• Jan 9th 2007, 03:18 PM
symmetry
Going into the final exam, which will count as 2/3 of the final grade, Terri has to test scores of 86, 80, 84 and 90.

a) What score does Terri need on the final in order to earn a B, which requires an average score of 80?

b) What does she need to earn an A, which requires an average of 90?
• Jan 9th 2007, 06:13 PM
topsquark
Quote:

Originally Posted by symmetry
Going into the final exam, which will count as 2/3 of the final grade, Terri has to test scores of 86, 80, 84 and 90.

a) What score does Terri need on the final in order to earn a B, which requires an average score of 80?

b) What does she need to earn an A, which requires an average of 90?

I am going to presume each test is out of 100?
The final counts as 2/3 the grade so the other 4 tests count 1/3. Thus:
$g = \frac{ \frac{1}{3} \cdot (86 + 80 + 84 + 90) + \frac{2}{3} \cdot (f) }{ \frac{1}{3} \cdot (4 \cdot 100) + \frac{2}{3} \cdot (100) } \cdot 100$
where g is the overall grade and f is the score on the final.

Let me explain this beast a little. This is something called a "weighted" average. We are weighting the first 4 scores with a value of 1/3 and the last score with a value of 2/3. The numerator represents the total number of points actually acquired, and the denominator represents the total number of points possible. That last 100 I tacked on at the end merely converts the average to a percentage.

a) We wish to find a value of f for g = 80.
$80 = \frac{ \frac{1}{3} \cdot (86 + 80 + 84 + 90) + \frac{2}{3} \cdot (f) }{ \frac{1}{3} \cdot (4 \cdot 100) + \frac{2}{3} \cdot (100) } \cdot 100$

Let's get the denominator first, since this is just a number:
$80 = \frac{ \frac{1}{3} \cdot (86 + 80 + 84 + 90) + \frac{2}{3} \cdot (f) }{ 200 } \cdot 100$

Multiply both sides by 200/100:

$160 = \frac{1}{3} \cdot (86 + 80 + 84 + 90) + \frac{2}{3} \cdot (f)$

$160 = \frac{1}{3} \cdot (340) + \frac{2}{3} \cdot (f)$

$480 = 340 + 2f$

$140 = 2f$

$f = 70$

So she needs to get a 70 on the final.

b) g = 90. Do the same thing. I get f = 100.

-Dan
• Jan 10th 2007, 12:54 PM
symmetry
ok
Do you we really have to use the fraction 1/3 to solve the question?

In other words, can we do it this way:

80 + 84 + 86 + 90 + x(2/3) = 90?

Can us the above set up to find what grade she needs to earn an A, which requires an average of 90?

Why is 1/3 needed to find the final exam score?
• Jan 10th 2007, 01:52 PM
topsquark
Quote:

Originally Posted by symmetry
Do you we really have to use the fraction 1/3 to solve the question?

In other words, can we do it this way:

80 + 84 + 86 + 90 + x(2/3) = 90?

Can us the above set up to find what grade she needs to earn an A, which requires an average of 90?

Why is 1/3 needed to find the final exam score?

Well technically we can set it up any way we want, as long as the final counts for twice what the sum of the other exams count. For example we can also write the grade formula as:
$g = \frac{1 \cdot (80 + 84 + 86 + 90) + 2 \cdot (f)}{1 \cdot (4 \cdot 100) + 2 \cdot (100)} \cdot 100$
where I have inserted the factors of 1 for clarity.

But generally the sum of the numerical factors (here 1 and 2, previously 1/3 and 2/3) are set such that they add to 1 as I had it in the original equation. I don't suppose there is a specific reason for this, just a habit to make sure the weighting factors are correctly calculated.

In case I didn't answer your question directly enough, the 1/3 is needed to "balance" the weight of 2/3 of the class that the final represents. If the final counts as 2/3, then the rest of the class only counts for 1/3 of the grade.

-Dan
• Jan 10th 2007, 03:42 PM
symmetry
ok
Dan,

I absolutely understand why you included 1/3 and 2/3.

Thanks.