How to make this an equation?

Question

Suppose a professor counts the final exam twice as much as the other test. If a student has these grades : 76,82,66,80, what must the student make on the final exam in order to have an average of 80 in the class? I must have an equation to show how the problem is worked.

Answer 1

(76+82+66+80+2n)/5 = 80

or

n = (400-80-76-66-80)/2

You can do the arithmetic.

Answer 2

There are 4 tests, each equal to 1 part. The 5th test is equal to 2 parts. Altogether there are 6 parts. To average 80...the total must be 6*80 = 480

So 480 = 76 + 82 + 66 + 80 + 2x
480 -76 - 82 - 66 - 80 = 2x
176 = 2x
88 = x

EDIT: To all the people below me using 5 for the average...it needs to be 6 since the last test is worth twice as much. Think of it this way...say he had 80,80,80, and 80 on the first 4 tests...and still wanted an 80 average. Well he'd need 80 on the last test obviously too right? My way (80+80+80+80 +80*2)/6 = 80. Your way...(80+80+80+80+ 2*80)/5 = 96? /EDIT

He needs 88 on the final exam to get an 80% average overall.

Check
[76+82+66+80 + 88(2)]/6 = 80
80 = 80

Answer 3

79,82,66,80 and a test score
to find the average, you must add up the test scores and divide it by the number of test scores to get 80.
Let n = test score
equation:
79+82+66+80+n
------------------------- = 80
5
.
307+n ____ 80
---------= ----------
5 ________ 1
this is a proportion.
307+n=400
solve: n=93

93 is your answer. you can check by adding 93+76+82+66+80=400
400/5 = 80

okayyss. i checked it already: 93 is your answer. your equation is the one on the top. have fun and i hope you get the math!♥

Answer 4

(76+82+66+80+2(x))/5=80, solve for x, that will be your solution.

Answer 5

I don't know if it'll work but my math teacher gave us an equation and it was

the grade you want minus your current grade times .8 divided by .2

Thats an actual equation, but I don't know if it'll work in this cae...

Answer 6

muhaaha thank you