Difference between revisions of "1987 AJHSME Problems/Problem 24"
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==Problem== | ==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 <math>0</math> for each unanswered question. John's score on the examination is <math>48</math>. What is the maximum number of questions he could have answered correctly? | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>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 <math>0</math> for each unanswered question. John's score on the examination is <math>48</math>. What is the maximum number of questions he could have answered correctly?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
<math>\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16</math> | <math>\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16</math> | ||
Revision as of 17:51, 27 March 2015
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 
.
See Also
| 1987 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
