Example of linear equations, Algebra

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In a certain Algebra class there is a total 350 possible points. These points come through 5 homework sets which are worth 10 points each and 3 hour exams that are worth 100 points each.  A student has attained homework scores of 4, 8, 7, 7, & 9 and the first two exam scores are 78 & 83.  Supposing that grades are assigned according to the standard scale and there are no weights assigned to any of the grades is it probable for the student to attain an A in the class and if so what is the minimum score on the third exam which will give an A? What about a B?

Solution

Let's begin by defining p to be the minimum needed score on the third exam.

Now, let's remember how grades are set.  As there are no weights or anything on the grades, the grade will be set by first calculating the following percentage.

                            actual points            / total possible points  =  grade percentage

As we are using the standard scale if the grade percentage is 0.9 or higher the student will get an A.  Similarly if the grade percentage is among 0.8 & 0.9 the student will get a B.

We know that the overall possible points is 350 and the student contain a total points (by including the third exam) of,

                                 4 + 8 + 7 +7 +9 + 78 + 83 + p = 196 + p

The smallest possible percentage for an A is 0.9 and thus if  p is the minimum needed score on the third exam for an A we will have the given equation.

                                                  196 + p/350 = 0.9

It is a linear equation which we will need to solve for p.

196 + p = 0.9 (350)= 315                  ⇒          p = 315 -196 = 119

Thus, the minimum needed score on the third exam is 119.  It is a problem as the exam is worth only 100 points.  In other terms, the student will not be getting an A in the Algebra class.

Now let's verify if the student will get a B.  In this case the minimum percentage is 0.8.  Thus, to determine the minimum required score on the third exam for a B we will have to solve,

                                   196 + p /350 = 0.8

Solving out this for p gives,

                                 196 + p = 0.8 (350) =280           ⇒        p = 280 -196 =84

Thus, it is possible for the student to get a B in the class. All that the student will have to do is get at least an 84 on the third exam.


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