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The problem: A student scores 75, 86, and 72 on three 100-point exams. If the final is worth 150 points, find the range of the final exam scores S that gives the student an overall percentage of 80% or higher.

The question: What's the proper way to attack this problem?

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    I'm not sure this is linear algebra – 2017-02-23
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    @joeb Its out of an intermediate algebra book in the linear equations and inequalitys section if its not I apologize. – 2017-02-23
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    Also, are there $350$ possible points for the class, with the first and second exams counting for $100$ points each and the third exam counting for $150$ points? You need to clarify – 2017-02-23
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    @joeb I was confused about this aswell, that's the problem straight out of the book. Can't clarify because I don't quite understand it myself. – 2017-02-23
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    Ah looking again I think that there are four tests, the first three worth $100$ points each and the fourth exam (or the final) worth $150$ points. I am assuming then that $S$ can be a score from $0$ to $150$. If this the case, out of a total of $450$ possible points, the student scores $75+86+72+S$. Then the following ratio should satisfy $0.80 \leqslant \frac{75+86+72+S}{450} \leqslant 1.00$ – 2017-02-23
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    @joeb That's assuming that S is a score and not a percentage but I see what you saying, just the answer doesn't seem to make sense to me. Are you sure thats the right method to solve this problem? – 2017-02-23
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    What about it doesn't make sense. All I did was find the ratio of all points actually scored ($75+86+72+S$) to the total of points that could have been scored ($450$). I should correct myself though and say forget the $...\leqslant 1.00$ part in the last comment. Just solving $0.80 \leqslant \frac{75+86+72+S}{450}$ and combine what you find with the fact that $S \leqslant 150$. – 2017-02-23
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    @joeb sorry I understand now what your saying and that does work what im confused about is wouldn't you have to make all of the test scores weighted because of the 150 test score? – 2017-02-23
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    It is not linear algebra, but you can certainly be forgiven for not knowing that. Someone at your stage should not be expected to know the difference. I'll correct the tags. – 2017-02-24
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    The final being worth points than the other tests is not a reason to somehow weight the other tests. It just means that more points are being made available at this time than were made available at each of the earlier times. There is nothing to indicate that the points themselves will be worth more or less than points in the other test, nor is there any reason I can see to think they should be. The final covers more material than the other tests, so of course more points would be available. joeb is correct on how to solve it. – 2017-02-24
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    The reason why you do not have to do a weighted average here is because the points all weigh the same. While the final exam has more points on it, a point on the final exam is worth the same as a point on the first exam. If we were looking at the percentages, we would need a weighted average because one percent on exam 1 is worth less than one percent on the final exam. – 2017-02-24

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The total number of points that the student scores is $75+86+72+S$. The total number of points that the student could have scored is $100+100+100+150 = 450$. Therefore, we must have

$\frac{75+86+72+S}{450} \geqslant 0.80 \,\, \Longrightarrow \,\, S \geqslant 127$

in order for the student to score an overall grade of at least eighty percent. Combining this with the fact that $S \leqslant 150$, we deduce $127 \leqslant S \leqslant 150$.

Why does this work - there was no wieghting involved? In a way there was, but to convince ourselves we got it correct, let's solve the problem again from a different perspective.

Let $p_1 = p_2 = p_3 = \frac{100}{450}$ and $p_4 = \frac{150}{450}$ so that $p_1 + p_2 + p_3 + p_4 = 1$. These are the respective weights for the four exams.

Moreover, we let $X_1,X_2,X_3,X_4$ be the respective percentages scored for the four exams - in particular,

$X_1 = \frac{75}{100}, \,\, X_2 = \frac{86}{100}, \,\, X_3 = \frac{72}{100}$, and $X_4 = \frac{S}{150}$.

Then the solution should be gotten from

$0.80 \leqslant X_1p_1 + X_2p_2 + X_3p_3 + X_4p_4$,

which I claim is the exact same inequality we arrived at before. Indeed,

$X_1p_1 + X_2p_2 + X_3p_3 + X_4p_4$

$ = \left( \frac{75}{100}\right)\left( \frac{100}{450} \right) + \left(\frac{86}{100}\right) \left(\frac{100}{450}\right) + \left(\frac{72}{100}\right) \left(\frac{100}{450} \right) + \left(\frac{S}{150}\right) \left(\frac{150}{450} \right)$

$= \frac{75 + 86+72 +S}{450}$.