If I am currently making a 92.19% and have to take a test that has a maximum student score of 320, what is the lowest score I can make to stay above a 90% ?

Question

reemii · Answer

What is the weight that will be given to the "92.19" and the next score? is it 50-50?
I think something is missing in the statement.

anonymous · Answer

This is what my grade book looks like, I know I worded it wrong, but I'm not sure how to tell from what I have here. http://24.media.tumblr.com/380de86fc168c9e8f1c790c573b2aadc/tumblr_mlxj4duMBV1sp7n2ho1_500.jpg

anonymous · Answer

If you can't see it, its says:
student score: NA/720
percentage: 0
points possible: 320
points earned: 0

I haven't taken the test yet.

reemii · Answer

Oh, the number 720 is what we need.

reemii · Answer

You got some scores and it have added up to $n/720$ which is $92.19/100$ (92.19 percent).
This means you had more or less $n=664$. Now you will have a new test and obtain a new score: $b/320$.
You need to find $b$ so that $(n+b)/(720+320)=\frac{90}{100}$

reemii · Answer

$b = 1040\frac{90}{100} - n = 936-664=272$.
Maybe with the approximation ($n=$663,...) it is better to say "at least 273".

anonymous · Answer

Oh okay, thank you so much!

reemii · Answer

yw