Difference between revisions of "1987 AJHSME Problems/Problem 24"
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Since we want to maximize the value of <math>c</math>, we try to find the largest multiple of <math>7</math> less than <math>88</math>. This is <math>84=7\times 12</math>, so let <math>c=12</math>. Then we have <cmath>7(12)+2b=88\Rightarrow b=2</cmath> | Since we want to maximize the value of <math>c</math>, we try to find the largest multiple of <math>7</math> less than <math>88</math>. This is <math>84=7\times 12</math>, so let <math>c=12</math>. Then we have <cmath>7(12)+2b=88\Rightarrow b=2</cmath> | ||
− | Finally, we have <math> | + | Finally, we have <math>w=20-12-2=6</math>. We want <math>c</math>, so the answer is <math>12</math>, or <math>\boxed{\text{D}}</math>. |
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==See Also== | ==See Also== | ||
− | [[ | + | {{AJHSME box|year=1987|num-b=23|num-a=25}} |
+ | [[Category:Introductory Algebra Problems]] |
Revision as of 08:02, 31 May 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
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
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 |