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18 Apr 2018, 18:25
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Difficulty:
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Question Stats:
72% (02:52) correct
28%
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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
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
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
There are a lot of ways to do the problem. One approach: straightforward weighted average.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
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
General Discussion
Arro44
Joined: 04 Jun 2018
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23 Jun 2019, 06:34
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generis
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.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
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
...
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
