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by Night reader » Sun Oct 31, 2010 4:50 pm
A student has obtained the scores 89 and 95 on his two midterm exams for a certain class. The student's final exam score receives 2.5 the weight of each of his two midterm exam scores; the course grade is based on the midterm and final exam scores only. What must be the student's score on the final exam for his course grade 92?

(A) 95
(B) 94
(C) 92
(D) 93
(E) 97
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by limestone » Sun Oct 31, 2010 6:52 pm
Hi,

The average score for his midterm exams is : (95+89) /2 = 92
His course grade is 92,
thus, regardless of the weight of the final test,
the final test must have a score of 92 to keep the average unchanged.

Pick C.
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by Night reader » Sun Oct 31, 2010 7:28 pm
limestone wrote:Hi,

The average score for his midterm exams is : (95+89) /2 = 92
His course grade is 92,
thus, regardless of the weight of the final test,
the final test must have a score of 92 to keep the average unchanged.

Pick C.
Like your short-cut. Below I post algebraic solution for such kind of problems (in case, average on midterms would not be equal to the course grade)

(89+95+2.5*F) / 4.5 (or 2 midterms and 2.5 weight) = 92
184+2.5F = 92*4.5
92*4.5- 184 = 2.5F, F=92

Now try different weight e.g. 2 (not in problem)
(89+95+2F)/4 = 92
184+2F = 92*4, F=92

The problem essentially tested knowledge of weighted averages.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by diebeatsthegmat » Sun Oct 31, 2010 11:57 pm
Night reader wrote:
limestone wrote:Hi,

The average score for his midterm exams is : (95+89) /2 = 92
His course grade is 92,
thus, regardless of the weight of the final test,
the final test must have a score of 92 to keep the average unchanged.

Pick C.
Like your short-cut. Below I post algebraic solution for such kind of problems (in case, average on midterms would not be equal to the course grade)

(89+95+2.5*F) / 4.5 (or 2 midterms and 2.5 weight) = 92
184+2.5F = 92*4.5
92*4.5- 184 = 2.5F, F=92

Now try different weight e.g. 2 (not in problem)
(89+95+2F)/4 = 92
184+2F = 92*4, F=92

The problem essentially tested knowledge of weighted averages.
can you explain more about the knowlegde of weighted average?
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