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A student is trying to achieve an overall average of at least 90.0 to 18 Apr 2018, 18:25
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55% (hard)

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69% (02:50) correct 31% (03:01) wrong based on 213 sessions

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A student is trying to achieve an overall average of at least 90.0 to earn a grade of A in a college course. In determining the overall average for the course grade, the instructor calculates a weighted average based on the four major, 100-point exams in the course. The mean (arithmetic average) of the scores on the first three exams counts as 60 percent of the overall average, and the fourth (and final) exam counts as 40 percent of the overall average. The student has scores of 80, 95, and 92 on the first three major exams. Which of the following scores on the fourth major exam would result in an overall average that would earn the student an A in the course?

I. 90
II. 91
III. 92

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
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C
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A student is trying to achieve an overall average of at least 90.0 to 18 Apr 2018, 19:54
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Bunuel
A student is trying to achieve an overall average of at least 90.0 to earn a grade of A in a college course. In determining the overall average for the course grade, the instructor calculates a weighted average based on the four major, 100-point exams in the course. The mean (arithmetic average) of the scores on the first three exams counts as 60 percent of the overall average, and the fourth (and final) exam counts as 40 percent of the overall average. The student has scores of 80, 95, and 92 on the first three major exams. Which of the following scores on the fourth major exam would result in an overall average that would earn the student an A in the course?

I. 90
II. 91
III. 92

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
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There are a lot of ways to do the problem. One approach: straightforward weighted average.

Weighted average equation is
(Mean of first 3 tests)*(.60) PLUS
(Final exam score) * (.40) =
Overall Average

The mean score of first three exams counts for 60% of the whole grade.

1) Find the mean of the the first three exams

\(\frac{80+92+95}{3}=\frac{267}{3}=89\)

2) Multiply by .6 to get the actual number of points he has so far: \((89 * .6) = 53.4\)

3) Desired overall average is 90. Points needed? He has 53.4

\((90 - 53.4) = 36.6\) points needed

That is, to earn a 90 overall, the product of
(final exam score) * (.4) must be \(\geq 36.6\)

4) Check Options
Multiply each option's score by (.4) to find the actual point value. He needs 36.6 points

Final exam score of
I. 90. NO
Mental math: \((90*.4) = 36\)

II. 91 ALSO NO
\(91= 90 + 1\)
\((90 * .4) = 36\)
\((1 * .4) = .4\)
\(36 + .4 = 36.4\)

If you're short on time, mark C and move on. Options I and II do not work.

III. 92 - YES
\(92 * 4 = 36.8\)

He needed 36.6 to get an overall average of at least 90. With a final score of 92, he made it:
\(36.8 + 53.4 = 90.2\)

Answer
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C
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A student is trying to achieve an overall average of at least 90.0 to 23 Jun 2019, 06:34
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Bunuel
A student is trying to achieve an overall average of at least 90.0 to earn a grade of A in a college course. In determining the overall average for the course grade, the instructor calculates a weighted average based on the four major, 100-point exams in the course. The mean (arithmetic average) of the scores on the first three exams counts as 60 percent of the overall average, and the fourth (and final) exam counts as 40 percent of the overall average. The student has scores of 80, 95, and 92 on the first three major exams. Which of the following scores on the fourth major exam would result in an overall average that would earn the student an A in the course?

I. 90
II. 91
III. 92

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
There are a lot of ways to do the problem. One approach: straightforward weighted average.

Weighted average equation is
(Mean of first 3 tests)*(.60) PLUS
(Final exam score) * (.40) =
Overall Average

The mean score of first three exams counts for 60% of the whole grade.

1) Find the mean of the the first three exams

\(\frac{80+92+95}{3}=\frac{267}{3}=89\)

2) Multiply by .6 to get the actual number of points he has so far: \((89 * .6) = 53.4\)

3) Desired overall average is 90. Points needed? He has 53.4

\((90 - 53.4) = 36.6\) points needed

...
Show more

Up to this point, my calculation approach was similar, afterwards I tried a shortcut.

We can see that the student needs to get atleast 36.6 points, thus the units digit of his score in the 90´s does atleast have to be 1.5.
Therefore, 92 is the only option with a units digit that is large enough, hence III is the only valid score.

Regards,
Chris
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A student is trying to achieve an overall average of at least 90.0 to 25 Jun 2019, 06:21
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By the looks of it, this question might seem difficult because it’s so verbose. But really, I think this is just a moderately difficult question on Weighted average.

The overall average that the student wants to achieve is 90 to obtain a grade A. But a bigger share of this average comes from the average of the first three exams and a smaller share comes from the average of the fourth exam (which is, of course, the marks obtained in the fourth exam).

The average marks of the first three tests is,
\(\frac{80 + 95 + 92}{3}\) = \(\frac{267}{3}\)

60% of this average contributes to the overall average. This means,
\(\frac{3}{5}\) * \(\frac{267}{3}\) = \(\frac{267}{5}\) = 53.4 is the contribution to the overall average.

To reach an overall average of 90, we need another 36.6 to be added to 53.4.

If he scores 90 on the fourth exam, his contribution to the average will be \(\frac{2}{5}\) * 90 = 36 (40% = \(\frac{2}{5}\)). Clearly, 90 is not sufficient to reach an overall average of 90. Answer options A and E can be eliminated.

If he scores 91 on the fourth exam, his contribution to the average will be \(\frac{2}{5}\) * 91 = 36.4, which is also insufficient to reach 90. Options B and D can be eliminated. The only option left is C, this HAS to be the answer.

If he scores 92 on the fourth exam, his contribution to the average will be \(\frac{2}{5}\) * 92 = 36.8. Now,
53.4 + 36.8 = 90.2
This way, we have also mathematically proved that a 92 on the fourth exam will result in an overall average of at least 90.

In hindsight, this has turned out to be a question on solving an equation than testing you on weighted average. But yes, for you to frame that equation, you will have to know that you have to take the weighted average of the two different groups.

Hope this helps!
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A student is trying to achieve an overall average of at least 90.0 to 29 Jun 2025, 01:17
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Bunuel
A student is trying to achieve an overall average of at least 90.0 to earn a grade of A in a college course. In determining the overall average for the course grade, the instructor calculates a weighted average based on the four major, 100-point exams in the course. The mean (arithmetic average) of the scores on the first three exams counts as 60 percent of the overall average, and the fourth (and final) exam counts as 40 percent of the overall average. The student has scores of 80, 95, and 92 on the first three major exams. Which of the following scores on the fourth major exam would result in an overall average that would earn the student an A in the course?

I. 90
II. 91
III. 92

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
Show more
The average of the first 3 scores is 89.
The weightage is distributed as 6:4 between the first three scores and the last one

(89*6 + x*4)/10 >= 90

x >= 91.5

Option 3
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