Function Posted March 16, 2014 Share Posted March 16, 2014 (edited) Hello In July, I'm doing the approval exam for Medicine school. The exam is a multiple-choice-exam, which consists out of 2 parts: "knowledge and insight in sciences" and "gathering and processing information". The first part consists out of 40 questions (10 maths, 10 physics, 10 biology & 10 chemistry), with each 4 possible answers. A good answer (let's call the number of good answers [math]a[/math]) results in +1 point, a bad answer (let's call the number of bad answers [math]b[/math]) in -1/3 point (correction for guessing) and 0 points if you leave the question open (the number of questions to which no answer is given would then be [math]40-(a+b)[/math]. I'd think that the total score, on 40 points, would thus be: [math]T=a-\frac{b}{3}[/math] or [math]T=40-\frac{b}{3}-(40-(a+b))[/math] I think I've already made a mistake, for if these two should be equal, it would result in the solution that [math]b=0[/math] or that 2 = -1. Now let's first work with the first equation: according to me, it says that for 1-3 bad answers, at least 21 good answers must be given, for 4-6 bad answers, 22 good answers, and so on. Would this be correct? I'd think so, for if 0 bad answers are given, at least 20 good answers must be given, in order to have at least a score of 20/40. (Other than that, one can have a maximal of 15 bad answers. However, if one makes 15 mistakes, he has to give 25 good answers, leaving no other question open.) Don't know why the second equation isn't right, though... Could someone help me? Thanks! F. Edited March 16, 2014 by Function Link to comment Share on other sites More sharing options...

John Posted March 16, 2014 Share Posted March 16, 2014 (edited) The second equation assumes a = 40, which means all questions were answered correctly. Setting your two expressions equal to each other, we have [math]\begin{array}{rcl}a - \frac{b}{3} & = & 40 - \frac{b}{3} - (40 - (a + b)) \\ a &= & 40 - 40 + (a + b) \\ a & = & a + b\end{array}[/math] thus b = 0 as expected. I'm not sure what reasoning leads you to conclude from this that 2 = -1. Edited March 16, 2014 by John Link to comment Share on other sites More sharing options...

Function Posted March 16, 2014 Author Share Posted March 16, 2014 (edited) This is how I got -1=2 [math] \begin{array}{lrcl}&a-\frac{b}{3}&=&40-\frac{b}{3}-(40-(a+b))\\ & & & \\ \Leftrightarrow&a-\frac{b}{3}&=&-\frac{b}{3}+a+b\\ & & & \\ \Leftrightarrow&a-\frac{b}{3}&=&\frac{2b}{3}+a\\ & & & \\ \Leftrightarrow&-1&=&2\end{array} [/math] Edited March 16, 2014 by Function Link to comment Share on other sites More sharing options...

John Posted March 16, 2014 Share Posted March 16, 2014 (edited) Alright. Continuing on, we have that a = a + b, so b = 0. In short, to reason that -1 = 2 from what you have above requires that b is nonzero, which we can't assume, and which is ultimately false. Edited because I originally misread your post. Edit 2: The array syntax only works for me when I have the entire thing on one line. Edited March 16, 2014 by John Link to comment Share on other sites More sharing options...

Endy0816 Posted March 16, 2014 Share Posted March 16, 2014 Is this what you are looking for? [latex] T=a-\frac{b}{3}-0(40-(a+b)) [/latex] Link to comment Share on other sites More sharing options...

Function Posted March 16, 2014 Author Share Posted March 16, 2014 Alright. Continuing on, we have that a = a + b, so b = 0. In short, to reason that -1 = 2 from what you have above requires that b is nonzero, which we can't assume, and which is ultimately false. Edited because I originally misread your post. Edit 2: The array syntax only works for me when I have the entire thing on one line. Ah yes, of course.. Once again I had overseen that if 1*0 = 2*0, 1 [math]\neq[/math] 2... My bad. Thanks. Link to comment Share on other sites More sharing options...

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