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Difference between revisions of "2004 AMC 12A Problems/Problem 2"

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==Problem==
 
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On the AMC 12, each correct answer is worth <math>6</math> points, each incorrect answer is worth <math>0</math> points, and each problem left unanswered is worth <math>2.5</math> points. If Charlyn leaves <math>8</math> of the <math>25</math> problems unanswered, how many of the remaining problems must she answer correctly in order to score at least <math>100</math>?
  
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<math>\text {(A)} 11 \qquad \text {(B)} 13 \qquad \text {(C)} 14 \qquad \text {(D)} 16\qquad \text {(E)}17</math>
  
 
==Solution==
 
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She gets <math>8*2.5=20</math> points for the problems she didn't answer. She must get <math>\left\lceil \frac{100-20}{6} \right\rceil =14 \Rightarrow \text {(C)}</math> problems right to score at least 100.
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{{AMC12 box|year=2004|ab=A|num-b=1|num-a=3}}
  
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{{AMC12 box|year=2004|ab=A|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:14, 3 July 2013

Problem

On the AMC 12, each correct answer is worth $6$ points, each incorrect answer is worth $0$ points, and each problem left unanswered is worth $2.5$ points. If Charlyn leaves $8$ of the $25$ problems unanswered, how many of the remaining problems must she answer correctly in order to score at least $100$?

$\text {(A)} 11 \qquad \text {(B)} 13 \qquad \text {(C)} 14 \qquad \text {(D)} 16\qquad \text {(E)}17$

Solution

She gets $8*2.5=20$ points for the problems she didn't answer. She must get $\left\lceil \frac{100-20}{6} \right\rceil =14 \Rightarrow \text {(C)}$ problems right to score at least 100.

See Also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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