Going into the final exam, which will count as 2/3 of the final grade, Terri has to test scores of 86, 80, 84 and 90.
a) What score does Terri need on the final in order to earn a B, which requires an average score of 80?
b) What does she need to earn an A, which requires an average of 90?
Do you we really have to use the fraction 1/3 to solve the question?
In other words, can we do it this way:
80 + 84 + 86 + 90 + x(2/3) = 90?
Can us the above set up to find what grade she needs to earn an A, which requires an average of 90?
Why is 1/3 needed to find the final exam score?
Well technically we can set it up any way we want, as long as the final counts for twice what the sum of the other exams count. For example we can also write the grade formula as:
Originally Posted by symmetry
where I have inserted the factors of 1 for clarity.
But generally the sum of the numerical factors (here 1 and 2, previously 1/3 and 2/3) are set such that they add to 1 as I had it in the original equation. I don't suppose there is a specific reason for this, just a habit to make sure the weighting factors are correctly calculated.
In case I didn't answer your question directly enough, the 1/3 is needed to "balance" the weight of 2/3 of the class that the final represents. If the final counts as 2/3, then the rest of the class only counts for 1/3 of the grade.
I absolutely understand why you included 1/3 and 2/3.