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March 11th, 2014, 01:32 AM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  Hideki and History exam
On Hideki’s history exams, he gets 4 points for each problem answered correctly, he loses 2 points for each incorrect answer, and he gets 0 points for each question left blank. On a 25question test, Hideki received a score of 70. a. What is the largest number of questions that he could have answered correctly? b. What is the fewest number of questions that he could have answered correctly? c. What is the largest number of questions that he could have left blank? My attempt: Let the no. of right answers = r, the no. of wrong answers = w and unattempted questions = u. We get the equation 4r  2(25  r) = 70 = 4r 50 + 2r = 70 > 6r = 120 > r = 20 and w = 25  r = 5 (a) There is the possibility that Hideki left some questions unanswered. If this were the case, she would need fewer correct answers. Since question (a) is asking for the maximum no. of questions answered correctly, this is 20. (b) The fewest no. of right answers will require the minimum deduction. Since 70 is not a multiple of 4, we conclude that she has to have at least 1 wrong answer. Having more wrong answers will mean an increasing number of right answers. So 4 x 18 = 72. Minus the 1 wrong answer(2 points) and we get 70. Hideki can't get less than 18 right answers. Otherwise she can't make it to 70. Unattempted questions don't matter, because they contribute nothing to her score. So the minimum no. of right answers = 18. (c) Hideki has to have a minimum of 18 right answers and 1 wrong answer to make her score 70. That leaves us with 6 unanswered questions. So the largest number of unanswered questions = 6. Is my solution correct? Is there a better way? Is there an equation or inequality that can encapsulate this thought process? Thanks. 
March 11th, 2014, 08:50 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,636 Thanks: 2621 Math Focus: Mainly analysis and algebra  Re: Hideki and History exam
I think that's a very good, clear and simple answer.

March 12th, 2014, 12:36 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,474 Thanks: 2039 
The integers r and u satisfy 3r + u = 60 [1] r + u ? 25 [2] u ? 0 [3] r ? 0 [4] Equation [1] implies u is a multiple of 3, but I don't use that or inequality [4]. (b). By subtracting [2] from [1], 2r ? 35, so r ? 18 (as r is an integer). (a). By subtracting [3] from [1], 3r ? 60, so r ? 20. (c). As u is maximized when r is minimized, the maximum value of u is 60  3(1 = 6. Alternatively, plot r against u. This makes the questions very easy to answer. By the way, Hideki politely requests that you don't refer to him as "she". 

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