Difference between revisions of "1986 AHSME Problems/Problem 12"
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John scores <math>93</math> on this year's AHSME. Had the old scoring system still been in effect, he would score only <math>84</math> for the same answers. | John scores <math>93</math> on this year's AHSME. Had the old scoring system still been in effect, he would score only <math>84</math> for the same answers. | ||
| − | How many questions does he leave unanswered? (In the new scoring system one receives <math>5</math> points for correct | + | How many questions does he leave unanswered? (In the new scoring system that year, one receives <math>5</math> points for each correct answer, |
| − | <math>0</math> points for wrong | + | <math>0</math> points for each wrong answer, and <math>2</math> points for each problem left unanswered. In the previous scoring system, one started with <math>30</math> points, received <math>4</math> more for each correct answer, lost <math>1</math> point for each wrong answer, and neither gained nor lost points for unanswered questions.) |
| − | one started with <math>30</math> points, received <math>4</math> more for each correct answer, | ||
| − | lost | ||
| − | |||
<math>\textbf{(A)}\ 6\qquad | <math>\textbf{(A)}\ 6\qquad | ||
| Line 15: | Line 12: | ||
==Solution== | ==Solution== | ||
| − | + | Let <math>c</math>, <math>w</math>, and <math>u</math> be the number of correct, wrong, and unanswered questions respectively. From the old scoring system, we have <math>30+4c-w=84</math>, from the new scoring system we have <math>5c+2u=93</math>, and since there are <math>30</math> problems in the AHSME, <math>c+w+u=30</math>. Solving the simultaneous equations yields <math>u=9</math>, which is <math>\boxed{B}</math>. | |
== See also == | == See also == | ||
Latest revision as of 18:32, 1 April 2018
Problem
John scores 
on this year's AHSME. Had the old scoring system still been in effect, he would score only 
for the same answers.
How many questions does he leave unanswered? (In the new scoring system that year, one receives 
points for each correct answer,
points for each wrong answer, and 
points for each problem left unanswered. In the previous scoring system, one started with 
points, received 
more for each correct answer, lost 
point for each wrong answer, and neither gained nor lost points for unanswered questions.)
Solution
Let 
, 
, and 
be the number of correct, wrong, and unanswered questions respectively. From the old scoring system, we have 
, from the new scoring system we have 
, and since there are problems in the AHSME, 
. Solving the simultaneous equations yields 
, which is 
.
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
