Difference between revisions of "1986 AHSME Problems/Problem 12"

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(Fixed the problem statement and added a solution with explanation)
 
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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 answers,
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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 answers, and <math>2</math> points for unanswered questions. In the old system,  
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<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 one point for each wrong answer, and neither gained nor lost points for unanswered questions.
 
There are <math>30</math> questions in the <math>1986</math> AHSME.)
 
  
 
<math>\textbf{(A)}\ 6\qquad
 
<math>\textbf{(A)}\ 6\qquad
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==Solution==
 
==Solution==
 
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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 17:32, 1 April 2018

Problem

John scores $93$ on this year's AHSME. Had the old scoring system still been in effect, he would score only $84$ for the same answers. How many questions does he leave unanswered? (In the new scoring system that year, one receives $5$ points for each correct answer, $0$ points for each wrong answer, and $2$ points for each problem left unanswered. In the previous scoring system, one started with $30$ points, received $4$ more for each correct answer, lost $1$ point for each wrong answer, and neither gained nor lost points for unanswered questions.)

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ \text{Not uniquely determined}$

Solution

Let $c$, $w$, and $u$ be the number of correct, wrong, and unanswered questions respectively. From the old scoring system, we have $30+4c-w=84$, from the new scoring system we have $5c+2u=93$, and since there are $30$ problems in the AHSME, $c+w+u=30$. Solving the simultaneous equations yields $u=9$, which is $\boxed{B}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
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

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