Calculating Weighted Averages in a Course

[tfhub]

the question statement is2. In a course, a student's final exam is weighted twice as heavily as his midterm exam. If a student receives a score of 84 on his final exam , 90 on his midterm, what is the average for his courses?

now i am pretty confused should I have to know the total marks for the exam. I can keep final score 84 and half the midterm to 45 and then average them. is this the right way to do it?

and i got an answer of 64.5

 

[Mentallic]

If the final is weighted twice as heavily as the midterm, then you can think of it as being 3 exams in total with all equal weighting, where 2 of them are 84, then find the average of those.

So you have

$$\frac{2*84+90}{3}=\frac{2}{3}84+\frac{1}{3}90=86$$

Your approach was almost there, but you missed the detail that since you had 1 score of 84 and then half a score of 90 (in a sense) then to average these you would've had to have divided by 1 and a half.

 

[SteamKing]

the question statement is2. In a course, a student's final exam is weighted twice as heavily as his midterm exam. If a student receives a score of 84 on his final exam , 90 on his midterm, what is the average for his courses?

now i am pretty confused should I have to know the total marks for the exam. I can keep final score 84 and half the midterm to 45 and then average them. is this the right way to do it?

and i got an answer of 64.5

The 'average' should wind up somewhere between the highest and lowest scores, whether there are two scores or twenty scores.
Under your scheme, a student whose individual scores were solid Bs (84 and 90) wound up failing the course. (64.5) :L

 

[mpx86]

You could also solve it in your head using : since the weights are in the ratio 1:2 therefore 2 times decrease in 90 corresponds to 1 time increase in 84
Therefore,
84-----90
85------88
86------86(which is the desired answer)