Big Buckin' Longhorn Posted May 29, 2009 Share Posted May 29, 2009 My son and I came across this math problem on a quiz, but neither of us really know how to work it. It has to do with weighted grades (which I don't understand which is why I don't use them, :D). The correct answer is noted to be B-48. I'm just not sure how they came up with that. Any help on exactly how to work this out would be so much appreciated. Thanks in advance! A student has received scores of 88, 82, and 84 on three quizzes. If tests count twice as much as quizzes, what is the lowest score the student can get on the next test to achieve an average score of at least 70? A. 13 B. 48 C. 70 D. 96 Quote Link to comment Share on other sites More sharing options...
Matryoshka Posted May 29, 2009 Share Posted May 29, 2009 It just means the test is worth twice what each quiz is worth, so you'd add it twice (I'm not sure if that explains it very well). Make "t" the test score: (88 + 82 + 84 +2t)/5=70 The divide by 5 is to get the average (3 quizzes plus test two times) - 70 is the average you're aiming for. 88 + 82 + 84 + 2t = 70 x 5 254 + 2t = 350 2t = 350-254 = 96 t = 96/2 = 48 Quote Link to comment Share on other sites More sharing options...
Big Buckin' Longhorn Posted May 29, 2009 Author Share Posted May 29, 2009 Thank you! Thank you! Thank you! It may be simple to some, but to us non-mathy folks, it's a bear, LOL. Thanks again! Quote Link to comment Share on other sites More sharing options...
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