Doublecheck homework percentages, and equations

[late347]

Homework Statement

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.)

Homework Equations

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 failedIt 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.

 

[Mark44]

Homework Statement

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.)

Homework Equations

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

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.

(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 failedIt 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.

 

[late347]

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.

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.

 

[Mark44]

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.

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.

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.

 

[late347]

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.

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.

 

[Mark44]

It turned out that you were overthinking it and my assumption was correct.

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.

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 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.