Difference between revisions of "1987 AJHSME Problems/Problem 24"
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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>. | 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== | ||
+ | [[2010 AMC 10B Problems/Problem 15|2010 AMC10B Problem 15]] | ||
{{AJHSME box|year=1987|num-b=23|num-a=25}} | {{AJHSME box|year=1987|num-b=23|num-a=25}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 09:56, 11 June 2024
Contents
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
Solution 1
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.