Question

Worded probem leading to simultaneous equation.
Question:
In a certain Algebra class there is a total of 350 possible points. These points come from 5 homework sets that are worth 10 points each and 3 one hour exams that are worth 100 points each. A student has received homework scores of 4,8,7,7, and 9 and the first two exams scores of 78 and 83. Assuming that grades are assigned according to the standard scale (B:80-89, and A:90-100) and there are no weights assigned to any of the grades;
1) is it possible for the student to receive an A in the class? If so, what is the minimum score required for an A?
2) What about a B?

Answer 1

Let's start by determining what the `minimum number of points` the student can receive to get an A.

\(\large \dfrac{x}{350}=.9\)

Let me interpret this for you.
We're saying that some amount of unknown points `x`, divided by 350 (the total number of possible points) equals a grade of 90%. The percent we wrote as a decimal.

From here we'll solve for x. Multiplying both sides by 350,\[\large x=.9(350)\]So we find that in order to get an A in the class the student needs at least \(\large 315\) total points.

Add up all the points the student has so far, and subtract that amount from 350. You can determine if the student, with his/her one remaining exam to take, can achieve an A in the class.

If the amount needed is more than 100 points, the student is unable to get an A in the class.

Answer 2

If you're confused by any of that, let me know c:

Answer 3

I am not confused by your interpretation, but i am trying to look at the questions' anwser from your point of view. But, how would i work part 2?

Answer 4

So far the student has accumulated \(\large 4+8+7+7+9+78+83\) points, which sums to \(\large 196\) points.

To get an A in the class the student needs \(\large 315\) points.

Meaning they need \(\large 315-196\) points MORE.
Which turns out to be \(\large 119\). Since the final exam is only worth 100 points, the student wont be able to get an A in the class, right?

Answer 5

For a \(\large \text{B}\) in the class, they would want to setup this proportion instead. \(\large \dfrac{x}{350}=.8\) This fraction is telling us that a grade of 80% is going to be some number of unknown points divided by 350, the total number of points.

Solving for x gives us, \(\large x=280\).
The student needs `at least` \(\large 280\) points for a \(\large \text{B}\) in the class.

Answer 6

This is actually a really good question :) If you take college courses these are the things you'll really want to be aware of late in the semester.

Answer 7

How many points you need, and such :D

Answer 8

Zepdrix, i am so astonished, you know i never thought the solution to this question was anything like what your saying and what your saying makes perfect sense

Answer 9

Based on the title "Worded probem leading to simultaneous equation." There is probably an algebraic way to represent this problem, with a system of equations or something. That's probably how they wanted you to approach it. But I think that's more confusing personally :P

We used the 1 equation at the start, to solve for the points needed for each grade value. But it's not too bad after you get that number! :)

Answer 10

Well, that's why i said that i was trying to see the answer from your point of view because i did not expect the response you gave i was expecting to see two equations, but i could'nt have gotten a better response than yours. I thank you very much for your time and he.

Answer 11

\c:/ glad i could help!

Answer 12

i am very happy that you could, blessings