Difference between revisions of "1987 AJHSME Problems/Problem 24"
5849206328x (talk | contribs) (New page: ==Problem== A multiple choice examination consists of <math>20</math> questions. The scoring is <math>+5</math> for each correct answer, <math>-2</math> for each incorrect answer, and <m...) |
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Finally, we have <math>b=20-12-2=6</math>. We want <math>c</math>, so the answer is <math>12</math>, or <math>\boxed{\text{D}}</math>. | Finally, we have <math>b=20-12-2=6</math>. We want <math>c</math>, so the answer is <math>12</math>, or <math>\boxed{\text{D}}</math>. | ||
| + | |||
| + | (As a side note, the extra work done at the end was to make sure the given situation is possible for c=12) | ||
==See Also== | ==See Also== | ||
[[1987 AJHSME Problems]] | [[1987 AJHSME Problems]] | ||
Revision as of 16:07, 13 March 2009
Problem
A multiple choice examination consists of 
questions. The scoring is 
for each correct answer, 
for each incorrect answer, and 
for each unanswered question. John's score on the examination is 
. What is the maximum number of questions he could have answered correctly?
Solution
Let 
be the number of questions correct, 
be the number of questions wrong, and 
be the number of questions left blank. We are given that
Adding equation 
to double equation 
, we get ![\[7c+2b=88\]](http://latex.artofproblemsolving.com/b/a/2/ba2f40dabea06c60495bd8b35bc4a2a5ab97bf57.png)
Since we want to maximize the value of , we try to find the largest multiple of 
less than 
. This is 
, so let 
. Then we have ![\[7(12)+2b=88\Rightarrow b=2\]](http://latex.artofproblemsolving.com/5/3/f/53f49ae376b755c4e81b04db3f37cff32cdb95d5.png)
Finally, we have 
. We want , so the answer is 
, or 
.
(As a side note, the extra work done at the end was to make sure the given situation is possible for c=12)
