# Computing Grades-getting average from a weighted test

Ok well I have this basic problem in which there are grades (4 grades). There's an upcoming final that is weighted to be some fraction toward the final grade (2/3). I have to find what the final grade has to be to get an average grade of 80. and then 90. I completly forgot the procedure as to how to tackle this problem. Anybody have any hints/tricks for me to start me off?

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If this is homework, you should add the homework tag to make that clear. – Oliver Sep 13 '11 at 23:13
If I am understanding the question correctly then perhaps thinking of it as two things weighing in on the final grade will help you: 2/3 the final grade is the score on the final and 1/3 is the average of the other 4 grades. So you would have 80(or 90) = 2/3(F) + 1/3(avg(other grades)) – Deven Ware Sep 13 '11 at 23:13
thanks! it was very helpful – Ronnie.j Sep 13 '11 at 23:17
@Ronnie.j no problem :) – Deven Ware Sep 13 '11 at 23:19
@Deven Ware: you could enter that as an answer so it can be accepted. It would be good to make the fractions clear: 2F/3 so everybody knows F is not in the denominator. Better yet, use $\LaTeX$ by putting it in dollar signs: \frac{2F}{3} gives $\frac{2F}{3}$ – Ross Millikan Sep 13 '11 at 23:21

The key to solving this problem is to realize that there are essentially two components that will go into the final grade :

1) The average of the previous four tests

2) The grade on the final

Thus we can set it up as follows : Let $G =$ Grade in the class, $a =$ average score from the previous 4 tests, and $f =$ final exam score. \begin{align*} G = \frac{2f}{3} + \frac{a}{3} \end{align*}

Using you're numbers you can solve for whatever possibilities you need.

EDIT: you can also use this approach for any different weightings by simply changing the fractional amounts that $a$ and $f$ are worth, for example if you want the final $f$, to be worth $3/4$ the final grade then it would reflect as: \begin{align*} G = \frac{3f}{4} + \frac{a}{4} \end{align*}

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Awesome, i shall forever remember this for future upcoming questions similar to this one. – Ronnie.j Sep 13 '11 at 23:47