Question 1203629: A student received scores of 84, 73, and 71 on their midterm algebra exams. If the final exam counts twice as much as a midterm, what score must the student make on their final exam to get an average score of 80? (Assume that the maximum possible score on each test is 100.)
Found 3 solutions by Theo, math_tutor2020, josgarithmetic:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i feel like something is missing in this problem.
what i think is missing is how much a midterm counts.
the way the problem is posed, the midterm doesn't count any more than a regular test.
if that's true, then the final counts twice as much as any test, not just the miderm, since the midterm doesn't count any more than a regular test.
with that in mind, i calculated as follows.
you have 3 regulars test that counts for 1 each.
the final counts as twice the midterm, which effectively means that it counts twice as much as any test, not just the midterm.\
if you let x equal the score of the midterm, then you get:
the sum of the scores on the 3 other test plus twice the score on the final should be equal to the sum of all the scores.
that gets you:
84 + 73 + 71 + 2x = 5 * 80 = 400
simplify to get:
228 + 2x = 400
subtract 228 from both sides of that equation to get:
2x = 172
solve for x to get x = 172/2 = 86
you require an 86 on the final to get an overall average of 80, assuming the final score counts twice as much as the scores on the other 3 tests.
228 + 2 * 86 = 400 / 5 = an overall average of 80, confirming the final test value is good.
the total number of scores to consider are 3 from the other tests plus 2 from the final.
Answer by math_tutor2020(3816)  (Show Source):
Answer by josgarithmetic(39613) (Show Source):
You can put this solution on YOUR website! This is not right:
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A student received scores of 84, 73, and 71 on their midterm algebra exams.
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One student, all the same class? Nonsense.
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