Calculating Weighted Averages in a Course

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To calculate the weighted average for a student's final and midterm exam scores, the final exam score of 84 is weighted twice as heavily as the midterm score of 90. The correct method involves treating the final as two scores of 84 and one score of 90, leading to the formula (2*84 + 90) / 3, which results in an average of 86. Some confusion arose regarding the need for total marks, but the focus should be on the weights assigned to each score. The average must logically fall between the two scores, and using incorrect calculations can lead to misleading results, such as an average of 64.5. The accurate weighted average for the course is 86.
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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
 
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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.
 
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tfhub said:
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
 
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)
 
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