university from 0-5 grades)

    (furthermore using common sense it is concluded that when a course is fulfilled, ergo it means that the course was not in that case failed. Ergo, the grade was not zero.)

    2. Relevant equations


    3. The attempt at a solution


    A.)

    This was little bit more difficult problem than the second one.

    Not much information was strictly able to be known immediately. To my understanding this problem was not really solvable unless you use simultaneous equation. (even though we did not cover that topic in math class, when this homework was given.) I was a little bit confused initially about that one, but apparently this was suppsoed to be an extra assignment for our class.

    The parts of information that could be gathered were essentially that
    1st part time + 2nd part time = total time
    (60km/h * 1st part time) + (100km/h*2nd part time) = 350km

    let x= 1st part time
    let y= 2nd part time

    ##x+y=4##
    ##60*x +100y= 350##

    ##\begin{cases} x+y=4 \\ 60x+100y=350 \end{cases}##

    x= 1,25
    y=2,75
    1,25 hours was driven at 60km/h probably....
    that would be 1h and 15min.

    doublechecking it would seem that this solution is true because

    60*1,25= 75km travelled
    100*2,75= 275km travelled

    75+275 = 350km
    and
    1,25+2,75= 4hours

    B.)

    100% of students existed in beginning

    75% of students fulfilled course by midterms--> ergo they all passed (fulfilled) the course

    25% of students attempted test. Of those 50% passed and 50% failed.


    25% of students attempted exam. Of those, 50% passed course, and the rest of them failed


    It was known that those who failed, were in fact 7 students in total.

    let x= total students in course

    (0.25x / 2)= 7
    0,125x=7
    x=56

    doublechecking
    56 students in course
    56- (0,75*56)=14
    14/ 0,25 = 56
    14 took part in the exam. And of those, half failed and half passed.
    14-(0,5*14)= 7
    7 failed.
     
  2. jcsd
  3. Oct 30, 2016 #2

    Mark44

    Staff: Mentor

    The equation above might not be accurate. The wording of the problem was "Some part of the distance was driven at speed of 60km/h, and some part of the distance was drivent at speed 100km/h."
    I recognize that English is not your first language, so it's possible some meaning was lost in translation. What you wrote above doesn't mean that "the rest of the distance was driven at 100 km/hr.


    In that case, the equation would be ##t_{60} + t_{100} + t_{\text{remaining}} = t_{\text{total}}##
    Here, the first two variables represent the time intervals during which the car was going 60 km/hr and 100 km/hr, respectively, and the third variable represents the remaining time.

    If the entire trip was driven at one or the other of the two speeds, then your equation is correct.
     
  4. Oct 30, 2016 #3
    I don't see how the problem would be easily solvable without assuming that to be true. Maybe it would be solvable even as you suggesst.
    There was not much information given
    There were only two speeds given, and the parts of the time, which corresponded to those two speeds. And there was the total time. And the total distance.

    I tried to do the translation as faithfully as possible but I could only seemingly assume that only two speeds existed. Because only two speed were really given to exist, as beginning information. I suppose we shall see by tomorrow when these are checked by my teacher.
     
  5. Oct 30, 2016 #4

    Mark44

    Staff: Mentor

    Your interpretation seems fine to me. I was just pointing out that the wording of the problem suggested that part of the trip was driven at some third speed. That's probably not what was intended in the problem -- the wording is not the clearest.
     
  6. Oct 31, 2016 #5
    It turned out that you were overthinking it and my assumption was correct. I suppose in a real life situation the driver has to physically accelerate the car from 60km/h ->100km/h but the overall "error" in this simplification of the two speeds (60km/h and 100km/h) is quite small over the entire distance 350km. It does not take too many seconds of time to accelerate to that speed in many typical cars. I was little bit apprehensive about that idea first also, but I could not really see a way around the problem so the best way seemed to assume only two speeds existed.
     
  7. Oct 31, 2016 #6

    Mark44

    Staff: Mentor

    I wouldn't say that I was "overthinking" here. The problem is worded in such a way that many people (and probably most mathematicians) would infer that there is at least a third speed involved here, and I don't mean the change in speed from 60 km/hr to 100 km/hr.

    If I said, "I have a basket with some apples in it and some oranges in it." One cannot conclude from this statement that the basket contains only apples and oranges. That was my point. IMO, the problem was worded strangely, at least in the English translation.

    Emphasis added by me.
     
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