For the issue of exams get over 42% tommorow and you've got a C get any less and you've got a D (but no lower).
Quote:
0.77 * 80 + 0.60 *20 = 73.6

Not so fast...
You've produced the correct result but have been inconsistent with you use of percentages, your statement is true but it is not the problem at hand rather a problem that helps solve the problem at hand. You may know what you are doing but I would hope to explain what is being done in a way that is consistent for other potential users. Do you see how it could be confusing, you've divided by 100 only some of the time for percentages, whereas I would do it all the time.
How to use percentages when doing maths if you don't already know
a) If you are doing maths, y% means y divided by 100. So when you've got a percentage and you want to do maths with it, divide it by 100 and remove the percent sign. Dividing by 100 is easy. 1% is 0.01, 3% is 0.03, 42% is 0.42, 60% is 0.6, 100% is 1 , 234% is 2.34 and 0% is 0.
b) Likewise when you finish your maths and for whatever cruel reason you want to stick in a percent sign you multiply your value by 100 and stick a percent sign on the end. Multiplying by 100 is also easy. 0.45 becomes 45%, 2.1 becomes 210%, 0.70 becomes 70% etc.
since you start your maths you've already done a).
then its
(0.77 * 0.8) + (0.60 *0.2) = 0.736
Notice how 0.736 means 73.6% which you find out at the end using b).
Okay problem solved but if you want to go further its possible to see how this is known to be true.
Let X be your total number of marks and T be the total number of marks available. When you look for the total percentage you look for X/T.
If a course is divided into r many parts then you get a (scaled) mark for each part x(i) for i from 1 to r, each of which is out of a total (scaled) mark of t(i). But you don't know these instead in each part you know the percentage mark you got (which is x(i)/t(i)) and what fraction each it can count towards the full mark (which is t(i)/T). But if you multiply the two, because of multiplicative associativity, you get x(i)/T and then if you sum all these bits together from every part, because multiplication is distributive, you get the (sum of x(i)) /T which is X/T.